Cidnbrtdp ndM In jJimmJ wtdnfUn
Multidimensional
Real Anal/sis 11
Integration
J. J. DUISTERMAAT
AND] A.C.KOLK

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CAMBRIDGE STUDIES IN
ADVANCED MATHEMATICS 87
EDITORIAL BOARD
B. BOLLOBAS, W. FULTON, A. KATOK, F. KIRWAN,
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MULTIDIMENSIONAL REAL
ANALYSIS II:
INTEGRATION

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11 J.L. Alperin Local rcprcsentaliou theory
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13 A. Pictsch Eigenvalues and s-mimbers
14 S.J. Paiienion An introduction to the theory of the Riemann zeta-function
15 HJ. Bancs Algebraic homotopy
16 V.S. Varadarajan Intioduction to harmonic analysis on semisiniple Lie groups
17 W. Dicks & M. Dunwoody Groups acting on graphs
IS L.J. Corwin & F.P. Gti^nlit^' Representations of nilpotent Lie groups and their applications
Id R. Fritsch & R. Piccinini Cellular structures in topology
20 H. Klingcn Introductory lectures on Siegel modular fonns
21 P. Koosis The logarithmic integral II
22 M.J. Collins Representations and characters of finite groups
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25 P. Wqjiaszczyk Banach spaces for analysts
26 J.E. Gilbert & M.A.M. Murray Clifford algebras and Dime operators in harmonic analysis
27 A. Frohlich & M.J. Taylor Algebraic number theoiy
2S K.. Gocbel & W.A. Kirk Topics in metric fixed point theory
29 J.E. Humphreys Reflection groups and Coxeter groups
30 DJ. Benson Representations and cohomology I
31 D J. Benson Representations and cohomology II
32 C Allday & V. Puppc Cohomological methods in transformation groups
33 C Soule ct al Lectures on Arakelov geometry
34 A. Ambrosciti & G. Prodi A primer of nonlinear analysis
35 J. Palis & F. Takeits Hyperbolicity, stability and chaos at homoclinic bifiircations
37 Y. Meyer Wavelets and operators I
3S C Weibel An introduction to homological algebra
39 W. Bruns & J. Herzog Cohen-Macaulay nngs
40 V. Snaidi Explicit Brauer induction
41 G. Laumon Cohomology ofDnnfield modular varieties I
42 E.B. Davies Spectral theoiy and differential operators
43 J. Diestcl, H. Jarchow & A. Tonge Absolutely swnming operators
44 P. Mattila Geometry of sets and measures in euclidean spaces
45 R. Pinsky Positive hannonic functions and diffiision
46 G. Tencnbaum Introduction to analytic and probabilistic number theory
47 C Peskine An algebraic introduction to complex projective geometry I
4S Y. Meyer & R. Coifman Wavelets and operators II
49 R. Stanley Enumerative combinatorics I
50 1. Portcous Clifford algebras and the classical groups
51 M. Audin Spirming tops
52 V. Jurdjevic Geometric control theoiy
53 H. Voelklein Groups as Galois groups
54 J. Lc Poiier Lectures on vector bundles
55 D. Bump Automoiphic forms
56 G. Laumon Cohomology ofDrinfeld modular varieties II
57 D.M. Clarice & B.A. Davey Natural dualities for the working algebraist
59 P. Taylor Practical foundations of mathematics
60 M. Brodmann & R. Sharp Local cohomology
61 J.D. Dixon, M.P.F. Du Sautoy, A. Mann & D. Segal Analytic pro-p groups, 2nd edition
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MULTIDIMENSIONAL REAL
ANALYSIS II:
INTEGRATION
J.J. DUISTERMAAT
J.A.C. KOLK
Utrecht University
Translated from Dutch by J. P. van Braam Houckgeest
H CAMBRIDGE
UNXVEHSITY FRE S S

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To Saskia and Floortje
With Gratitude and Love

Contents
Volume n
Preface xi
Acknowledgments xiii
Introduction xv
6 Integration 423
6.1 Rectangles 423
6.2 Riemann integrability 425
6.3 Jordan measurability 429
6.4 Successive integration 435
6.5 Examples of successive integration 439
6.6 Change of Variables Theorem: formulation and examples .... 444
6.7 Partitions of unity 452
6.8 Approximation of Riemann integrable functions 455
6.9 Proof of Change of Variables Theorem 457
6.10 Absolute Riemann integrability 461
6.11 Application of integration: Fourier transformation 466
6.12 Dominated convergence 471
6.13 Appendix: two other proofs of Change of Variables Theorem . . 477
7 Integration over Submanifolds 487
7.1 Densities and integration with respect to density 487
7.2 Absolute Riemann integrability with respect to density 492
7.3 Euclidean rf-dimensional density 495
7.4 Examples of Euclidean densities 498
7.5 Open sets at one side of their boundary 511
7.6 Integration of a total derivative 518
7.7 Generalizations of the preceding theorem 522
7.8 Gauss' Divergence Theorem 527
7.9 Applications of Gauss'Divergence Theorem 530
8 Oriented Integration 537
8.1 Line integrals and properties of vector fields 537
8.2 Antidifferentiation 546
8.3 Green's and Cauchy's Integral Theorems 551
vii

viii Contents
8.4 Stokes' Integral Theorem 557
8.5 Applications of Stokes'Integral Theorem 561
8.6 Apotheosis: differential forms and Stokes' Theorem 567
8.7 Properties of differential forms 576
8.8 Applications of differential forms 581
8.9 Homotopy Lemma 585
8.10 Poincare's Lemma 589
8.11 Degree of mapping 591
Exercises 599
Exercises for Chapter 6 599
Exercises for Chapter 7 677
Exercises for Chapter 8 729
Notation 779
Index 783
Volume I
Preface xi
Acknowledgments xiii
Introduction xv
1 Continuity 1
1.1 Inner product and norm 1
1.2 Open and closed sets 6
1.3 Limits and continuous mappings 11
1.4 Composition of mappings 17
1.5 Homeomorphisms 19
1.6 Completeness 20
1.7 Contractions 23
1.8 Compactness and uniform continuity 24
1.9 Connectedness 33
2 Differentiation 37
2.1 Linear mappings 37
2.2 Differentiable mappings 42
2.3 Directional and partial derivatives 47
2.4 Chain rule 51
2.5 Mean Value Theorem 56
2.6 Gradient 58
2.7 Higher-order derivatives 61
2.8 Taylor's formula 66
2.9 Critical points 70
2.10 Commuting limit operations 76

Contents ix
3 Inverse Function and Implicit Function Theorems 87
3.1 Diffeomorphisms 87
3.2 Inverse Function Theorems 89
3.3 Applications of Inverse Function Theorems 94
3.4 Implicitly defined mappings 96
3.5 Implicit Function Theorem 100
3.6 Applications of the Implicit Function Theorem 101
3.7 Implicit and Inverse Function Theorems on C 105
4 Manifolds 107
4.1 Introductory remarks 107
4.2 Manifolds 109
4.3 Immersion Theorem 114
4.4 Examples of immersions 118
4.5 Submersion Theorem 120
4.6 Examples of submersions 124
4.7 Equivalent definitions of manifold 126
4.8 Morse's Lemma 128
5 Tangent Spaces 133
5.1 Definition of tangent space 133
5.2 Tangent mapping 137
5.3 Examples of tangent spaces 137
5.4 Method of Lagrange multipliers 149
5.5 Applications of the method of multipliers 151
5.6 Closer investigation of critical points 154
5.7 Gaussian curvature of surface 156
5.8 Curvature and torsion of curve in R^ 159
5.9 One-parameter groups and infinitesimal generators 162
5.10 Linear Lie groups and their Lie algebras 166
5.11 Transversality 172
Exercises 175
Review Exercises 175
Exercises for Chapter 1 201
Exercises for Chapter 2 217
Exercises for Chapter 3 259
Exercises for Chapter 4 293
Exercises for Chapter 5 317
Notation 411
Index 412

Preface
I prefer the open landscape under a clear sky with its depth
of perspective, where the wealth of sharply defined nearby
details gradually fades away towards the horizon.
This book, which is in two parts, provides an introduction to the theory of vector-
valued functions on Euclidean space. We focus on four main objects of study
and in addition consider the interactions between these. Volume I is devoted to
differentiation. Differentiable functions on R" come first, in Chapters 1 through 3.
Next, differentiable manifolds embedded in R" are discussed, in Chapters 4 and 5. In
Volume II we take up integration. Chapter 6 deals with the theory of n-dimensional
integration over R". Finally, in Chapters 7 and 8 lower-dimensional integration over
submanifolds of R" is developed; particular attention is paid to vector analysis and
the theory of differential forms, which are treated independently from each other.
Generally speaking, the emphasis is on geometric aspects of analysis rather than on
matters belonging to functional analysis.
In presenting the material we have been intentionally concrete, aiming at a
thorough understanding of Euclidean space. Once this case is properly understood,
it becomes easier to move on to abstract metric spaces or manifolds and to infinite-
dimensional function spaces. If the general theory is introduced too soon, the reader
might get confused about its relevance and lose motivation. Yet we have tried to
organize the book as economically as we could, for instance by making use of linear
algebra whenever possible and minimizing the number of e-^ arguments, always
without sacrificing rigor. In many cases, a fresh look at old problems, by ourselves
and others, led to results or proofs in a form not found in current analysis textbooks.
Quite often, similar techniques apply in different parts of mathematics; on the other
hand, different techniques may be used to prove the same result. We offer ample
illustration of these two principles, in the theory as well as the exercises.
A working knowledge of analysis in one real variable and linear algebra is a
prerequisite; furthermore, familiarity with differentiable mappings and
submanifolds of R", as discussed in volume I, for instance. The main parts of the theory
can be used as a text for an introductory course of one semester, as we have been
doing for second-year students in Utrecht during the last decade. Sections at the
end of many chapters usually contain applications that can be omitted in case of
time constraints.
This volume contains 234 exercises, out of a total of 568, offering variations
and applications of the main theory, as well as special cases and openings toward
applications beyond the scope of this book. Next to routine exercises we tried
also to include exercises that represent some mathematical idea. The exercises are
independent from each other unless indicated otherwise, and therefore results are
xi

xii Preface
sometimes repeated. We have run student seminars based on a selection of the more
challenging exercises.
In our experience, interest may be stimulated if from the beginning the
student can perceive analysis as a subject intimately connected with many other parts
of mathematics and physics: algebra, electromagnetism, geometry, including
differential geometry, and topology. Lie groups, mechanics, number theory, partial
differential equations, probability, special functions, to name the most important
examples. In order to emphasize these relations, many exercises show the way in
which results from the aforementioned fields fit in with the present theory; prior
knowledge of these subjects is not assumed, however. We hope in this fashion to
have created a landscape as preferred by Weyl,' thereby contributing to motivation,
and facilitating the transition to more advanced treatments and topics.
' Weyl, H.: The Classical Groups. Princeton University Press, Princeton 1939, p. viii.

Acknowledgments
Since a text like this is deeply rooted in the literature, we have refrained from giving
references. Yet we are deeply obliged to many mathematicians for publishing the
results that we use freely. Many of our colleagues and friends have made important
contributions: E. P. van den Ban, F. Beukers, R. H. Cushman, J. P. Hogendijk,
W. L. J. van der Kallen, H. Keers, E. J. N. Looijenga, T. A. Springer, J. Stienstra,
and in particular D. Zagier. We were also fortunate to have had the moral support
of our special friend V. S. Varadarajan. Numerous small errors and stylistic points
were picked up by students who attended our courses; we thank them all.
With regard to the manuscript's technical realization, the help of A. J. de Meijer
and F. A. M. van de Wiel has been indispensable, with further contributions coming
from K. Barendregt and J. Jaspers. We have to thank R. P. Buitelaar for assistance
in preparing some of the illustrations. Without MIjeX, Y&Y TeX and Mathematica
this work would never have taken on its present form.
J. P. van Braam Houckgeest translated the manuscript from Dutch into English.
We are sincerely grateful to him for his painstaking attention to detail as well as his
many suggestions for improvement.
We are indebted to S. J. van Strien to whose encouragement the English version
is due; and furthermore to R. Astley and J. Walthoe, our editors, and to F. H. Nex,
our copy-editor, for the pleasant collaboration; and to Cambridge University Press
for making this work available to a larger audience.
Of course, errors still are bound to occur and we would be grateful to be told of
them, at the e-mail address kolk@math.uu.nl. A listing of corrections will be made
accessible through http://www.math.uu.nl/people/kolk.
xni

Introduction
Motivation. Analysis came to life in the number space R" of dimension n and its
complex analog C". Developments ever since have consistently shown that further
progress and better understanding can be achieved by generalizing the notion of
space, for instance to that of a manifold, of a topological vector space, or of a
scheme, an algebraic or complex space having infinitesimal neighborhoods, each of
these being defined over a field of characteristic which is 0 or positive. The search
for unification by continuously reworking old results and blending these with new
ones, which is so characteristic of mathematics, nowadays tends to be carried out
more and more in these newer contexts, thus bypassing R". As a result of this
the uninitiated, for whom R" is still a difficult object, runs the risk of learning
analysis in several real variables in a suboptimal manner. Nevertheless, to quote F.
and R. Nevanlinna: "The elimination of coordinates signifies a gain not only in a
formal sense. It leads to a greater unity and simplicity in the theory of functions
of arbitrarily many variables, the algebraic structure of analysis is clarified, and
at the same time the geometric aspects of linear algebra become more prominent,
which simplifies one's ability to comprehend the overall structures and promotes
the formation of new ideas and methods".^
In this text we have tried to strike a balance between the concrete and the abstract:
a treatment of integral calculus in the traditional R" by efficient methods and using
contemporary terminology, providing solid background and adequate preparation
for reading more advanced works. The exercises are tightly coordinated with the
theory, and most of them have been tried out during practice sessions or exams.
Illustrative examples and exercises are offered in order to support and strengthen
the reader's intuition.
Organization. This is the second volume, devoted to integration, of a book in
two parts; the first volume treats differentiation. The volume at hand uses results
from the preceding one, but it should be accessible to the reader who has acquired
a working knowledge of differentiable mappings and submanifolds of R". Only
some of the exercises might require special results from Volume I.
In a subject like this with its many interrelations, the arrangement of the material
is more or less determined by the proofs one prefers to or is able to give. Other ways
of organizing are possible, but it is our experience that it is not such a simple matter to
avoid confusing the reader. In particular, because the Change of Variables Theorem
in the present volume is about diffeomorphisms, it is necessary to introduce these
initially, in Volume I; a subsequent discussion of the Inverse Function Theorems
then is a plausible inference. Next, applications in geometry, to the theory of
differentiable manifolds, are natural. This geometry in its turn is indispensable for
the description of the boundaries of the open sets that occur in this volume, in the
Theorem on Integration of a Total Derivative in R", the generalization to R" of the
^Nevanlinna, F., Nevanlinna, R.: Absolute Analysis. Springer-Verlag, Berlin 1973, p. 1.
XV

xvi Introduction
Fundamental Theorem of Integral Calculus on R. This is why differentiation is
treated in the first volume and integration in this second. Moreover, most known
proofs of the Change of Variables Theorem require an Inverse Function, or the
Implicit Function Theorem, as does our first proof. However, for the benefit of
those readers who prefer a discussion of integration at an early stage, we have
included a second proof of the Change of Variables Theorem by elementary means.
We have stuck to the (admittedly, old-fashioned) theory of Riemann integration.
In our department students take a separate course on Lebesgue integration, where
its essential role in establishing completeness in many function spaces is carefully
discussed. For the topics in this book, however, the Lebesgue integral is not needed
and introducing it would cause an overload. In the applications considered, Arzela's
Dominated Convergence Theorem, for which we give a short proof, is an effective
alternative for Lebesgue's Dominated Convergence Theorem.
On some technical points. We have tried hard to reduce the number of e-S
arguments, while maintaining a uniform and high level of rigor.
Even for linear coordinate transformations the Change of Variables Theorem
is nontrivial, in contrast to the corresponding result in linear algebra. This stems
from the fact that in linear algebra the behavior of volume under invertible linear
transformations is usually part of the definition of volume. In analysis the notion
of volume relies on the Riemann integral, and for the latter only invariance under
translations is an immediate consequence of the definition.
The (i-dimensional density on a rf-dimensional submanifold in R" is considered
from two complementary points of view. On the one hand, the tangent space of the
manifold can be mapped onto R'' ~ R'' x {ORn-rf} C R" by means of a suitable
orthogonal transformation; pulling back the rf-volume on R'' under this mapping
one then finds a (i-density on the manifold. On the other hand, one can supplement
the basis B^ for the tangent space by a set of mutually perpendicular unit vectors
all of which are perpendicular to the tangent space, to form a basis B„ for R". Next
one defines the rf-volume of the span of B^ to be the n-volume of the span of B„ (in
other words, area equals volume divided by length). Both ways of thinking lead to
the same formalism, which unifies the many different formulae that are in use.
Vector analysis should look familiar to students in physics: therefore we have
chosen to center on the notion of vector field initially and on that of differential form
only later on. Leitmotiv in our treatment of vector analysis is the generalization of
the Fundamental Theorem of Integral Calculus on R to a theorem on R". There are
two aspects to the Fundamental Theorem of Integral Calculus on R: the existence
of an antiderivative for a continuous function; and the equality of the integral of
a derivative of a function over an open set with the integral of the function itself
over the boundary of that set. By generalizing the former aspect one arrives at
the infinitesimal notions in vector analysis, like grad, curl, div; and at Poincare's
Lemma, and its relation with homotopy. Likewise, the latter aspect leads to the
global notions, like the integral theorems, and their relations to homology.
This generalization to R" begins with the Theorem on Integration of a Total

Introduction xvii
Derivative, for which an easy proof is offered, by means of a local substitution of
variables that flattens the boundary. All other global theorems are reduced to this
theorem.
The existence of an antiderivative (or potential) for a vector field on R" with
n > 1 requires integrability conditions to be satisfied. That is, one needs the
vanishing of an obstruction against integrability, viz. of Af, twice the anti-adjoint
part of the total derivative Df of the vector field /. In R^ and R^, Af essentially
is the curl of /. Furthermore, Af approximately equals the sum of the values of /
at the vertices of a parallelogram, and that sum in turn is a Riemann sum for a line
integral of / along that parallelogram. Globalization of this argument leads to a
rudimentary form of Stokes' Integral Theorem: a relation between the circulation
of / and a surface integral of Af, i.e. an integral of the obstruction.
Vector analysis in R" is not a study of partial derivatives of components of vector-
valued functions, leading to a coordinate-dependent formulation and a "debauche
d'indices". Rather, it is an investigation of these functions and of their total
derivatives in their entirety, which is greatly facilitated by linear algebra, especially by the
decomposition of the derivative into self-adjoint and anti-adjoint parts using adjoint
linear operators.
The definition of positive orientation of a curve is an infinitesimal one. In
concrete examples it is often easy to verify whether it is satisfied without an appeal
to geometric intuition. The global definition, which is current in many elementary
texts, is less rigorous and may lead to cumbersome formulations and/or proofs, of
Green's and Stokes' Integral Theorems in particular.
Although formally the theory of differential forms receives an independent
treatment, the stage for it is in fact set by much of the preceding material. The main
result in the theory is Stokes' Theorem, and the whole discussion aims at proving
that theorem at the earliest possible moment. Therefore we have adopted a definition
of exterior derivative whereby we achieve this, and the proof of Stokes' Theorem
itself is then presented as a direct generalization of the proof of the rudimentary
form mentioned previously. The amount of multilinear algebra required for this
has been reduced to a minimum. In particular, the general differential k-foua is
introduced by means of determinants instead of exterior multiplication of forms of
lower order, which usually requires a laborious definition.
Exercises. Quite a few of the exercises are used to develop secondary but
interesting themes omitted from the main course of lectures for reasons of time, but which
often form the transition to more advanced theories. In many cases, exercises are
strung together as projects which, step by easy step, lead the reader to important
results. In order to set forth the interdependencies that inevitably arise, we begin an
exercise by listing the other ones which (in total or in part only) are prerequisites as
well as those exercises that use results from the one under discussion. The reader
should not feel obliged to completely cover the preliminaries before setting out to
work on subsequent exercises; quite often, only some terminology or minor results
are required.

xviii Introduction
Notational conventions. Our notation is fairly standard, yet we mention the
following conventions. Although it will often be convenient to write column vectors as
row vectors, the reader should remember that all vectors are in fact column vectors,
unless specified otherwise. Mappings always have precisely defined domains and
images, thus / : dom(/) —> im(/), but if we are unable, or do not wish, to specify
the domain we write / : R" D-> R'' for a mapping that is well-defined on some
subset of R" and takes values in R''. We write No for {0} U N, Ncx, for N U {oo},
and R+ for {xeR|x>0}. The open interval {x eR \ a < x <Z>}inRis
denoted hy]a,b[ and not by (a,b), in order to avoid confusion with the element
(a, b) e R2.
Making the notation consistent and transparent is difficult; in particular, every
way of designating partial derivatives has its flaws. Whenever possible, we write
Djf for the j-th column in a matrix representation of the total derivative Df of a
mapping / : R" —> R^. This leads to expressions like Djfj instead of Jacobi's
classical ^, etc. The convention just mentioned has not been applied dogmatically;
in the case of special coordinate systems like spherical coordinates, Jacobi's notation
is the one of preference. As a further complication, Dj is used by many authors,
especially in Fourier theory, for the momentum operator ' ^
1 3«V ■
We use the following dictionary of symbols to indicate the ends of various items:
□ Proof
O Definition
ik Example

Chapter 6
Integration
In this chapter we extend to R" the theory of the Riemann integral from the calculus
in one real variable. Principal results are a reduction of n-dimensional integration
to successive one-dimensional integrations, and the Change of Variables Theorem.
For this fundamental theorem we give three proofs: one in the main text and tvi^o
in the appendix to this chapter Important technical tools are the theorems from
Chapter 3 and partitions of unity over compact sets. As applications we treat
Fourier transformation, i.e. the decomposition of arbitrary functions into periodic
ones; and dominated convergence, being a sufficient condition for the interchange
of limits and integration.
6.1 Rectangles
Definition 6.1.1. An n-dimensional rectangle B, parallel to the coordinate axes, is
a subset of R" of the form
B = {x€R" laj < xj < bj (1 < ;■ < n)}, (6.1)
where it is assumed that aj, bj e R and aj < bj, for 1 < j < n, compare with
Definition 1.8.18.
The n-dimensional volume of B, notation vol„(B), is defined as
voi„(B)= n (^j-'^j)-
Note that vol,, (B) = 0 if there exists a j with aj = bj, that is, if B is contained in
an (n — l)-dimensional hyperplane in R", of the form {x e R" | Xj = aj}.
423

424
Chapter 6. Integration
A partition of a rectangle B is a finite collection <S = {B, \ i £ / } (here / is
called the index set of <S) of n-dimensional rectangles B, such that
B = [jBi; Bj nBj =Q or vol„(B,- n Bj) =0 if i ^ j.
iel
(6.2)
Let <S and £' be partitions of a rectangle B, then £' is said to be a refinement
of £ if for every Bj e <S the Bj e <S' with Bj C B, form a partition of B,. O
Proposition 6.1.2. Assume {B; \ i e I} is a partition of a rectangle B C R". Then
vol„(B) = 2^vol„(B,).
1 ; 1 1 1
I I ] ± .
1 j A !_.^ , L
I I I I ' i_
Illustration for the proof of Proposition 6.1.2
Proof. We first prove two auxiliary results.
(i). Assume B as in (6.1); and for 1 < j < n, let tj e [aj,bj] be arbitrary.
Consider
B' = {x e R" I aj < Xj < tj, and a^ < Xj^ < bj^, fork ^ j },
B" = [x € R" I tj < Xj < bj, and ak < xj, < bj,, fork y^ j }.
Because bj — aj = (bj — tj) + (tj — aj), it follows straight away that vol„(B) =
vol„(B') + vol„(B").
(ii). Assume next that for every 1 < y < n the segment [ aj, bj ] is subdivided by
the intermediate points
a,=tr'<
< t
{NU)) _
= bj.
(6.3)
'J - -J - - -J
Then we have, for every n-tuple
a = {a(\),..., a(n)) e N" where 1 < a(j) < N(j), (6.4)

6.2. Riemann integrability 425
a rectangle
B„ = {X e R" I rf "^-'^ < xj < ff ^^'^ (1 < y < n)}.
Together the Ba, with a as in (6.4), form a partition of B; and successive application
of assertion (i) now gives
vol„(B) = 5;]vol„(B„). (6.5)
a
Finally, let {B-,} be the given partition of B. For every 1 < J < n, collect
the endpoints of the y-th coordinate interval of all rectangles B-,, and arrange these
points in increasing order, as in (6.3) (see the illustration). Thus one obtains a
partition {B„} of B for which (6.5) holds. Likewise, the B^ with B„ C B; form a
partition of Bj for which
vol„(B,-)= Yl vol„(B„).
Therefore
Y^ vol,, {Bi) = Y^ Yl v°i'' iB^) = Yl ^°i" (^«) = ^°i« (^)•
Note that in the second summation vol,, (B„) may occur more than once, specifically
whenB„ C Bi and when £„ C Bj,fori ^ j. But this need not concern us, because
then vol„(Ba) = 0 on account of (6.2). □
Lemma 6.1.3. Any two partitions £' and £" of a rectangle B possess a common
refinement.
Proof. Analogous to the proof of the foregoing proposition. For every 1 < j <n,
collect the endpoints of the y-th coordinate interval of all subrectangles. □
6.2 Riemann integrability
Throughout this section we shall assume B to be an n-dimensional rectangle and
/ : B —> R to be a bounded function.
Definition 6.2.1. For every partition <S = {Bj | i e /} of B we define the lower
sum and the upper sum of f determined by <S, notation S_(f, £) and S(f, £),
respectively, by
S(f m = y] inf fix) vol„(B,), S(f S) = V sup f(x) vol„(B,). O
7^'^^^ 7^'^B'

426 Chapter 6. Integration
Lemma 6.2.2. (i) If S' is a refinement of S, then S_(f S) < S_(f, S') <
S(/, S') < S(f S).
(ii) For any two partitions £ and £' of B one has S_(f, £) < S(f £').
Proof. If Bj C Bi, then infj^g^, f(x) < inf ,.gB'. f(^)- Consequently one has, by
Proposition 6.1.2,
S(f£) =J2 inf fix) vol„(Bi) = J2mf fix) Y] vol„(B'.)
iei iei {j\B'j(zBi)
<T T, inf /Wvol„(B;) = Vinf /(x)vo1„(B;.)
lei {j\B'.(zBi) .' jeJ J
= Sif £'),
and this proves (i). Assertion (ii) follows from (i) and Lemma 6.1.3. □
Definition 6.2.3. The lower Riemann integral f fix) dx and the upper Riemann
—B
integral j^fix) dx of f over B are defined by
/ fix) dx = sup{ Sif S) I S partition of B },
J—B
/ fix) dx = inf {S(/, <S) | <S partition of B }. O
From the preceding lemma follows, for every partition £ of B,
Sif m< f f(x)dx < f fix)dx<'Sif £).
J „ J B
Definition 6.2.4. Let B be an n-dimensional rectangle. A function / : B —> R is
said to be Riemann integrable over B if
/ is bounded on B and / fix)dx= I fix)dx.
J g' J B
The common value of the lower and upper Riemann integrals is called the integral
of / over B, written as
/ fix)dx.
Jb
O

6.2. Riemann integrability 427
Proposition 6.2.5. Let f : B ^>- Rbe a bounded function. Then the following are
equivalent.
(i) / is Riemann integrable over B.
(ii) For every e > Q there exists apartitionS of B suchthatS{f, S) — S_{f, £) <
€.
Proof, (i) ^ (ii). For every e > 0 there is a partition £' of B such that S(f, £') <
/^/(x)rfx+|, and another partitions "of B such that S(/, £") > f^ f(x)dx—^.
For the common refinement <S of £' and £" it follows immediately that
S(f m-S(f m < S(/, £')-S(f S") < e.
(ii) ^ (i). The existence of such a partition <S implies f gf(x) dx — f f(x) dx < e;
and this holds for every e > 0. Hence follows (i). □
Now let / : R" —> R be a function satisfying
/ is bounded on R" and zero outside a bounded subset of R". (6.6)
Then there exists a rectangle B C R" with
fix) =0 if x^B. (6.7)
If / is Riemann integrable over B, the number f^ f(x)dx is independent of the
choice of B, provided that (6.7) is satisfied. Indeed, let B' be another such rectangle.
Then B = (BHB') U(B\ B'), where BHB'is an n-dimensional rectangle, while the
closure of B\B' can be written as a finite union of rectangles Bi, for / e /, satisfying
(6.2). Consequently, {B (1 B'} U {B,- \ i e /} is a partition of B. One furthermore
has fg f(x) dx =0 for i e /, because f(x) = 0 for x e int(B,) C R" \ B'. Here
int(B), the interior of a rectangle B, is obtained by only allowing the inequality
signs in (6.1). Therefore
/ f{x)dx= / f{x)dx= / f(x)dx.
Jb JBnB' Jb'
We need the following:
Definition 6.2.6. The support of a function / : R" —> R is defined as the set, see
Definition 1.2.9,
supp(/) = {xeR«|/(x)^0}. O
Note that supp(/) is compact (see the Heine-Borel Theorem 1.8.17) if supp(/)
is bounded. If x ^ supp(/), then there exists a neighborhood U of x such that
/(>') = 0,forall>'ef/.

428 Chapter 6. Integration
Definition 6.2.7. Let ^(R") be the collection of functions / : R" ^^ R which
satisfy (6.6) and which are Riemann integrable over B, for a rectangle B C R" as
in (6.7). The elements from .^(R") are called Riemann integrable functions with
compact support. Define, for every / e .^(R"),
/ fix)dx= / fix)dx= / fix)dx. O
^R" " J ' Jb
We recall the definition of the characteristic function 1^ of a subset A C R",
with
Ia(x) = 1 if X € A, IaM = 0 if x ^ A.
It follows immediately that the characteristic function 1 ^ of an n-dimensional
rectangle B is a Riemann integrable function with compact support. In addition
L
IsMdx = vol„(B).
R"
Theorem 6.2.8. We have the following properties for ^(R").
(i) .^(R") is a linear space with pointwise addition and multiplication by a
scalar
(ii) The mapping I : ^(R") —> R with f i-> f^„ f(x)dx is linear, and
monotonic, that is, if f < g (that is, f(x) < g(x), for all x e R"), then
1(f) < 1(g).
(iii) / e .^(R") if and only if f+ and /_ e ^(R"), where
f± = \(\f\±f)>0, f = f+-f-, 1/1 = /+ + /-;
and one has 1(f) = /(/+) — I(f-). In particular we have |/| £ ^(R") if
f e Sl(W), while
f f(x)dx\< f \f(x)\dx.
Jr" ' ^R"
(iv) If f, g e .^(R"), then fg e ^(R"), with fg defined by pointwise
multiplication.
Proof. Parts (i) and (ii) readily follow. With respect to (iii) we note that, for a
rectangle B,
sup fg - inf fg < sups / sups g - Mb f Mb g
= (sup / - inf /) sup g + M f (sup g - inf g).

6.3. Jordan measurability 429
to see this, distinguish the cases 0 < inf ^ /, inf ^ / < 0 < sup^ / and sup^ / < 0.
It follows that/ e .^(R")ifandonlyif/+ e .^(R")and/_ e .^(R"); and if this is
the case, then / (/) = / (/+) — / (/_). The particular case of |/| now follows from
I/I = /+ + /_; and the inequality of integrals is seen to result from the estimates
/ < I/I and"-/ < |/|.
(iv). Because of fg = f+g+ — /_ g+ — /+ g_ + /_ g_, it suffices to prove the
assertion for the case where / > 0 and g > 0. We then have
sup fg - inf fg < sups / sups g - Mb f inf b g
B B
= (sup / - inf /) sup g + inf f (sup g - inf g).
B B g B g B
The Riemann integrability of fg now easily follows. □
6.3 Jordan measurability
Definition 6.3.1. Let A C R" be bounded. Let / : R" —> R be a function and
assume that / is bounded on A. Then / is said to be Riemann integrable over A if
/ 1a is a function in the space ^(R"). If such is the case we write
/ f(x)dx:= I IaM fix)dx,
J A Jr"
'A JR"
and we speak of the Riemann integral or simply the integral of / over A.
The lower and upper Riemann integrals of 1 ^ are said to be the inner and outer
measures, respectively, of A. If 1^ is a function in the space ^(R"), then A is said
to be Jordan measurable, while
vol„(i4) := I dx = I lA(x)dx
J A Jr"
is said to be the n-dimensional volume or the n-dimensional Jordan measure of A.
In particular, therefore, an n-dimensional rectangle B is a Jordan measurable set.
The set A is said to be negligible in R" if vol,, {A) =0. O
Let A C R" be a bounded subset. Let B C R" be a rectangle with A (Z B and
let <S = {B; I z e /} be a partition of B. Define
leixtenor) = { Z £ / | B; H A ^ 0 }, Ii{„tenor) = { Z £ / | B; C A }.
Since sup^. 1^ = 1 if z € h, and inf b; 1^ = 1 if z € /,, it follows that
5(1a, m - SCIa, S) = j;] vol„(B,.) - 5;] vol„(B,.)
16/<r 16/;
Y^ vol„(B,-).
{;6/|B,nA5^0 £md Bi\Ai^9l]
(6.8)

430 Chapter 6. Integration
Theorem 6.3.2. Let A C R" be a bounded subset. Then the following assertions
are equivalent.
(i) A is Jordan measurable.
(ii) dA is negligible.
Proof, (i) ^ (ii). Let e > 0 be arbitrary. According to Proposition 6.2.5 there
exists a partition <S = {B,- \ i e /} of B such that (6.8) leads to
5^vol„(B,) - 5^vol„(B,) < ^.
vol„(i^,) <
Since U,g/^.B,- is closed, being a finite union of closed sets, and since A is the smallest
closed set containing A, it follows that
Ac[jBi. (6.9)
ieh.
Replacing the rectangles B; with / e /,■ by similar rectangles Bi whose edges are
of somewhat smaller length, we can arrange that
Ad[JB,d[J int(B,) D y Bi (6.10)
16/; 16/; I6/i
and
J2 vol,,(Bi) - J2 vol" (B'i) < e. (6.11)
isle isl.
Because U,-g/j int(B,-) is open, being a union of open sets, and because int(i4) is the
largest open set contained in A, (6.10) leads to
mt(A)D\jBi. (6.12)
ieii
From (1.4), (6.9) and (6.12) we now obtain
dA=J\mt(A)c[jB!\[jBi. (6.13)
;6/<: ,'6/;
But (6.13) and (6.11) together imply that the outer measure of dA is smaller
thane.
(ii) ^ (i). For this proof we also apply Proposition 6.2.5. Hence let e > 0 be
arbitrary. Then we can find a partition {Bi \ i e 1} of B such that
9i4 C 1)int(B,) and 2_] (sup 1a — inf 1a) vo1„(B,-) < e.
iei iei •
For every X e A\9 A, we can select a rectangle Bj: satisfying x e \n\.{Bx) C B.v C A.
Note that sup^ 1a — ™^b^ 1a = 0. Because A is compact, the open covering

6.3. Jordan measurability
431
Illustration for Formula (6.13)
[B; I i e I}D [Bx I X e A\^A} of A admits a finite subcovering, say <S',
on the strength of the Heine-Borel Theorem 1.8.17. Next we use the endpoints of
the coordinate intervals of the rectangles in S' to define a partition S of B; then
each of the rectangles in <S' is a union of rectangles in <S. It is immediate now that
5(/, S) - S(/, S) < e. □
Corollary 6.33. Assume A and B C R" are bounded and Jordan measurable.
Then the following sets are also bounded and Jordan measurable:
Ar\B, AUB, A\B, mt(A),
A.
Proof. Note that Iahb = ^a^b, ^aub = 1a + 1b - ^ahb, and 1a\b = I a - Iahb-
Now use Theorem 6.2.8.(i) and (iv). According to Formula (1.4) one has int A =
A\dAdsiA~A = A\JdA. □
Remark. In what follows we shall often be dealing with compact subsets of R".
Such sets are not necessarily Jordan measurable (see Exercise 6.1 for an example
inR).
Definition 6.3.4. For A C R" we denote by ${A) the collection of compact and
Jordan measurable subsets of i4. O

432
Chapter 6. Integration
The following theorem gives a most useful criterion for the Riemann integra-
bility of a function over a set.
Theorem 6.3.5. Let K e ^(R") and let f : K ^>- R be a continuous function.
Then we have the following properties.
(i) / is Riemann integrable over K.
(ii) Next, assume that g : K ^ Viis also continuous and that g(x) < f(x),for
allx e K. Then L e |(R"+') if
L = {(x,y)e R"+' \xeK, g(x)<y< f(x)};
and vol„+, (L) = /^^ (f(x) - g(x)) dx.
Illustration for the proof of Theorem 6.3.5.(ii)
Proof. Considering Theorem 6.2.8.(iii), it is sufficient to prove (i) for the case
f(x) > 0 (x e K). (6.14)
Further, let c = minj^g^: g(^)- Since c < g(x) < f(x), for x e K, we have
L ={(x,y)eR"+^ \x sK, c<y<f(x)}
~ (6.15)
\{(x,>')€R"+' \xeK, c<y<g(x)}.
In view of this we first prove (ii) for the case where
L = {(x, y) e R"+' I X e ii:, 0<y< f(x)}. (6.16)

6.3. Jordan measurability 433
Under the assumptions (6.14) and (6.16) we will now simultaneously prove both
(i) and (ii). Let B C R" be a rectangle with ii: C B and let <S = [B, | z e /} be a
partition of B. Define f{x) = 0, if x £ B\K. Then
S(U/, S) = 5^supU(x)/(x)vol„(B,-)= Yl sup/vol,,(B,),
S(lKf, S) = V inf lK(x)f(x)vol„(B,) = Y] inf/vol„(B,).
'xeBi ^—^ Bi
iel Bi<zK
Further, we note that L C B := B x [ 0, sup^j- / ], where B is a rectangle in R""*"'.
In addition
^C U B,x[0, sup/], (6.17)
and, if B; C K,
B; x[0, inf/]cL. (6.18)
Bi
Let <S be a partition of B such that it is a common refinement of all rectangles
Bi X [ 0, sup5_, / ], for Bir[K i^ 0, and all B,- x [0, infg, / ], for B/ C K. From
(6.17) and (6.18), respectively, one then finds
s(1l, S)< J2 sup/vol,,(b,) = suk/, m,
S(lL,m> Yl inf/vol„(B,) = S(U/ S).
Bi
Bi(zK
Accordingly, (i) and (ii) result if we prove that the following difference becomes
small upon a suitable choice of the partition <S:
S(\Kf, m-SilKf, m= Yl sup/ vol,,(B,)- Y, inf/vol„(B,)
BiHK^Id ^' BidK '
= 2^(sup/-inf/)vol„(B,-)+ Yl sup/vol„(B,).
By Theorem 1.8.15, / is uniformly continuous on the compact set K, that is, for
every ?j > 0 there exists a ^ > 0 such that sup^, / — inf g. / < ?j, if B, C K and B;
has edges of length < S. This may be arranged for all B; by making them smaller
where necessary. Because of Formula (6.8), the Jordan measurability of K implies
Y2 vol„(B,) < r],
BinK^a, Bi\K^0
for suitably chosen <S. And hence, for such a <S,
SilKf, m-SilKf, m < r,YolniK)+maxfix)r,.
xeK

434 Chapter 6. Integration
We return to the general case for (ii). This is a consequence of the foregoing and
of (6.15), once we know that the following sets M and M' have the same (n + 1)-
dimensional volume, where
M := {(x, y) e R"+' \x€K, c<y <g(x)}
and
M' := {(x, y) e R"+' \x sK, c<y < g(x)}.
But this is the case, because vol„+i(M \ M') = 0 on the basis of the inclusion
M \M' C dM and the preceding theorem. □
Definition 6.3.6. We denote the space of continuous functions R" —> R with
compact support by
Cc(R") = C°(R") = {/ : R" ^^ R I / continuous and supp(/) compact}. O
Corollary 6.3.7. Cc(R") C .^(R"), that is, the linear space of continuous
functions on R" with compact support is contained in the linear space of the Riemann
integrable functions on R" with compact support.
Proof. Let B C R" be a rectangle with supp(/) C B, then / is continuous on
B. Hence, according to (i) of the preceding theorem, / is Riemann integrable over
B. □
Corollary 6.3.8. LetK e $(R''),ford < n, and let f : K —> W^"^ be a continuous
mapping. Then
graph(/) = {(x, f{x))\x€K}
is a negligible set in R". More generally, finite unions of this kind of graph are
negligible in R".
Proof. The boundedness of the continuous function x i-> (/i(Ji:),..., /„_rf_i(x))
on iiT implies there is a rectangles C R""''"' such that (/i(x),..., /„_rf_i(x)) e B
if X e K. Defining the continuous function / : AT x B —> R by f{x, t) = fn~d{x),
we have
graph(/) ={(x, f{x))€W\x€K}
C {{x, t, fix, t)) e R" \{x,t)^K^B} = graph(7).
Now let c = v[m\(^j)^KxB 7i.x, t). Then
graph(/) C 9{ (x, t, y) e R" \ {x,t) € K x B, c < y < f(x, t)},
and so the assertion follows from the two preceding theorems. □

6.4. Successive integration 435
Example 6.3.9. (i) The interval [ — 1, 1 ] belongs to ^(R), because it is a rectangle
inR.
(ii) The circular disk B^ = {x e R^ | ||x || < 1} belongs to $(R^). Indeed
B^ = {X e R2 I xi e [-1, 1], -fix,) <X2< fix,)},
where / : [—1, 1] —> R is the continuous function with fix,) = Jl — x^. The
assertion therefore follows from (ii) of Theorem 6.3.5.
(iii) The subset Kofa solid cone in R^ defined by
A: = {x eR^ |0<X3 < 1, xf+x^ < il-xs)^}
is compact; indeed, for x e K one has x^ + x| < 1, therefore K C {x e R^ \
\^i I < 1 (1 < 2 < 3)}. Furthermore, AT is a Jordan measurable set in R^. This is so
because (x,,X2,X3) e AT implies that (x,,X2) £ B^. Conversely, with (xi,X2) £ B^
fixed, the inequalities Jx^ + x^ < 1 — X3 and 0 < X3 imply that then X3 may still
vary as follows: 0 < X3 < 1 — Jx^ + x^. In other words
a: = {x e R^ I (xi, X2) e B^ 0 < X3 < /(xi, X2)},
where / : B —> R is the continuous function with /(xi,X2) = 1 — Jxj+X2.
Thus, again by (ii) of Theorem 6.3.5, the assertion follows. ik
6.4 Successive integration
We now formulate results that will enable us to reduce an n-dimensional integration
to successive lower-dimensional integrations. This method is most effective if
reduction is possible to one-dimensional integrations which then can be performed
by computing antiderivatives.
Consider a function / : R^ x R'' —> R. If / is continuous, the function
Z H> fiy, z) is a continuous function on R'', for every y e R''. Remarkably, for
Riemann integrability there is no valid analogous result. If / is Riemann integrable,
it does not necessarily follow, for every y e R'', that the function z i-^- fiy, z) is a
Riemann integrable function on R^. This explains the formulation of Theorem 6.4.2
below.
Definition 6.4.1. A bounded function / : R" —> R is said to be a step function
if there exist a rectangle B C R" and a partition <S = {B,- | i e 1} of B such
that fix) = 0, for x ^ B and such that /IjntCBV ^°^ ^^^ ^ ^ /, is a constant
function, while, for every 1 < j < n, the function Xj i-> / (x 1, .., xj,.., x„) is left-
continuous. If these conditions are met, / is said to be a step function associated
with £. Note that a step function is Riemann integrable over R". O

436 Chapter 6. Integration
Theorem 6.4.2. Let f be a Riemann integrable function with compact support, on
W+^. Then
y^ f fiy,z)dz and y^ f f(y,z)dz (y sR")
are Riemann integrable functions with compact support on R^, and
f f(x)dx= f f f(y,z)dzdy= f f f(y,z)dzdy.
JRP-I-'I Jrp J Ry Jrp J Rl
Proof. Let B C R''"'"^ be a rectangle such that / vanishes outside B, let S = [Bj |
i e /} be a partition of B, and let g_, g+ be step functions associated with <S such
that
g^(x) < fix) < g+(x) (X e B).
Then for every y e R^ the functions ^ i-> g^iy, z)andz i-> g+(y, z) are step
functions on R'' dominated by ^ i-> f(y, z), and dominating z i-^- f(y, z), respectively.
Accordingly one has, for every y e R'',
f g-iy,z)dz< I fiy,z)dz< I f(y,z)dz< I g+(y, z)dz.
Jri J—Ri •' ^'' ''R'
But y i-> /jj, g±(y, z) dz in turn are step functions on R^; and we obtain
f g-{x)dx =f f g-iy,z)dzdy< f f f(y,z)dzdy
Jrp-'-i Jrp Jri J_RpJ_Rq
<f f f(y,z)dzdy<f f f(y,z)dzdy
J RPJ ^a J RPJ Rl
</ / g+(y,z)dzdy= g+(x)dx.
Jrp Jri Jrp->-i
Because the supremum of the left-hand side and the infimum of the right-hand side,
both taken over all possible partitions <S, equal /j^p.,., f(x) dx, it follows that
f f(x)dx= f f f(y,z)dzdy= f f f(y,z)dzdy,
JrP-I-1 •L.RPlL.RI J RPJ R"
which proves the assertion about y h> f f(y,z)dz. □
Remark. The formulation of the preceding theorem may not be simplified, as the
following example demonstrates. Let f(x, y) = -, if x = ^ with p,q e N, where
p and q are relatively prime, p < q, and y e Q D [0, 1 ]. Let /(x, y) = 0, in all
other cases. Then / is Riemann integrable over R^ with vanishing integral. But for
every x = ■^ as above, y ^-^ f(x,y) is not Riemann integrable: / f(x,y)dy = 0,
while J^fix, y) dy = ^.

6.4. Successive integration 437
Remark. If in the preceding theorem the roles of y e R^ and ^ e R'' are
interchanged, it follows that the functions Z\-^ j f(y, z) dy and z i-^- f^pfiy, z) dy
are also Riemann integrable over R'', and that their integrals over R^ both equal
/r/'-I"; f{x)dx. This constitutes another proof, different from the one of
Theorem 2.10.7.(iii), that the order of integration may be interchanged. In particular, we
have obtained the following:
Corollary 6.4.3 (Interchanging the order of integration). Let f : R''+^ —> R
be a continuous function with compact support. Then, for every y e R'', the
integral f^^ f(y, z)dz is well-defined, and in addition, for every z € R'', the integral
Irp /(y^ 2) ^y ^^^ the functions thus defined are Riemann integrable with compact
support on W and W, respectively. Furthermore,
I f(x)dx= f f f(y,z)dzdy= f f f(y,z)dydz.
jRP-i-l Jrp Jri Jri Jrp
Example 6.4.4. The requirement of continuity (or of boundedness) of / does play
a role, as becomes apparent from the following. One has
If the integrals
/ / -dydx and / / -dxdy
Jo Jo (x + y? Jo Jo (x + y?
were both well-defined, one would, on symmetry grounds, expect them to be equal.
The interchangeability of the order of integration would then imply that the original
integrals vanish. But we have, for x > 0, which causes the convergence,
/" x-y . _ rix-jx + y) _ /"/ 2x 1 X
Jo (x + yy ^ Jo (x + yy ^ Jo Hx + yy (x + yrJ^^
(X + yr
1 y'='
1 X
+ r +
(X + 1)2 X + l X2 X (X + 1)2'
Consequently,
Jo Jo (x + y? ^ "" Jo (x +
—::dx =
1)2
1 -■'
X + l
--- 1--
Interchanging the roles of x and y, we obtain
/ / ~—^dxdy = -, thatis / / ~—'^—;dxdy = --. ilV
Jo Jo ix + yf 2' /o Jo ix + yf 2

438 Chapter 6. Integration
Theorem 6.4.5. Let ai, Z>i e R with ai < bi, and write Ki = [a\,bi ]. Assume
that there have been defined, by induction over 2 < j <n,
continuous functions a j, bj : Kj-i —> R with aj < bj;
and sets Kj C R-' given by
{x e R-' I (xi,...,Xj^i) e Kj^i, ajixi, ...,xj_i) <Xj < bj(xi,... ,Xj^i)}.
Then the following assertions hold.
(i) The sets Kj belong to $(RJ).
(ii) For every continuous function / : if „ —> R one has
I f(x) dx
Jk„ '
nbi nh2{xi) nb„{xi,...,x„.i)
= 11 •••/ f(xi,--.,x„..i,x„)dx„---dx2dxi.
Jai Ja2{xi) Ja„{Xi,...,x„-i)
Proof. Assertion (i) follows by induction over j and Theorem 6.3.5. Write
x' = (xi,... ,x„_i) e R''-^ whence X = (x', x„) e R". Note that 1a:„(x) =
1a:„_,(-^') h"„(x'),b„(x')](x„). For every x' e K„_i the function x„ i-^^ (Ik„ f)(x\ x„)
is continuous on the interval [fl„(x'), &„ (x')]; and so, by Theorem 6.3.5.(i), it is Rie-
mann integrable over the said interval. That is, for every x' e K„_i,
r ri>n{x')
/ (1a:„ f)(x', x„)dx„ = Ik„.i(x') j f(x\ x„)dx„.
Jr Ja„(x')
-bnix')
la„(x')
Applying the preceding corollary, one finds
i-b„{x')
I f{x)dx =1 {\K„f){x)dx=\ ^.K„.M') I fix', Xn)dx„dx'
JK„ Jr" Jr"-' Ja„{x')
r rhn{x')
= I I fix', x„) dxn dx'.
The proof of assertion (ii) can now be completed by induction over n, provided we
know
rb„(x')
x'^ / fix',x„)dx„ (6.19)
Ja„{x')
to be a continuous function on if„_i. In order to verify this, define
nb
7 : R^ X K„_y ^^ R by /(a, b, x') = / fix', x„) dx„.
J a

6.5. Examples of successive integration 439
Then
\Iia,b,x')-I{a,b,x')\ < \I(a,b,x') - Iia,b,x')\
+ \I(a, b,x') - I (a, b,x')\ + |/(^, b,x') - I {a, b,x')\
na I nb
/ f{x',Xn)dXn + / fix',X„)dXn
J a ' Jb
I /■*
+ / {.fix', X„) - f(x', X„))dXn
Because / is uniformly continuous on if,,, there exists, for every ?; > 0, a ^ > 0
such that, uniformly in x„,
\f(x', x„) - fix', x„)\ < TO i\\x' - x'W < S).
From this follows, for \a — a\ < min{?j, 1}, \b — b\ < min{?j, 1}, ||x' — x'|| < S,
\Iia, b,x')-Iia, b,x')\ <?j(2max|f| + |fl| + |Z>|+2),
K„
proving the continuity of I in (a, b, x'). But this implies the continuity of the
function in (6.19). □
Remark. In the proof above the Continuity Theorem 2.10.2 is not directly
applicable since the interval of integration J = [a,b]of the variable x„ is also allowed
to vary.
6.5 Examples of successive integration
Example 6.5.1. Let K be the compact set in R^, bounded by the planes
{x sR^ \xi+X2+X3 = a} (fl > 0), {xeR^|x,=0} (1 < i < 3).
Then
/.
II Ii2j '^^
\\x\\ ax = —.
20
We have
AT = {x e R^ I 0 <xi, 0 <X2, 0 <X3, xi +X2+X3 <a}.
Therefore x e AT means that 0 < while X2 > 0 and X3 > 0, that
is Xi e [0, fl ]. Once xi e [ 0, a ] has been fixed, 0 < X2 < a — xi — X3 and X3 > 0
imply that X2 may still vary within [ 0, a — Xi ]; and for (xi, X2) thus fixed, X3 may
still vary within [0, a — xi — X2 ]. Consequently
A" = {X e R^ I 0 < Xi < fl, 0 < X2 < fl — Xi, 0 < X3 < fl — Xi — X2 }.

440
Chapter 6. Integration
Illustration for Example 6.5.1
Hence we have, in the notation of the preceding theorem,
Ki = [0,al K2 = {ixi,X2)€R^\xi€Ki, 0<X2<a-Xi}, K^ = K.
As a consequence
I \\x\\'^dx= II I (X'^ + X2+X^)dX3dX2dXi
Jk Jo Jo Jo
= \xfx3+X^X3 + Y
xi=a—xi—X2
dX2 dXi
.V3=0
X2)^
= I I (xy(a - xi — X2) + X2ia — xi — X2) H ) dx2 dxi
Jo Jo J
= / x'^(a — X
x^x| (fl—xi)x| X2 (a — xi — X2)'^
dx\
2 -^1
-xiY-
x}(a — xi) (a—Xi) (a — xi) (a — Xi)
12 /
Jo \ 2 6 / ^' 20'
1^

6.5. Examples of successive integration
441
Example 6.5.2 (Intersection of two cylinders). The intersection of the two solid
cylinders in B? given by {x e R^ | x^ + x| < 1} and {x e R^ | x^ + x| < 1},
respectively, has volume y.
Proof. I. From x\ + x\ < 1 and x\ + x\ < 1 we see that xi may vary in the entire
interval [ — 1, 1 ]. We set b{x) = Vl — x^. Given xi, we see from x| < 1 — x^ that
X2 may still vary in [—Z>(xi), Z>(x,)]; likewise, X3 may vary in the same interval
[ —Z>(xi), Z>(x,) ]. The intersection, therefore, is given by
{x e R^ I -1 < X, < 1, -b{x{) < X2 < bixi), -bixi) < X3 < Z>(x,)}.
The desired volume is
.1 pbiXi) nb{xi)
-fe(.v,) J-b{xy)
/ / dx^dx2dxi =S I I I dx^dx2dxi
-I J-b(xi) J-b(xi) Jo Jo Jo
-'If
Jo Jo
1 pb{xi)
b(xi)dx2dx^
= 8 /" b(x
Jo
ydxi = 8
xi
= 8(1--). □
Proof. II.Fromx^+x| < 1 it follows that X2 may vary in the entire interval [—1,1 ].
Given X2, we see from x^ < 1—x| that xi may vary in the interval [—Z>(x2), b(x2)].
Finally then, X3 may vary in [ —Z>(xi), b(xi) ]. Thus the intersection is given by
{x e R^ I -1 < X2 < 1, -b(x2) < X, < Z>(X2), -b(xi) < X3 < b(Xi)}.
And so the desired volume is
/I /'b{X2) /'b{xi) n\ t'b{X2) /'b{xi)
I I dxj, dxi dx2 = S I I I dxs dxi dx2
-1 J-b{X2) J-b{xi) Jo Jo Jo
= 8 / / J\—x\dxidx2.
Upon computation of an antiderivative of J1 — x, = .—L this becomes
m
l-x? +
arcsmx,
X\=b{X2)
;c]=0
dX2
— Xrj ) ClXn
= 4 / (x2y 1 — X2 + arcsin ^ 1 — x|) i
= 4 / X2J\ — x| dx2 + 4 / arccosx2 dx2.
Jo Jo

442
Chapter 6. Integration
And now
• 1
/
Jq
f
Jq
X2J I — Xy dX;
2 ax2 =
(1 - x|)3/2
Jo 3'
arccos X2 dx2 = [ X2 arccos X2 ]
= arccos 1
Jo
dX2
□
Illustration for Example 6.5.2: Intersection of two cylinders
Example 6.5.3 (Sharpening a pencil). A hexagonal pencil, not sharpened, has
a thickness of 7 mm between two parallel faces. The pencil is sharpened into an
exact conical point such that at the tip the angle between two generators on the
conical surface is at most ^ radians. The volume of the material removed equals
0.389 • • • cm^, if the sharpening has been done in the most economical way.
Indeed, choose the origin of the coordinate system for R^ at the tip of the
pencil, the negative direction of the xs-axis along the axis of the pencil, the xi-axis
perpendicular to a face of the pencil. Let a = 3.5. Because of the sixfold symmetry
of the pencil about the xs-axis, the planes through the xs-axis and those edges of the
pencil that lie in the plane {X e R^ | Xi = a } form an angle of ^ radians. Therefore,

6.5. Examples of successive integration 443
these planes are defined by the equations X2 = ±tan|-xi = ±|V3xi. LetC C R^
be the conical surface, part of which bounds the pencil. Note that for given (xi, X2)
there exists a unique x^ = x°(xi, X2) such that X3 < 0 and (xi, X2, x°) e C. Now
the volume removed equals six times that of the set
[x sR^ \0 <xi <a, --VSxi <X2< -VSx,, x°(xi, X2) < ^3 < 0}.
Next we derive an equation for C. To calculate tan -^ we use the formula for the
tangent of the double angle:
1 /- 71 2 tan T^ , 7J- 6 TT
-V3 = tan — = ^^—, that is, 1 — tan — = —= tan —.
3 6 1 - tan2 ^ 12 V3 12
The positive root of this quadratic equation equals tan ^ = |(—2-s/3 + Vl2 + 4) =
2 - V3, so
/^ ^ \ 1 1 ^ /x
tan I 1 = = = 2 + V3.
\2 12/ tan§ 2-V3
Therefore, x e R^ lying on one of the two generators of the conical surface C
contained in the plane {x e R^ | X2 = 0} satisfies X3 = — tan(|- — f^)xi =
— (2+-s/3)xi. And C itself has the equation X3 = —(2+-s/3)»/x^+x|, in particular
x° = x°ixi, X2) = -(2 + 73)^x2 + xj.
Consequently, the desired volume is given by
' / /
Jo J-^V3xi J-Q
dXs dX2 dXi
(2+V3)^.Vj-+^|
= 12(2 + V3)/ /' Jx^ + x^dx2dxi.
2, 2
Using an antiderivative of .L-i^ with respect to X2, we find, modulo the factor
V-'^r+jt2
12(2 + V3)
I I jy/4+^f + y log(^2 + 7^1 + ^?
rfxi
= ^(4 + 31og3)(3.5)^ = 8.689---. tV

444 Chapter 6. Integration
6.6 Change of Variables Theorem: formulation and
examples
In the integral calculus on R an important result is the Fundamental Theorem 2.10.1,
/
f'(x)dx = f(b)-f(a).
which gives a relation between integration and differentiation on R. This result has
a direct analog in the calculus of integrals on R" (see Theorem 7.6.1), in the form
of a relation between integration and total differentiation on R". On R the Change
of Variables Theorem
/ fix) dx= (/ o ^)(y)^'(y) dy, (6.20)
for ^ : [a,b] —> R continuously differentiable and / : ^([a,b]) —> R
continuous, is a direct consequence of the Fundamental Theorem. Indeed, let F
be an antiderivative of / on ^([a, b]), then both sides of (6.20) are equal to
F o^ (b) — F o^ (a). Hence, the essence is to recognize the function / as a
derivative; on R" with n > 1, however, this is not possible, because a derived function
on R" does not take values in R, but in Lin(R", R) ~ R". Nonetheless, a Change
of Variables Theorem does exist for R" with n > 1, but its proof is distinctly more
complicated than that for R. The proof for R" is obtained by means of a reduction
to R, and requires some technical tools: the Implicit Function Theorem 3.5.1, and
results which enable us to localize an integral, that is, to study an integral near
a point. For this reason we prefer to begin by merely stating the theorem, and
then illustrating it by some examples. Section 6.7 contains the results relating to
localization, while the proof can be found in Section 6.9.
Other proofs. The appendix to this chapter, in Section 6.13, contains two more
proofs of the Change of Variables Theorem which consider the problem from
different perspectives.
Theorem 6.6.1 (Change of Variables Theorem). Assume U and V to be open
subsets o/R" and let ^ : V —> U be aC^ diffeomorphism. Let f : U ^>- Rbe a
bounded function with compact support. Then f is Riemann integrable over U if
and only if the function
jh^(/ovI/)(j)|detDvI/(j)|
is Riemann integrable over V. If either condition is met, one has
f fix) dx= f fix) dx= f if o vl/)(j)| det D^iy)\ dy.
^*(V) Ju Jv

6.6. Change of Variables Theorem: formulation and examples 445
An immediate consequence of this theorem and its proof is the following:
CoroUary 6.6.2. IfL e %iV), then *(L) e |(f/), while
vol„(vI/(L))= /" U^L){x)dx= f \detD^(y)\dy.
Ju Jl
If moreover, ^ satisfies the condition \ det D'^{y)\ = I, for all y e L, we have
vol„(vI/(L)) = vol„(L).
In view of this corollary a C' diffeomorphism ^ : V —> f/ with the property
I det D^\ = 1 on V is said to be volume-preserving.
Remarks. Theorem 6.6.1 tells us how an integral over U transforms under a C'
regular coordinate transformation x = ^(y) onU. Specifically, the integral over a
transformed set of a function equals the integral over that set of the product of the
transformed function and the absolute value of the Jacobian of the transformation.
Here we recall the terminology Jacobi matrix of ^ at y to describe D^(y); and the
Jacobian of ^ at y is defined as
detD^Cy).
Consider the case n = \. For a C' diffeomorphism ^ : V —> f/ where V,
f/ C R, we have, on account of Example 2.4.9, I>*0') = "^'(y) y^ 0, for all
y S V. However, the proof of the Change of Variables Theorem (6.20) on R did
not assume ^'(y) ^ 0, for all y e ]a,b[; thus the "old" version of the Change
of Variables Theorem on R is stronger than the "new" one. We now consider the
"new" version of the Change of Variables Theorem on R. If ^ : ] a, Z> [ —> R is a
C' diffeomorphism, then by the Intermediate Value Theorem 1.9.5 the continuity
of *' and the fact that *'(>') y^ 0, for all j e ] a, Z> [, imply that either ^'Cy) > 0,
or ^'(y) < 0, for all y e ]a,b[. But this means that ^ is strictly monotonic,
either increasing or decreasing, on [a, b], and either ^([a, b]) = [^(a), ^(b)]
or *([fl,Z>]) = [*(Z>), *(«)], respectively. In the latter case IdetD^Cy)! =
— ^'(y), and so the Change of Variables Theorem 6.6.1 then yields
f " f(x)dx = - f (/ovI/)(y)vI/'(>-)rf>-= f\foyiJ)(y)yiJ'(y)dy.
J>ii(b) J a Jb
The formulation of Theorem 6.6.1 as given above is more or less dictated by
the fact that under these conditions the structure of the proof remains reasonably
transparent. Thus, the present formulation of the theorem is for open sets in R",
because these are the natural domains of diffeomorphisms. Sometimes, the
mapping ^ : V —> f/ which would seem the most natural, cannot be extended to a
diffeomorphism V —> f/, for example because det D^ vanishes at points of dV.

446 Chapter 6. Integration
Furthermore, the condition on / ensures that no convergence problems will crop
up.
Nevertheless, in many cases Corollary 6.6.2 in the form given above cannot be
used to calculate the volume
= f Ik(x)
VOln(K) = I lK(x)dx
Jr"
of a compact and Jordan measurable set K C R", for the following reason. The
calculation requires finding a C^ diffeomorphism ^ : V —> f/ and a set L e $(V)
with ^(L) = K. The problem often involved in this is that the obvious change of
variables ^ becomes singular on dL; consequently, ^ is not a C' diffeomorphism
on an open neighborhood of L in such cases.
Therefore the proof of the Change of Variables Theorem is followed, in
Section 6.10, by limit arguments which justify calculating vol,, (if) by means of the
theorem after all: on account of Theorem 6.3.2 one has vol,, (if) = vol„(int K)
for Jordan measurable sets K, and the characteristic function of int(if) may be
approximated from within by admissible functions. Accordingly, in the examples
we will use the Change of Variables Theorem for calculating the volumes of
"reasonable" sets, meaning sets in R" bounded by a finite number of lower-dimensional
C' manifolds.
Example 6.6.3 (Yolume of parallelepiped). LetB c R" be a rectangle and assume
^ : R" —> R" is a linear mapping. Then ^(B) is ai parallelepiped in R", that is, a
set of the form
[a+Y. O^^MO<0< 1(1 <;•<«)};
here a is one of the vertices of ^(B) and the Z>^^\ for 1 < j < n, are the direction
vectors of the edges that originate from a. By the Change of Variables Theorem
(see Proposition 6.13.4 for a direct proof) it follows that
vol„(vI/(B))=|detvI/|vol„(B),
that is, vol,, {^(B))/ vol„(B) is independent of the set B, whereas it does depend
on the linear mapping ^.
Note that in linear algebra this transformation property usually is one of the
defining properties of volume in an n-dimensional vector space; in our
development of analysis in R" the Change of Variables Theorem is required to obtain this
result. This comes about because in Section 6.3 we have chosen a definition of
n-dimensional volume which is more elementary, one in which rectangles play a
fundamental role. ik

6.6. Change of Variables Theorem: formulation and examples 447
Example 6.6.4 (Polar coordinates). Let f/ = R^\(]—oo, 0]x {0}) be the plane
excluding the nonpositive part of the x,-axis. Then define V and ^ by
V = R+ X ] —7Z, JT [ C R^ and ^(r, a) = r(cosa, sina).
On account of Example 3.1.1, ^ is a C' diffeomorphism, while
det D*(r, a) = det . ) = r > 0 ((r, a) e V).
\ sin a rcosa /
Using the Change of Variables Theorem, together with Corollary 6.4.3, we obtain,
for continuous functions / that vanish outside a compact subset contained in U,
f f(x)dx= f(foyiJ)(y)\dctD^(y)\dy
7r2 Jv
= / '" / f (r cos a, r sin a) da dr = I I rf(r cos a, r sin a) dr da.
v/r.). J-tt J-tt Jr+
Consider —tz < a^ < a2 < tz and (p e C(] ai, a2[), and define
V = {(r, a) e V I a, < a < 012, 0 < r < (j>(a)}.
By way of application we compute the area of the bounded open set U' = ^(V),
that is, U' is the bounded open set in R^ that in polar coordinates is bounded by the
lines with equation a = ai and a = a2 and the curve with equation r = <p{a). It
follows that
area(f/') = / / rdrda = - I (p(a)^da.
Jd] Jq ^ Jai
In particular, the area of the bounded open set in R^ that in polar coordinates is
bounded by the lemniscate r^ = a^ cos 2a with a > 0 (see Exercise 5.18), equals
a f^
area(t/') = 2— / cos 2a; da = a^. i^
2 J_!L
Example 6.6.5 (Volume of truncated cone). Let G C R^ be an open set. The
truncated cone with apex (0, 0, t) where 0 < ^, of height h where Q < h <t, and
with Jordan measurable basis G C R^, is defined as the open set t/ C R^ with
U = {{{!- s)y, st)€'R^\y€G, 0 < ^ < - }.
t
We now define V C R^ and * : V ^^ f/ by
V = Gxlo,-[ and ^{y,s) = {{\ - s)y, st).

448 Chapter 6. Integration
Then it is easy to see that ^ : V —> f/ is a bijective C' mapping, and that
/ 1 - ^ 0 -yi \
det D^Cy, ^) = det 0 I - s -y2 \ = (I - sft > 0 {{y, s) e V).
V 0 0 t I
Using the Global Inverse Function Theorem 3.2.8 we see that ^ : V —> f/ is a C'
diffeomorphism, and by the Change of Variables Theorem and Theorem 6.4.5 we
find
VOl3(f/)
= f dx= f t(l- sfdyds = t f dy /"' (1 - sf
Ju Jv Jg Jq
= f areaCG)!"-^^^^
^=|area(G)(l-(l-^
In particular, if G is a square with edges of length a, the upper face of U is also
a square, with edges of length b = a(l — j). Furthermore, in that case ^ = ■^.
Hence the volume of a truncated pyramid of height /z, having lower and upper faces
with edges of length a and b, respectively, equals
--{a^-b^) = -ia^ + ab + b^). ik
3 a 3
Example 6.6.6. It is sometimes easier to give the mapping <1> := ^""^ : f/ —> V,
and there may be no need for an explicit description of ^ : V —> U. Assume we
have, as in Application 3.3.A,
f{x) = \\xf, f/ = {x e R2 I 1 < xf -x^ < 9, 1 < 2xiX2 < 4},
V = {y e R2 I 1 < J, < 9, 1 < yz < 4}, <D(x) = (xf -x|, 2xiX2) =: y.
According to Application 3.3.A, <1> : U —> V is a C' diffeomorphism, with inverse
1
4|b'll'
-9 /.4
^ := <!)-' and det D^(y) = j^ and we have /(x) = \\y\\. Therefore
/ f(x)dx= / \\y\\-—-—dy = - dy = - dyi dy2 = 6. ik
Ju Jv M\y\\ "^Jv 4y, JI
Example 6.6.7 (Nevrton's potential of a ball). The Newton potential of set A e
$(R^) is the function 0^ : R^ \ A ^^ R, defined by
1 r 1
(pAix) = ---- / -dy.
^^ J A \\x - y\\

6.6. Change of Variables Theorem: formulation and examples 449
Now assume A to be the closed ball about the origin, of radius R. In view of the
rotation symmetry of A we may then assume x = (0, 0, a), with a = \\x\\ > R.
Therefore
(t>A(x) = -— / —=
4^ J\\y\\sR Jy2
+ H + (ys - ay
Now introduce cylindrical coordinates y = ^(r,a,y^) = (rcosa, rsina, y^)
(see, for example. Exercise 3.6). Then detD^(r,a, y^) = r. Because of this
determinant the resulting integrand in cylindrical coordinates is given by
s/r^+iys-a)^'
The antiderivative with respect to r of this function can then be readily obtained;
hence the following. Because ||j||^ = r^ + yf < R^, for y e A, it follows that
f/ := {y e R^ I llyll < R} = *(V), with V the solid half cylinder
V := {(r, a, y^) ^R^ \-n <a <n, -R < y^ < R, 0 <r < J R'^ - j|}.
On application of Theorem 6.4.5 this gives
1 r^ r^ fy/^^-yl r
-<PaM =— III dr dy^ da
An j_, j_R Jo yr2+(>-3-«)2
1 r^
= -/ (s/R^ + a^-lays-lys-aDdys
^ J —R
= ^ \-^^(R2 + a^-2aysr-ays + ^
4nR^ I vol (A) 1
V3 = i?
y3,=-R
3 471 a An \\x\\'
See Exercise 6.50.(viii) for the computation of the volume of the ball. Hence
vol(yl) 1
An \\x\\
Newton's potential therefore behaves as if the total "mass" of the ball were
concentrated at the origin. The fundamental ideas for this calculation are first found in
the literature,' in geometrical form, in Proposition LXXIV in Newton's Principia
'See also: Z,iK/eivoo(i\Mi\ce//a«y, Cambridge University Press, 1986, p. 169.

450 Chapter 6. Integration
Mathematica, London, 1687. In Exercise 7.68 one finds two other proofs of this
result.
In the Newtonian theory of gravitation the potential (p^ plays the following role.
Consider a mass of unit density occupying the space of the solid A, that is, of total
mass M = vol(A). Then the force F exerted by this mass M on a point mass m
placed at a point x e R^ is given by
mhd X
Fix) = - grad m <Pa(x) = —-— -—-r.
Thus we obtain Newton's law of gravitation: the mass M exerts on the point mass
m an attractive force of magnitude inversely proportional to the square of the
distance ||x|| between m and the center of gravity of M. (See Exercise 3.47.(i)
for the relation between conservation of energy and the description of force as
minus the gradient of potential energy.) Because of the minus sign in the formula
for F, we have to consider negative potentials in order to get an attractive force
law. Conversely, the potential at x equals the work done by the force F when we
displace a unit mass from the point x to infinity; in particular, the potential equals
0 at infinity. In fact, in gravitation this work is independent of the path y taken.
Writing y : / = [0, 1 ] —> R^ with y(0) = x and y(\) = oo, we obtain for this
work the following line integral (see Section 8.1):
{F(s),dis} :=- (gTad(<j)Aoy),Dy)(t)dt = - —{(t)Aoy){t)dt
= 4>a(,x.).
In the theory of electrostatics the forces between like charges are repulsive, as a
consequence, in that discipline, potentials are not defined with a minus sign as in
gravitation. ik
Example 6.6.8 (Kepler's second law). The notation is that of Application 3.3.B. In
particular f i-> x(^) is a C* curve in R^, J C R an open interval, V := ] 0, 1 [ x 7 C
R^
xii -.V ^ P]\= *(V) C R^ defined by *(r, t) = rx(t).
From Application 3.3.B we know that ^ : V —> P/ is a C* diffeomorphism for
J suitably chosen, and that detD^(r, t) = r det(x(f) x'(t)). By means of the
Change of Variables Theorem one finds
(Pj) = dx = \det D<if(r,t)\drdt
J^{V) Jv
= f r f \det(x(t) x'(t))\dtdr = - /" |det(x(f) x'(t))\dt.

6.6. Change of Variables Theorem: formulation and examples 451
Now let k >2. Without loss of generality it may be assumed that 7 = ] 0, .s [. Then
the following holds:
the area of Pj is proportional to the length of J, (6.21)
if and only if there exists a constant / e R such that, for all admissible s,
|det(x(0 x'(t))\dt =ls.
f
The, permitted, calculation of the second derivative with respect to s of this identity
gives, on account of the chain rule and Proposition 2.7.6,
det(x'(f) x'(t)) + det(x(f) x"(t)) = det(x(0 x"(0) = 0.
It follows that (6.21) is true if and only if
x(s) and the acceleration x"(s) are linearly dependent vectors.
In physical terms: the acceleration of a point particle orbiting in a plane is radial
if and only if the particle obeys Kepler's second law. its radius vector sweeps out
equal areas in equal times. ik
Example 6.6.9 (Transport equation). The notation is that of Section 5.9. Let
(^OreR be a one-parameter group of C^ diffeomorphisms of R" with associated
tangent vector field i/r : R" ^^ R", let L e g(R"), and let / e C'(R"). Then the
following transport equation holds:
— f f(x)dx= f {Dfix)fix) +fix)divfix))dx. (6.22)
Indeed, by continuity detD^'(j) > 0, for f e R and y e L. According to
the Change of Variables Theorem 6.6.1, the Differentiation Theorem 2.10.4 and
Formulae (5.22) and (5.31) we get
d_
dt J >!/((£,)
f fix) dx = ^ fifo vI/')(>') det Dvl/'OO dy
= j^[Df(^-(yM(^-(y)) + (f o vI/')(>') divfi^'iy))) dct D^-(y)dy.
And using the Change of Variables Theorem once again we obtain Formula (6.22).
Finally, choose / = 1, and conclude that the following assertions are equivalent.
(i) vol,, (*'(L)) = vol„(L), for all r e R and all sets L e $(R").
(ii) div^ = OonR". ik

452
Chapter 6. Integration
6.7 Partitions of unity
We introduce an important technical tool. Recall Definition 1.8.16 of an open
covering.
Definition 6.7.1. Let K C R" be a subset and let 0 = {<?,• | z £ /} be an open
covering of K. AC^ partition of unity on K subordinate to 0, where k e Nq, is a
finite collection of functions Xj : R" —> R with 1 < J < / such that
(i) Xj is a C^ function, for 1 < j < /;
(ii) 0<Xj(x) < l,forx e R";
(iii) for every j there exists O S 0 such that supp(xj) C O;
(iv) T,i<j<i. Xj(x) = 1, forx e K. O
Example 6.7.2. Let the notation be like the one above. Assume in addition K is
compact and Jordan measurable, and / : R" —> R is a given function. Further
assume that each of the functions fxj '■ R" —> R, with 1 < J < /, is Riemann
integrable over K (this may be regarded as local information). Then
(lKf)(x) = U(X)/(X) J2 XjM = Yl ^K(x)(fXj)(x).
But this yields (the global information) that / is Riemann integrable over K, while
f f(x)dx= J2 f f(x)Xj(x)dx. t\
Theorem 6.7.3. For every compact set K C R" and every open covering 0 of K
there exists a continuous partition of unity over K subordinate to 0.
Before proving the theorem we introduce auxiliary functions that play a role
in the proof. Assume a/ < a < b < b' and define the piecewise affine function
/ = fa\b',a,b :R^^ Rby
/W =
0
X -
a -
1
X -
b-
0
-a'
-a'
-b'
-b'
(x < a');
(a' <x < a);
(a <x < b);
(b<x< b');
(b' < x).
(6.23)

6.7. Partitions of unity 453
Then / is continuous on R. Let B and B' be rectangles in R", as in (6.1), such that
BcB' and a'^ < aj < bj < b] (1 < ; < n). (6.24)
Then
/ = fB', B : R" ^ R with f(x)= Yl fa'.,b'., aj, hj (Xj) (x € R")
(6.25)
is a continuous function such that
0</(x)<l (xeR"); supp(/) = B'; /(x) = 1 (x € B).
(6.26)
Proof. Because 0 is a covering of K, there exists, for every x e K, aset O £ 0
such that X e O; denote this set O by Ox- Since Ox is open in R", there exist
rectangles Bx and B'^ that satisfy both (6.24) and
X elx:= mt(Bx) CB'^cOx- (6.27)
The collection {Ix \ x e K} forms an open covering of the compact set K. By the
Heine-Borel Theorem 1.8.17 there exist finitely many points x,,..., X/ e K such
that
KC [J Ixj. (6.28)
Now define, in the notation of (6.25),
fj = /b'.,Bj, where B] = B'^,, Bj = B^, (1 < ; < /).
Then the ■^l/j : R" —> R are continuous functions satisfying (see (6.26) and (6.27)),
forx e Bj,
0<t/^j(x)<1 (xeR"); suppdrj) c B'j c O,/, t/^j(x) = 1. (6.29)
Further, let
X. = ^.; xj+i = (i-^.)(i-^2)---(i-^,)^j+. (1<;</-1).
(6.30)
It follows from (6.29) that the X\, ■ ■ ■, Xi satisfy the requirements (i)-(iii) for a
partition of unity subordinate to 0. The relation
I] X,-=1 - n (1 - ^'> (6-3i>
is trivial for _/ = 1. If (6.31) is true for j < I, then summing (6.30) and (6.31)
yields (6.31) for j + I. Consequently (6.31) is valid for j = I. If x e K, then
by (6.28), (6.27) and (6.29) there exists an i such that T/f,(x) = 1; thus follows
T,i<j<iXj(x) = i. □

454
Chapter 6. Integration
Remark. After Formula (6.29) one might be tempted to finish the proof by setting
Xj = ly—^~^- However, the zeros of the denominator then make it necessary to
take additional measures.
Illustration for the proof of Theorem 6.7.4
(a) g on [0, 1]; (b) g_2,-, on [-2, -1];
(c) /2_2,_i on [-2, -1]; (d) /2_2,2,-i,i on [-2, 2].
For clarity of display the scale in various graphs has been adjusted
With a view to applications in a subsequent section we further formulate:
Theorem 6.7.4. For every compact set K C R" and every open covering A of K
there is a C^ partition of unity on K subordinate to A.
Proof. The proof proceeds in a way analogous to that of Theorem 6.7.3, provided
we may replace the function / in (6.23) by a C°° function. To verify this we add
the following remarks.
(a) The function g : R —> R defined by
g{x) = 0 (X < 0); g{x) = e-''^
(x >0)
is a C°° function; this is also true at x = 0, where all derivatives vanish. In
fact, there exist polynomials pk such that g^''\x) = Pk(\)g(x), for x > 0.
In particular, limj^^^o g^'^^x) = limy^ooPit(>')^~^ = 0- Iii turn, this implies
g«(0)=0,foralljfceNo.

6.8. Approximation ofRiemann integrable functions 455
(b) Let a < b. The function ga/, : R —> R is a C°° function if
ga,b(x) =g(x-a)g(b-x) (x e R).
(c) The function /Iq^ /, : R —> R is a C°° function if
ha,b{x)=i ga,h{y)dyI I ga,b{y)dy.
while
ha,b{x) = 0 {x <a); 0 < ha,b{x) < 1 {a < x <b)\
ha,b(x) = 1 (b <x).
(d) Let a' < a < b < b' and define h = hgrh'aj, : R —> R by /2(x) =
ha',a{x) h-y^ _/,(—x), for X £ R. Then /z is a C°° function, and one has
/7,(x)=0 (x < fl'); 0 </j.(x) < 1 (fl'<x<fl);
/7,(x) = 1 {a <x <b); 0 < /j.(x) < 1 (Z>< x < b');
h(x)=0 (b'<x). ^
6.8 Approximation of Riemann integrable functions
The techniques from the preceding section enable us to prove that Riemann
integrable functions can be approximated, to arbitrary precision, by continuous
functions.
Lemma 6.8.1. Let K C R" be a compact subset. For ^ > 0, let
K^ = [y ^ R" I there existsx e K with ||>' — x|| < ^ }.
Then Ks is compact. IfUc R" is an open set with K C U, then S > 0 exists such
that Kg CU.
Proof. For every x e AT there exists ^(x) > 0 such that B{x; 2S(x)) C U,
see Definition 1.2.1. Furthermore AT C [J^^[^ B{x; ^(x)). On account of
Definition 1.8.16.(ii) there existxi,... ,xi e K with
/: C IJ Bixj- S(xj)).

456 Chapter 6. Integration
Let S = min{ ^(xj) | 1 < 7 < /} > 0, and let y e Ks. Then there is an x e K with
\\y ~ ^W < ^. and also 1 < 7 < / withx e B{xj; S(xj)). Consequently
\\y-xj\\ < \\y-x\\ + \\x-xj\\ < S(xj) + S(xj) = 2S(xj).
This gives y e B(xj; 2S(xj)) C f/; and hence Kg CU. □
Theorem 6.8.2. Let f : R" —> R Z>e a bounded function with compact support.
(i) r/zen / is Riemann integrable if and only if for every e > 0 there exist
functions g^, g+ e Cc(R"). the space of continuous functions with compact
support, such that
g-< f < g+ and / {g+{x) - g_(x)) dx < e.
And in this case one also has
I f(x)dx— I g±(x)dx
< e.
(ii) Let U C R" be an open subset and assume supp(/) C U. If f is Riemann
integrable, the functions g_ and g+ may then be chosen such that supp(g_)
and supp(g+) C U.
Proof, (i). Consider first an arbitrary rectangle B C R" and let e > 0 be arbitrary.
By choosing a somewhat larger rectangle B' with B C B' and using the function
fs', B from (6.25), we can see that there exists a g+ e Cc(R") with
Ib <g+. Be swpp(g+) = B', / g+(x) dx - vol„(B) < -.
Interchanging the roles of B and B' we obtain ag_ s Cc (R") with similar properties.
One finds
g- <Ib < g+, supp(g_) C B C supp(g+) C B',
/
ig+(x)- g-(x))dx < e.
Now assume / to be Riemann integrable and let B C R" be a rectangle with
supp(/) C B. According to Proposition 6.2.5 there exists a partition £ of B with
Y] ( sup / vol,, (Bi) - inf / vol,,(Bi)) < e. (6.32)

6.9. Proof of Change of Variables Theorem 457
In consideration of the equality f = f_^_ — /_ from Theorem 6.2.8.(iii) we may
then assume that / > 0. With e suitably chosen, apply the argument above to
the functions 1 b, , and multiply the g^^ and g^^ thus found by inf g,. / and sup^. /,
respectively. Assertion (i) follows with
supp(g_) C supp(g+) C y Bi.
(ii). According to the preceding lemma, there exists ^ > 0 such that suppC/)^ C U.
We may assume that diameter(B,') < S, for i e /, for which possibly <S in (6.32)
must be refined. But then B'j n supp(/) ^ 0 implies Bi C U; this proves the
assertion. □
6.9 Proof of Change of Variables Theorem
The proof proceeds in five steps. In Step II we show that for every y° e V the
restriction of the diffeomorphism ^ to a suitably chosen open neighborhood V (y°)
of y° in V can be written as a composition of "simpler" diffeomorphisms that
essentially behave like diffeomorphisms in one variable. The proof relies on the
Implicit Function Theorem 3.5.1. Indeed, this is an important reason for developing
the theory of this theorem prior to the theory of integration in R". In Step EI we
prove that the theorem follows if it is known to hold, for all y° e V, for functions
/ with supports contained in the open sets ^{V(y°)) in U. An essential element
in this proof is the theory from Section 6.7 concerning compact sets and partitions
of unity. In Step IV we show that the theorem is true for the composition of two
diffeomorphisms if we already have it for the individual diffeomorphisms. In Step V
the theorem is proved for the "simpler" diffeomorphisms by means of the Change of
Variables Theorem for R. Because the Change of Variables Theorem for R usually
is proved under the assumption that / is continuous, we prove in Step I that the
theorem is generally valid if it holds for continuous functions / with supp(/) C U.
Other proofs. The Change of Variables Theorem 6.6.1 can also be proved without
recourse to the Implicit Function Theorem 3.5.1 as in Step II. Since it is of
independent interest we give such a proof, called the second, in Appendix 6.13 to this
chapter. As a consequence Chapter 6 can be studied independently from Chapters 3
through 5, which might appeal to readers who prefer to make an early start with
the theory of integration. Which proof one prefers is mainly a matter of taste or
prerequisites. By way of justification of the procedure followed in this section it
may be remarked that the second proof uses the linear version of the decomposition
from Step II, as well as Steps I and IV. Furthermore, localization as in Step IE is an
extremely useful technique in analysis.
The appendix contains one more proof, called the third; it might be slightly
surprising yet it is quite efficient.

458 Chapter 6. Integration
Step I (Reduction to continuous /). Let / be a Riemann integrable function
with supp(/) C U. By Theorem 6.8.2.(ii) there exist, for every e > 0, continuous
functions g_, g_^ on R" such that
supp(g_), supp(g+) CU; g-< f <g+; / {g+M - g-.(x))dx < e.
Ju
Therefore (g_ o*) | det D*| < (/o*) | det D*| < (g+ o *) | det D*| on V; and
so
j (g-oyiJ)(y)\ det D^(y)\ dy < j
(g-oyiJ)(y)\ det D^(y)\dy< / (f o ^iJ)(y)\ det D^(y)\dy
V
< f (f oyiJ)(y)\ det D^(y)\dy< f (g+oyiJ)(y)\ det D^(y)\dy.
J V Jv
Under the assumption that the theorem is true for / replaced by the continuous
functions g_, g+ with supports in U, the first and last terms equal /^ g-(x) dx and
/y g+{x) dx, respectively; but these numbers differ by less than e, and moreover
their difference from /^ f(x)dx is smaller than e. Ergo, the upper and lower
Riemann integrals of (/ o ^) | det D^\ over V differ by less than e, and therefore
their difference from y^/(x)rfx is smallerthane. Since this is true for every e > 0,
it follows that (/o^) | det D^\ is Riemann integrable over V, with Riemann integral
f^f(x)dx.
Next, assume that conversely g := (f o ^) |detD^| is Riemann integrable
over V. One has
(/ o ^)(y) = g(y)\ det D^(y)r (y € V);
that is, for x e U,
fix) = (g o ^-')(x)\detD^i^-\x))\-' = (g o^-')(x)\detD^-'(x)\,
because the chain rule gives detD^(^"''(x)) detD^~^(x) = 1. Applying the
preceding argument, with / replaced by g and ^ by ^~^, we conclude that / is
Riemann integrable over U.
Step II (Reduction to the case of dimension one). Let us write ^(y) =
(*,(>'),..., *„(>'))inR". For the moment let y" e Vbe fixed. Because ^^(y") e
Aut(R"), there exists an index I < j < n such that Dj^„(j°) ^ 0. One may
assume j = n, that is, D„^„(j°) ^ 0, which may require prior permutation of the
coordinates of y. By virtue of the Implicit Function Theorem 3.5.1 this means that
the equation for y,,,
*«(>'i, •••,>'«) =-^H, (6.33)
with the >'!,..., y„-uXn as parameters (near (yj",..., j°_j, <i/„(y°,..., y°))), can
be solved near y°. Denote the solution by
y„ = 4>,>(>'i,..., >'«-! ,Xn); (6.34)

6.9. Proof of Change of Variables Theorem
459
it is C'-dependent on yi,..., y„-\, x„. But this implies that the C^ mapping S„
defined in an open neighborhood of y° in V by
SnCy) = (yi, . . . , y„-i, "^niy)) = (yi, . . . , y,,-!, X„),
is invertible on its image, with C' inverse
R
R
yn
1
(yu
k
• ••. y„-
"^niyi
-^W-'
0'i,---,>'«-iJt'
n-1
(6.35)
Illustration for Formula (6.35)
S„ (yi,..., yn-i,x„) = iyi,..., y„-i,(p„iyu ..., y„-i,x„)).
(6.36)
Consequently, 3„ is a (locally defined) C' diffeomorphism leaving the first n — 1
coordinates invariant. In addition we have, on account of (6.36), (6.33) and (6.34),
((* o 3,7')(>',,..., y,,-!, x„))„ = ^„(yi,..., y„^i, (p„(yi,..., y„-i,x„)) = x,„
while, for 1 < y < n — 1,
((* o a~^)(yi,...,y„-\,x„))j = <i/j(yi,...,yn^i,(t>„(y\,...,yn-\,Xn)).
That is, ^ o 3,7^ leaves the n-th coordinate invariant and acts like a C'
diffeomorphism (C'-dependent on x„) on the first n — 1 coordinates. Finally, in an open
neighborhood of y° in V,
vl/ = (vl/ o 3,7') o 3„.
Once more applying, mutatis mutandis, the foregoing procedure to ^ o 37', we
finally obtain that the restriction of the diffeomorphism ^ to a suitably chosen
open neighborhood V(y°) of y° in V can be written as a composition of finitely
many diffeomorphisms of a form similar to that in (6.35), save permutation of the
coordinates.

460 Chapter 6. Integration
Step m (Localization). Let K = supp(/) C U. Then if is a compact set. We
now demonstrate that the Change of Variables Theorem holds if for every x e K it
is possible to find a bounded open neighborhood 0^- of x in f/ such that the theorem
holds for continuous functions / additionally satisfying
supp(/) c 0,. (6.37)
Indeed, the collection 0 = {Ox \ x e if} is an open covering of K. On
account of Theorem 6.7.3 there exists a continuous partition Xi. • • •. X/ of unity on
K subordinate to 0. Consequently, for every 1 < y < / there is an x e K such
that supp(xj /) C supp(xj) C Ox', that is, Xj f satisfies (6.37). Because Xj is
continuous, Xj f is continuous. It follows that
f (Xj f)Mdx = f (Xj f)(^(y)) I det D^(y)\ dy.
Ju Jv
Hence
f TxM)f(^)dx = I Txji'^iy))m{y))\A^^D^{y)\dy.
Ju " Jv j
But this leads to the desired conclusion, because ^ ■ Xji^) = 1, for x e K =
supp(/).
Step rV (Composition). If the Change of Variables Theorem holds for a C'
diffeomorphism ^ : V —> f/ and for a C' diffeomorphism 3 : W —> V, then it
also holds for the composition ^ o 3 : W —> f/. Indeed, the chain rule gives
detD(vI/ o a)(z) = det(DvI/(3(z)) o DE(z)) = det D^(a(z)) det Da(z).
Therefore
f fi^ o Siz))\ det Di^o'B)iz)\dz
Jw
= f (f oyiJ)(S(z))\det DHS(z))\\ detDE(z)\dz
Jw
= f (f oyiJ)(y)\ det D^(y)\dy= f f(x)dx.
Jv Ju
Step V (Case of dimension one). The Change of Variables Theorem holds if ^
has the special form (compare (6.35))
'i^(y) = {yi,...,y„-i,i^(y)).

6.10. Absolute Riemann integrability
where i/f is a C' mapping. Indeed,
461
D^iy) =
( 1
0
0
^nfiy) 7
implies detD^Cy) = D„'tl/(y) ^ 0. Therefore y,, i-> i^iyi,. ■., yn) may be
assumed strictly monotonically decreasing or increasing. Using Corollary 6.4.3
one then finds
f (f o vl/)(3;)| det D^(y)\ dy = f f(y,,..., >-„_:, t(y)) \D„t(y)\ dy
Jv Jv
-f-if
fiyi,..., 3?„_i, fiyi,..., y,,)) \D„i/(yi,..., y„)\ dy„
0
In view of the monotony we may remove the absolute signs in the integrand and
account for them in the integration limits for the integral with respect to y„. To
the last term we may subsequently apply the Change of Variables Theorem on R,
which leads to
j (f o vl/)(3;)| det D^(y)\ dy = f ■' (f /()'.. • • •. ^«)^^«) '' • ^3'.
= / f(x)dx.
6.10 Absolute Riemann integrability
The proof of the Change of Variables Theorem as it stands assumes that the integrand
/ is bounded and vanishes outside a compact set. In applications one is often
interested in functions / like f(x) = log ||x ||, for 0 < ||x|| < 1, or f(x) = e^H^H",
for X e R"; that is, an / which either is itself unbounded or has an unbounded
domain. For such functions Corollary 6.10.7 below is of importance.
Definition 6.10.1. Let U C R" be open and let / : f/ —> R. The function / is said
to be locally Riemann integrable in U if for every x e f/ there exists a rectangle
B C U such that x e int(B) and / is Riemann integrable over B. O
Lenrnia 6.10.2. Let U C R" be open and let / : f/ —> R. The following assertions
are equivalent.

462 Chapter 6. Integration
(i) The function f is locally Riemann integrable in U.
(ii) For every function x ^ CcCR") with supp(x) C U the function x f i^
Riemann integrable over U.
Proof, (ii) ^ (i) follows by choosing a suitable rectangle B C U and subsequently
approximating 1^ by functions x ^ CcCR"). using Theorem 6.8.2.(ii).
(i) ^ (ii). Choose x ^ CcCR") with supp(x) C U. For every x e supp(x)
there exists a rectangle Bx C U with x e mt(Bx) and / Riemann integrable over
Bx. Applying Theorem 6.7.3 to the compact set supp(x) and the open covering
{int(B,.) I X e supp(x) },wefindxj € Cc(R") where 1 < 7 < /, with the following
properties. On supp(x) one has ^Xj = 1> ^nd for every index j there exists an
X e supp(x) with suppCx;) C int(B;,). Hence follows XjXf = (XjX)(^B,f)-
By means of Corollary 6.3.7 and Theorem 6.2.8.(iv) one concludes that XjXf is
Riemann integrable over U. But then xf = 5Zv XjXf is also Riemann integrable
over U. □
Lemma 6.10.2 immediately has the following consequence.
Lemma 6.10.3. Let U C R" be open and let / : f/ —> R. Then, with f = fj^ — /_
as in Theorem 6.2.8. (Hi), we have the following properties.
(i) / is locally Riemann integrable if and only if f+ and /_ are locally Riemann
integrable.
(ii) If f is continuous then f is locally Riemann integrable.
(iii) Iff is locally Riemann integrable then f is integrable over every K e $(U).
(iv) The collection of functions that are locally Riemann integrable over U forms
a linear space.
In preparation for a new definition we formulate the following:
Proposition 6.10.4. Let A C R" be bounded and let f \ A ^ VL be bounded and
Riemann integrable over A. Then, with f = f^ — /_ as in Theorem 6.2.8.(iii),
I f(x)dx= sup / f_^.(x)dx— sup / f^(x)dx.
J A ' K^${A) Jk Ke$iA) Jk '
Proof. In view of Theorem 6.2.8.(iii) it is sufficient to consider /+ and /_ separately,
hence we may assume / > 0. Suppose that B is an n-dimensional rectangle with
A C B and that e > 0. Consider a partition S = {B-, \ i € 1} of B with

6.10. Absolute Riemann integrability 463
S(/, S) - S(/, S) < e. Next let /' = {/ e / | B-, C A}, then K := U,-g// 5,- e
${A). Since Bj<^A implies inf^,. / = 0, we find
5(/, S) = S(/, S') < /" /(x)^x < S(/, S') < S(/, S).
Jk
On the other hand, we have the same inequalities with jj^ f(x)dx replaced by
Ia f(^) ^^- Accordingly
/ fix)dx-€< / fix)dx< / fix)dx. □
J A Jk J a
Definition 6.10.5. Let U C R" be open and let f : U ^ R. The function / is
said to be absolutely Riemann integrable over U if / is locally Riemann integrable
in U and if
sup / \f{x)\dx<oo.
k^${U)Jk
Since 0 < /± < |/| = /+ + /-, we have that /+ and /_ are absolutely Riemann
integrable if and only if / is so. For / absolutely Riemann integrable over U, we
define
/ f(x)dx= sup / f_^.(x)dx— sup / f^(x)dx. O
Ju ' K^${U) Jk ' K^${U) Jk
Proposition 6.10.4 shows that Definition 6.10.5 and Definition 6.3.1 agree for
functions for which both definitions apply. The collection of functions that are
absolutely Riemann integrable over U forms a linear space, but, in contrast with
Theorem 6.2.8.(iv), it is not closed under pointwise multiplication of functions.
For example, the function x i-> 4= is absolutely Riemann integrable over ] 0, 1 [,
V-V
whereas x i-> (4=) = - is not.
Let U C R" be open. In the following theorem we will consider sequences
{Kk)k^^ in SiU) with [JKk = U, KkCmi{Kk+^) (/: e N).
(6.38)
First we show how to construct such sequences. Define
C, = {X e f/ I ||x|| < /:, ||x - yll > I as >- ^ f/ } (k € N).
k
Then Ck is a compact subset of U on account of the Heine-Borel Theorem 1.8.17,
while Ck C Ck+\. To show that {C^ | A; e N} is a covering of f/, let x e f/ be
arbitrary. Since U is open, we have inf{ ||x — yH | y ^ f/ }>0; hence there exists
A; e N with X ^ Ck. Next, we note that the set
Vk+i = {x^U\ \\x\\ <k+\, \\x-y\\ >^as>-^f/} (/: e N)

464 Chapter 6. Integration
is open. Since Ck C Vk+\ C Ck+i, it follows that Ck C mt(Ck+i)- The sets
Ck are not quite the sets we want, since they may not be Jordan measurable. We
construct the sets Kk as follows. For each x e Ck choose a rectangle that is
centered at x and is contained in mt(Ck+\). The interiors of these rectangles cover
the compact set Ck', choose finitely many of them whose interiors cover Ck and let
their union be Kk. Since Kk is a finite union of rectangles, it belongs to $(U). Then
Ck C mt(Kk) C Kk C mt(Ck+i)- Then (Kk)keN satisfies the conditions in (6.38).
Next we give a useful criterion for the absolute Riemann integrability of a
continuous function over an open set.
Theorem 6.10.6. Let U C R" be open and let f : U ^>- R be continuous. Suppose
(Kk)keN i^ (^^ in (6.38). Then the following are equivalent.
(i) / is absolutely Riemann integrable over U.
(ii) ( /jj. |/(-^)| dx)i;^f^ is a bounded sequence in R.
If one of these conditions is satisfied, we have that ( f^^ \f(-x)\ (ix)^g[^ is monoton-
ically nondecreasing and
lim / f(x)dx= / f(x)dx. (6.39)
Proof, (i) ^ (ii). Since / is absolutely Riemann integrable over U if and only /_
and /+ are so, we may assume that / > 0. Obviously (ii) follows, since for every
jfceN,
/ f(x)dx < sup / f(x)dx = I f(x)dx.
JK^ K^${U) JK Ju
(ii) ^ (i). Let K e ^ (f/) be arbitrary, then K is covered by the increasing collection
of open sets {mi{Kk) \ k £ N}; hence, being compact, by finitely many of them,
and therefore by one of them, say int(ArAr). Accordingly
/ f(x)dx< / f(x)dx< lim / f(x)dx.
Jk ' Jkn ' ''^°° Jki,
It follows that / is absolutely Riemann integrable over U and that
/ f(x)dx= sup / f(x)dx< lim / f(x)dx.
Ju ' Ke$iU) Jk ' ''^°° Jk,,
Finally, we prove Formula (6.39). We have
lim / f(x)dx < sup / f(x)dx = I f(x)dx,
I'^°°Jk,, k^${U)Jk' Ju'
and the conjunction of the last two formulae proves the identity. □

6.10. Absolute Riemann integrability 465
Corollary 6.10.7. Assume U and V to be open subsets ofR" and let^ : V ^>- U be
a C' dijfeomorphism. Let f : U ^ Vibe a function. Then f is absolutely Riemann
integrable over U if and only if the function (f o ^) | det D^\is absolutely Riemann
integrable over V. If either of these conditions is met, then
I f(x)dx= f(foyiJ)(y)\dctD^(y)\dy.
Ju Jv
Proof. As follows from the preceding theorem and construction, we may
approximate / by Riemann integrable functions with compact supports. Then apply the
Change of Variables Theorem 6.6.1 to these functions. □
Example 6.10.8 (/r^-'' dx = ^ = 1.772453 850905516 •• • ). (See also
Exercises 2.73, 6.15, 6.41 and 6.50.(i)). The function f(x) = e^U^U" is continuous
on R^, and therefore / is locally Riemann integrable in R^. Define B(R) = {x e
R^ I lUII < ^ }. for R > 0. Using polar coordinates x = r(cosa;, sina) one finds,
for all R>0,
I f(x)dx= I I re '' dr da = 271
Jb(R) J-71 Jo
.-"
R
= 77:(1 - e~^^) < 71.
>Bm . „.„ L JO
(6.40)
Let K e ^(R^), then there exists a number R > 0 such that K C B(R). Because
/ is a positive function it follows that
/ f(x)dx < I f(x)dx<7z;
JK Jb(R)
that is, / is absolutely Riemann integrable over R^. On account of Corollary 6.4.3
one also has, for every i? > 0,
(j e-''~dxy=(f e-''dxMf £-^2^x2)=/" e-^^'+'^-^dx,
where C(R) = [-R, R] x [-R, R]. We have B(R) C C(R) C B{R-JT), and by
(6.40) therefore
-""l < I fix)
JC(R)
7z(l - e-"') < / fix) dx < 7zil - e-"-"');
IciR) '
from which
( e^""'dx) = lim / fix)dx = 7i. i^

466 Chapter 6. Integration
Remark. For a function / on R we have the notion of improper Riemann inte-
grability. This concept is of particular importance if / is Riemann integrable over
R, while I/I is not. An example of this was given in Example 2.10.11. Absolute
Riemann integrability is a more stringent condition on a function / on R", in that
integrability of |/| is required. Accordingly, cancellations due to oscillatory
behavior are not taken into account. For the Riemann integral in R", with n > 1,
there is no useful analog of the concept of improper Riemann integrability. This is
related to the fact that an unbounded set in R", with n > 1, can be approximated
from within by compact sets in many different ways.
6.11 Application of integration: Fourier transformation
We will show that there exists an ample class of functions that can be written as a
continuous superpositions of periodic functions of a simple type, see Formula (6.43).
We recall the notation from Formula (2.24). In particular we have, for a vector
X = (xi,..., x„) e R" and a multi-index a = (ai,..., an) £ NJj,
x" =x"^'---x"„", \a\=ai + --- + a„, D" = D^'• •-D"". (6.41)
Definition 6.11.1. We define the space -5(R") of Schwartz functions on R" as the
linear space of all C°° functions / : R" —> C such that, for all multi-indices a,
sup{ \x^(D"f)(x)\ I X e R" } < oo. O
Note that if / e -5(R"). then • ^/ : x i-> x^fix) and D"/ are also functions
in -5(R")- Furthermore, for every / e -5(R") there exists a constant c > 0 such
that, for all X e R",
|/(x)| < c(l + ||x||)-("+');
and therefore also, with § e R" arbitrary,
|e-'(^''^>/(x)| < c(l + ||x||)-("+^^
Consequently, the integrand in the following definition is absolutely Riemann
integrable over R".
Definition 6.11.2. For every / e -5(R'^) we define the Fourier transform f of f
as the function / : R" —> C with
/c
;?)= f e-'^'-^^f{x)dx. O

6.11. Fourier transformation 467
Theorem 6.113. (i) The Fourier transformation f \-^ f is an endomorphism of
-5(R")- For every multi-index a e NJj and^ e R"
(ii) CDV)(?) = (i^rm, (iii) cv)(?) = (iDrm.
Proof, (iii). Note that
\(iD^)"ie-"^'''^^f(x))\ = |x"e-'<^'^>/(x)| < |x"/(x)| (x e R"),
while •"/ again belongs to -5(R"). Therefore assertion (iii) follows from the
Differentiation Theorem 2.10.13, or rather, from its direct extension to the case of
integration over R". (See Exercise 6.82 for a direct proof of assertion (iii).)
(ii). Since for ;6 £ N^
(2?)^e-'(^-^> = (-D,)^e-'(^-^>),
we find from assertion (iii), by integration by parts,
(i^fdiDrm) = f (-D,)^e-'<^'^>)x"/(x)rfx
= f e-'(^'^>D^(x"/(x))^x.
(6.42)
< oo;
Note that this is allowed, and that the boundary terms have vanishing limits, because
/ e ■S(R'"). In particular, assertion (ii) follows from (6.42), by taking a = 0 £ Ng.
Finally we have, for all § e R",
\^^(D"m)\< f \D^(x''f(x))\dx
Jr"
which proves / e -5(R")-
Example 6.11.4. If g(x) = e-jH^H", then f = {InY^'^g (see Exercise 6.83 for a
different proof). Indeed, we have (Djg)(x) = —Xjg(x), for 1 < y < n. Therefore,
by (ii) and (iii) from the preceding theorem —^j'g'i^) = (Dj'g)(^), for 1 < j < n.
Solving this system of differential equations we obtain J"(§) = ce^j"^"", with
constant c ^ 0. Then, by Example 6.10.8,
c = f (0) = / g(x) dx= I e-ill-^ll" dx = (27z)i. i^
^R" ^R"

468 Chapter 6. Integration
Example 6.11.5 (Convolution). For /, g e -5(R'") we define the convolution
f*g:R"^Chy
f * gi-x) = / fi-x - y)g(y)dy.
Jr"
Note that, for all x.y e R"onehas|/(x—)')g()')| < \g(y)\ sup |/|; and this implies
that the integrand is absolutely Riemann integrable over R". In fact, /*g e -5(R^');
for the proof see Exercise 6.85. Moreover
f*g = fg-
This can be proved by noting that in the following integrals the order of integration
may be interchanged.
(7>7)(?) = I e~'^'-'^^ f f(x-y)g(y)dydx
Jr" Jr"
= f g(y) f e-'^''^^f{x-y)dxdy
JR" JR"
= f g(y) f e-'^'+y-^^f{x)dxdy
JR" JR"
= f e-'^y'^^g(y)dy f e''^^'^^f{x)dx = m^)- tV
Theorem 6.11.6 (Fourier Inversion Theorem). The Fourier transformation is an
automorphism of-$(VL'"). The inverse of Fourier transformation is given by
fix) = (In)-" f e'^''^^n^)d^ if e S(R''), x € R"). (6.43)
^R"
Proof. First we verify Formula (6.43) for x = 0, that is, we prove
f(0) = (27zr" f md^. (6.44)
JR"
As a starting assumption about / e -5(R") let /(O) = 0, then
f(x) = j -^{tx)dt= Y, ""jf Djf(tx)dt=: J2 ^Jgji^)'
l<j<n l<j<n
where 'gj e C°°(R") for 1 < j <n. Now use (6.26) combined with (d) in the proof
of Theorem 6.7.4 to find a C°° function x with compact support such that x = 1
in a neighborhood of 0. Then we have gj e -5(R") for 1 < j < n, if we take
gji^) = (gjx)(x) + Tf^(/(i - x))M (X € R");

6.11. Fourier transformation 469
and furthermore
f(x) = Y^ Xjgj(x).
Using (iii) from the preceding theorem one obtains / = / X!,^ j^,, Dj'gj. On the
strength of Corollary 6.4.3, the Fundamental Theorem of Integral Calculus 2.10.1,
applied to Xj i-> Dj'g'j(x), and the fact that gj e -^CR"^), it then follows that
/.
/(?)rf?=0. (6.45)
This proves (6.44) if /(O) = 0. Now let / e 4(R") be arbitrary; set h :=
/ - fiO)g, with g as in Example 6.11.4. One then has h e -5(R™), hiO) = 0 and
h = f — /(O)J". Thus, according to (6.45),
0= /" md^= f md^-f(o) f t(^)d^.
Jr" Jr" Jr"
Furthermore,
f fi^) d^ = (2n)i f £-211^11'rfx = (277:)".
^R" ^R"
Therefore (6.43) holds for x = 0. The formula follows for arbitrary x° e R" by
replacing the function / by /° : x i-> f(x +x°) in (6.44). Indeed
75(^)= f e-'<^-^>/(x+x°)rfx= f e-'(^-^°'^>/(x)^x=e'(^°'^>/(§). □
^R" ^R"
Remark. The Inversion Formula (6.43) tells us that a Schwartz function / on R"
can be written as a superposition over the different/re^Menc/e^ § e R" of the plane
waves X i-> e'(^'^> in R", where /(§) determines the amplitude of the wave of
frequency §.
The plane waves are characterized by the fact that they are exactly the bounded
eigenfunctions for the differential operator D, that is, if / e C^(R", C) satisfies the
eigenvalue equation Df(x) = A. f(x) with A. e C", and if / is a bounded function,
then X = i^ with § e R", and f(x) = /(0)e''<*'^'. The arbitrary function in
-5(R™) can therefore be written as a superposition of bounded eigenfunctions for
the differential operator D acting on C'(R", C). The formula (5^ oDo5^-^)/(§) =
(i §)/(§) shows that the differential operator D acting on S (R") can be diagonalized
by conjugation with the Fourier transformation 3^, and that the action in S (R™) of the
operator obtained by conjugation is that of multiplication by / and by the coordinate
Example 6.11.7. The heat equation (see Example 7.9.5 for more details) for a
function m : R" x R —> R, assumed to be differentiable sufficiently many times,
reads
Dru(x, t) = k A^uix, t) (jfc > 0, X e R", t e R), (6.46)

470 Chapter 6. Integration
where A,- is the Laplace operator or Laplacian,
A. = A = 5; Dj.
This is an example of a partial differential equation, relating different partial
derivatives of u. We want to solve the initial value problem for this equation, that is, we
look for solutions u of (6.46) satisfying the following additional condition, for ^ = 0:
u(x,0) = f(x) (xeR"), (6.47)
where / is a given function.
We are going to apply a Fourier transformation to the function x i-> u(x, t);
therefore we assume that «(., t) e SiR"), for all r e R, and that / e -5(R").
Define
u{^j)=j e~'^'''^'>u(x,t)dx.
Jr"
Assuming that on the right-hand side differentiation under the integral sign is
allowed, we obtain from (6.46) and (6.47), using Theorem 6.11.3.(ii),
D,u(^, t) = -m f"(?, t), u(^, 0) = m (? e R", r e R).
The function t h> £?(§, t) therefore satisfies a first-order ordinary differential
equation, with the solution
Application of Example 6.11.4 yields
£-'*"'= |J(?), with g,(x) = (471 ktyie-^. (6.48)
On account of Example 6.11.5 we therefore have
"(?, t) = mtri^) = (/^)(?).
It then follows from the Inversion Theorem that
u(x, t) = if * g,)ix) = / fix - y)g,iy) dy
Jr"
= i47zktyi / fix - y)e~'^ dy = n~i / fix - 2-v47>')e~"^"' dy.
JR" JR"
(6.49)
Because the calculation above makes several assumptions about the function u, it
has to be checked that the u from Formula (6.49) does indeed satisfy (6.46) and
(6.47). For this the reader is referred to Exercise 6.92.
Finally, assume K C R" to be a bounded set with the property supp(/) C K,
and / to be nontrivial with / > 0 on R". But for every x e R" and all t e R_|_ there
exists y e R" such thatx — iVkty e supp(/); and this implies fix — iVkty) > 0.
It follows, for all x e R" and t e R_|_, that m(x, t) > 0; that is, the solution u has
infinite speed of propagation. ik

6.12. Dominated convergence All
6.12 Dominated convergence
From analysis on R we recall the result that integration and taking a limit may
be interchanged for a sequence of functions that converges unifonnly on a closed
interval in R. We now prove that this interchange is also permitted if the sequence
of functions is boundedly convergent, that is, if it is pointwise convergent and
uniformly bounded.
The construction in the proof of Theorem 6.12.2 below leads to functions whose
Riemann integrability is not guaranteed. This explains why we encounter J, the
lower Riemann integral from Definition 6.2.3, in the next proposition.
Proposition 6.12.1. Let K e ^(R") and let (fk)keN be a sequence of functions on
K. Assume that the following properties hold.
(i) /i is bounded, and lim^^^oo fk{x) = O,for every x e K.
(ii) (fk)keN is monotonically decreasing, that is, fk+i{x) < fk(x), for all k e N
andx e K.
Then one has
im / fk(x)dx = 0.
lim
Proof. Select e > 0 arbitrarily. By applying Theorem 6.8.2.(i) to the step function
corresponding to a suitable lower sum of _/>, we can find a sequence (gk)ksfi of
continuous functions on K such that gk < fk and
j fk(x) dx< j gk(x)dx + ^ (k€ N).
Note that the inequality above remains valid if gj^ is replaced by {gk)+, hence we
may assume that 0 < gk- Now introduce the sequence (hk)keN of functions on K by
hi = gi and hk = mm(gk, hk-i). Then every hk is a continuous function on K (in
fact, min(/, g) = ^(/+g —|/ —g|),forany twofunctions/andg). The sequence
is monotonically decreasing and satisfies lim^^^oo hk{x) = 0 for all x e K, since
0 < hk < gk < fk- The identity — min(/, g) = max(—/, —g), which is valid for
any two functions / and g, implies
fk -hk = fk - minfe, h^^i) = max(/t - g^, fk - /Jat-i)
< fk- gk + fk - hk-i < fk -gk + fk-i - hk-i.
Accordingly, one proves by mathematical induction over A; e N
fk-hk< Yl^fj~^J^' ^° / fk(^)^^^ / hk(x)dx+ Y^ TJ+T-

472 Chapter 6. Integration
As the sequence (hk)^^^^ satisfies the conditions of Dini's Theorem 1.8.19 it
converges uniformly on K to the function 0. Therefore we can find k^ sN such that
for all k >kowe have
/ hk(x)dx < ^; hence 0 < / fk(x)dx <^+ Yl TTm < ^- ^
Theorem 6.12.2. Let K e ^(R"), and let the functions f and fk, for k s'N, be
Riemann integrable over K. Assume that the following properties hold.
(i) For every x e K one has lim^^^oo fk(x) = f(x).
(ii) There exists a number m > 0 such that \fk(x)\ < m, for every x e K and
jfceN.
Then
lim / fk(x)dx= I f(x)dx.
I'^°°Jk Jk'
Proof. The functions fk — f are Riemann integrable over K and have pointwise
limit 0; consequently there is no loss of generality in assuming / = 0. In addition,
by Theorem 6.2.8.(iii) we may assume that _/> > 0. For each A; e N set gj; =
sup{ fk+i I y e No}. (Note that the gk are not automatically Riemann integrable.)
Obviously, the sequence {gk)kej^ satisfies the conditions in Proposition 6.12.1, and
therefore
0 < lim / fk(x)dx< lim / gk(x)dx = 0. □
The generalization of the last theorem to the case of absolutely Riemann
integrable functions is the following:
Theorem 6.12.3 (Arzela's Dominated Convergence Theorem). Let U be an open
set in R", and assume f and fk'.U^ R, for k e N, to be absolutely Riemann
integrable functions over U. Suppose that we have the following.
(i) limk^oo fk(x) = f(x), for every X e U.
(ii) There exists a function g : t/ —> R that is bounded on U and absolutely
Riemann integrable over U such that \fk(x)\ < g(x), for all k e N and
x€U.
Then
k^
lim / fj^{x)dx= I f(x)dx.

6.12. Dominated convergence 473
Proof. Let e > 0 be arbitrary. Apply Theorem 6.10.6 with |/| and g, to find a set
K e |(f/) such that
/" \f{x)\dx<\, f \Mx)\dx<f g(x)dx<^- (ks-N),
Ju\K J Ju\K Ju\K J
respectively. According to the preceding theorem there exists Icq sN such that for
k > ko,
/ fix)dx - / fk{x)dx\
\Ju Ju I
= [ if(x) - fk(x)) dx+ f fix) dx- f fkix) dx
\Jk Ju\k Ju\k
< I f {fix) - Mx))dx\ + I \fix)\ dx+ f \Mx)\ dx
\Jk I Ju\K Ju\K
€ € €
By way of application of this result we now give a version of the Differentiation
Theorem 2.10.13 in the case of integration over (unbounded) sets in R^.
Theorem 6.12.4 (Differentiation Theorem). Let U C R" and V C R'' be open
subsets, and let f : U x V ^>- Rbe a function with the following properties.
(i) For every x e U, the function t i-> f(x, t) is absolutely Riemann integrable
over V.
(ii) The total derivative Di/ : f/ x V —> Lin(R", R) with respect to the variable
in U exists and, for every x e U, the mapping t h> Di f(x, t) is absolutely
Riemann integrable over V (here the integration is by components).
(iii) There exists a function g : V —> [0, oo [ which is bounded on V and
absolutely Riemann integrable over V, such that \\Dif(x, OIIeuci — SiO^f^^f <^ll
{x, t)eU X V.
Then F : f/ —> R, given by F(x) = fy f(x, t)dt, is a differentiable mapping
satisfying
DiF(x) = / Dif{x,t)dt {x e U).
Jv
Proof. Let a e U and suppose (hk)keN is a sequence of arbitrary vectors in R"
converging to 0. The derivative of the mapping: R —> R with s i-> f(a + shk, t)

474 Chapter 6. Integration
is given by ^ i-> Dif(a + shk, t) hk', hence we have, according to the Fundamental
Theorem 2.10.1,
a + hk, t) - f(a, t) = I Di f(a + shk, t) ds hk.
Ja
/c
Furthermore, as a function of ^ the right-hand side is absolutely Riemann integrable
over V. Therefore
Fia + hk)-Fia) = i (f(a + hk,t) - f(a,t))dt
Jv
V
1
Jv Jo
Dif(a + shk, t)dsdt hk =: ^{a + hk)hk.
where <p : U ^ Lin(R", R). Applying the Dominated Convergence Theorem
twice with the mappings t i-> Dif(a + shk, ^). which converge pointwise to t h>
Dif(a, t) for k —> oo, we obtain
lim(p(a+hk)= I I limDif(a+shk,t)dsdt= I Dif(a,t)dt = (p(a).
k^cxi Jy Jq k^oo Jy
On the strength of Lemma 1.3.3 and Hadamard's Lemma 2.2.7 this implies the
differentiability of F at a, with derivative /^ Di f(a, t) dt. □
We will extend Corollary 6.4.3 on changing the order of integration to
continuous functions / : f/ —> R that are absolutely Riemann integrable over the open set
U C R^+''.
Example 6.12.5. Consider / : R^ ^^ R given by /(y, z) = e'^ '"-'". In view of
Example 6.10.8
/ f(y,z)dz= , / / f(y,z)dzdy = ^ dy < oo,
JR V^ + y -^R-^R JRy/l + y'^
yet one has the divergent integral f^ f(y, 0)dy = f^ldy. On the other hand,
/r /(y. z) dy < oo, for all z ^ 0. For every K e $(R^) there is i? > 0 with
K C [—i?,i?]x[—i?,i?]. Using this fact and the substitution of variables (y, z) =
(y, —7^=) in R^, one can prove that Theorem 6.10.6.(ii) is satisfied. It follows that
f is absolutely Riemann integrable over R^. Examples like this one explain why
the conditions in the following two propositions are not automatically satisfied if /
is continuous and absolutely Riemann integrable. ik

6.12. Dominated convergence 475
Proposition 6.12.6. Let U C R'^+^ be an open set and let f : U ^ R be a
continuous function that is absolutely Riemann integrable over U. Further suppose
that
y^ fiy^z)dz= / f(y,z)dz,
Jr'I Ju"{y)
z^ f(y,z)dy= / f(y,z)dy
Jrp Ju'(z)
both are continuous functions (in particular, they assume finite values everywhere).
Here U"(y) = [z sR'' \ (y, z) & U } and U'(z) = [y sRp \ (y,z) & U }. Then
[ f(x)dx=[ I f(y,z)dzdy=[ I f(y,z)dydz.
Ju Jrp Jr'i Jr'i Jrp
Proof. Since / is absolutely Riemann integrable over U if and only if /+ and /_ are
so, we may suppose / > 0. Let the sequence (iffc)j;gN of sets in ^(f/) be as in (6.38)
and the subsequent construction, and define Kidy) = {z ^R'' \ (y, z) € K^ }, for
A; e N and y e R'\ The sets Kic(y) are Jordan measurable in R^. Therefore the
following functions gk and g : R^ —> R are well-defined.
gk(y)= f(y,z)dz= iK,{y)(z)f(y,z)dz,
jKtiy) Jri
g(y)= / iu"{y)(z)f(y,z)dz.
Jri
On the strength of Theorem 6.4.2,
f f(x)dx= f gk(y)dy (ksN).
JKt Jrp
{Kic(y))i^^f^ is a nondecreasing sequence of sets in ^(R^) with union equal to U"(y).
Applying Arzela's Dominated Convergence Theorem 6.12.3 to the sequence of
functions (fk)keN satisfying ,/> = lA:j(y)/(>', •) : R^ ^ R and limfc^oo fk = f(y, •),
and using f(y, •) as the majorizing function, we obtain that limfc^oo gkiy) = g(y),
for all y sR''. Since / > 0 implies that (gk)keN is a nondecreasing sequence, with
g as a limit and majorizing function, application once more of Arzela's Dominated
Convergence Theorem gives
/ f(x)dx= lim / f(x)dx =hm / gk(y)dy= / g(y)dy.
Ju ^^°°Jki, k^ooJs,,, Js,p
Interchanging the roles of y and z finally gives the equality of both iterated
integrals. □
In practice it might be difficult to establish absolute Riemann convergence of /
over U, whereas computation of an iterated integral might be feasible. Therefore
results of the following type are extremely important in analysis.

476 Chapter 6. Integration
Proposition 6.12.7. Let f : U ^>- R be continuous on the open set U C R''"'"^,
and assume that y i-> f^^ \f(y, z)\dz and z i-^- f^^p \f(y,z)\dy both are
continuous functions. Then y i-> f^^ f(y,z)dz and z i-^- f^^,, f(y,z)dy are
continuous. Further suppose that one of the iterated integrals for \f\ converges; say
Irp Iri 1/(3'' 2)1 '^2 '^y < °°- Then f is absolutely Riemann integrable over U and
both iterated integrals for f are equal to the integral of f over U, in other words
[ f(x)dx= [ I f(y,z)dzdy= [ I f(y,z)dydz.
Ju Jr)' Jri Jr'i Jrp
Proof. In order to prove the absolute integrability of / we verify that
Theorem 6.10.6.(ii) is satisfied. To this end we may suppose that every Ki^^, for A; e N,
is a finite union of rectangles {B^.,- | i e / }. For every 1 < j < n, collect the
endpoints of the j-th coordinate interval of all rectangles B^.,-, and arrange these
points in increasing order, as in the proof of Proposition 6.1.2. Thus one obtains a
finite number of nonoverlapping rectangles B^, C R'', with I e L^, and, for every /,
a finite number of nonoverlapping rectangles B^',„, C R^, with m e Mki, such that
Kk = \J{ B'„ X b;',„, I / e L,, m e M^,} (k e N).
Because |/| is continuous on K^ we obtain from Theorem 6.4.2, for A; e N,
f \f(x)\dx = j2 Y. i, „ i/wi^^
= EE/, /„ \fiy,z)\dzdy = J2f,Ylf„ \f(y,z)\dzdy
I m •'^'kl •'Kim I •'Kl m •'Kim
<y,f f \f(y,z)\dzdy< f f \f(y,z)\dzdy.
, JB[, Jri Jrp Jri
Hence / is absolutely Riemann integrable over U. Therefore application of
Proposition 6.12.6 implies that the iterated integrals of / both equal the integral of / over
U. ' " □
In the formulation above it is essential to work with the absolute value of /.
Indeed, the existence of an iterated integral, if it involves cancellation, need not
imply the existence of the integral of / over U. Observe that we have to deal with
only one of the two iterated integrals, which is fortunate since it is often the case
that one is easier to estimate.
Proposition 6.12.7 is a very special case of Fubini's Theorem, which is part of
the theory of Lebesgue integration. In this theory one is able to weaken the notion
of Riemann integrability to that of Lebesgue integrability and yet to integrate
functions belonging to this wider class. The property of being a Lebesgue integrable
(or measurable) function is preserved under the formation of partial functions and
integrals thereof, and as a consequence Fubini's theorem has a more natural
formulation than Proposition 6.12.7. Results like this make Lebesgue integration superior
to Riemann integration.

6.13. Appendix: proofs of Change of Variables Theorem All
6.13 Appendix: two other proofs of Change of Variables
Theorem
In this appendix we treat two other proofs of the Change of Variables Theorem 6.6.1,
which we will call the second and the third proof. Each of them highlights different
aspects of the problem: the second is intuitive, geometrical, but rather technical;
although less intuitive, the third proof is quite efficient.
Second proof. Contrary to the demonstration in Section 6.9 this proof does not
require the Implicit function Theorem 3.5.1; on the other hand, some more detailed
information from linear algebra is needed. As a consequence Chapter 6 can be
studied independently from Chapters 3 through 5, which might appeal to readers
who prefer to make an early start with the theory of integration. In this setup the
proof of the Change of Variables Theorem in full generality requires the validity of
the theorem in the special case of a diffeomorphism ^ : R" —> R" that belongs to
Aut(R"). The treatment of the latter case needs some linear algebra, in particular
Lemma 6.13.2 below.
We begin with a definition. We denote the standard basis vectors in R" by e^,
for 1 < 7 < n.
Definition 6.13.1. A transformation in End(R") is said to be basic if it is given by
one of the following formulae, for 1 < A;, / < n, A; ^ /, x e R" and A. e R \ {0}:
^wC-^) = -^ =•= x,ek, Mi(X){x) =x + (X- \)x,e,,
Ski(x) =x + (xi - Xk)(ek - ei). O
We note some properties of basic linear transformations. In view of
(F*)-' = F,7, Miixr' = M,(A-'), S,V = Sik,
every basic linear transformation is invertible, with an inverse that is basic too.
Next we study the compositions A B and B A, where A is an arbitrary linear and B
is a basic linear transformation of R". In doing so we denote a linear transformation
and its matrix by the same symbol. From
[ ej, j ^ /;
Fki (ej) = ^j ± ^ji^k = \
[ ei±ek, J = /,
we see that the matrix AF^ is obtained from the matrix A by the column operation
of adding/subtracting the k-th column vector of A to/from the l-th column vector of
A. From (F^)' = F* we obtain (F^A)' = A'(F^y = A'F*, writing A' for the
transpose matrix of A. Combining this with the preceding result we immediately

478 Chapter 6. Integration
see that F^A arises from A by the row operation of addition/subtraction of the /-th
row vector of A to/from the k-th row vector of A. Furthermore,
M,(X)(ej) = ej + (X-\)Sj,e, = \
[ Xei, j = I,
implies that AMi(X) arises from A when the /-th column vector of A is multiplied
by A., and that Mi(X)A is the result of the analogous row operation. Also, from
Skiiej) = ej + (Sji - Sjk)(ek - e,) =
ei, j = k;
Sk, j = I,
we see that A Ski and Ski A are obtained from A by interchanging the k-th and
the /-th column vectors and row vectors, respectively, of A. Thus right and left
multiplication of a matrix A by the matrix of a basic transformation amount to
performing on A one of the column and row operations, respectively, which are
well-known from linear algebra.
These results enable us to prove the following:
Lemma 6.13.2. Every element in Aut(R") can be written as a product of basic
linear transformations.
Proof. Let A e Aut(R") be arbitrary. Since the top row vector of A is different
from 0 there exist Bi and B[ e Aut(R"), both being products of basic linear
HID
transformations, with BiAB', = 1 I. But then we can find Bi and B', as
above such that
BiAB[ = I ^ " 1 with Ai e End(R"-')-
Because det(BiAB[) is a nonzero multiple of det A ^ Owe have det Ai ^ 0, that
is, i4i e Aut(R"~'). By induction over the dimension n we can therefore show
the existence of B and B' e Aut(R") such that both are products of basic linear
transformations and that BAB' = /; this implies A = B~^B'~^, which proves the
lemma. □
Remark. The decomposition into basic linear transformations is not unique. For
instance, F^ = M;(-1)F^|M,(-1).
Now we are sufficiently prepared to establish the Change of Variables
Theorem 6.6.1 in the special case of a mapping ^ : R" —> R" that belongs to Aut(R").
In fact, we shall need the proposition for a slightly more general class of mappings,
which is given in the following:

6.13. Appendix: proofs of Change of Variables Theorem 479
Definition6.13.3. Abijecdveaffinetransformation^ofR"isamappingR" —> R"
that can be written in the form ^(y) = x°+Ay, where x° e R"andi4 e Aut(R").0
Note that the bijective affine transformation ^ above is a C' mapping and that
its inverse ^~' : R" —> R" is given by
<if-\x) = -A-'x° + A-'x. (6.50)
Furthermore, x° and A are uniquely determined by ^ since x° = ^(0) and A =
* - *(0).
Proposition 6.13.4. Let x° e R" and A e Aut(R"), and denote by ^ the
corresponding bijective affine transformation ofR". Suppose f : R" —> R to be a
continuous function with compact support. Then
f f(x)dx= f (/ovI/)(y)|detDvI/(>-)|^>- = |detyl| f f(x° + Ay)dy.
Jr" Jr" Jr"
Proof. Using D^(y) = A, for all y e R", and Corollary 6.4.3 we reduce the
problem to the case of dimension one,
f (/ovI/)(3,)|detDvI/(3;)|^>- = |detyl| f f(x° + Ay)dy
JR" JR"
= \detA\ f ■■■([ f{x° + A(yi,..., y„))dyi\ ■ ■ ■ dy„
= I det i41 y^ • • • yj^ f(A(yu...,yn))dy,y--dy,.
Here we used the invariance under translation of the one-dimensional integration.
In view of the preceding Lemma 6.13.2 and Step IV on composition in Section 6.9
we may assume A to be a basic linear transformation. Now in the case of A = F^
we have det A = \ and we write the last integral as the iteration of an (n — 2)-
dimensional integral over R"~^ and of
nfiyi, ■■■,yi±yk,---,yn) dy, dyk= l / /(y,, ...,>';,...,>'„)dy,dyk,
: JRJR
employing again the invariance under translation of the one-dimensional integration.
Thus we obtain
/ ■ ■ ■ ( / /(^)d^A---dxn = I fix)dx.
\f A = Ml (A.) we use det A = A. and
/ \Mfiyu ...,Xyi,..., y„) dy, = / f(yi,... ,y,,... ,yn) dy,.
JR JR

480 Chapter 6. Integration
Finally, for A = Ski the result follows from Corollary 6.4.3. □
Proof of Change of Variables Theorem. On the basis of Step I on reduction to
continuous / in Section 6.9 we may assume that / is continuous with supp(/) C U.
The proof then proceeds in three steps. In Step I the volume of a small cube C is
compared with that of its image ^(C). To this end ^ is replaced by its first-order
Taylor polynomial at the point y° at which C is centered, that is, by the affine
mapping T = TyO^ : R" —> R" satisfying
Tiy) = ^iy°) + D^iy°)iy - y°) = ^iy°) - D^iy°)y° + D^iy°)y.
Thus we compare ^(C) with the parallelepiped T(C), and the error estimate is
a direct consequence of the definition of differentiability of ^. As T is bijective
because D^(y°) is, we may as well estimate the difference between (T~^ o ^)(C)
and C itself. It turns out that for every e > 0 there exists ^ > 0 such that
C cube of diameter less than S =^ (T~^ o *)(C) C C%
where C^ is the cube concentric with C whose sides are multiplied by 1 + e. As
Proposition 6.13.4 is applicable with ^ replaced by T we find the upper bound (6.56)
below. Technically it is convenient to compare (T~^o^)(C) and C, instead of ^(C)
and T(C), because estimating the volume of the set of points at a distance less than
e from a given set is easier if the latter set is a cube rather than a parallelepiped. A
complication in the argument is that we do not know right away whether ^ (C) is
Jordan measurable.
In Step II the upper bound (6.56) is used for majorizing the integral /^ f(x) dx
by (1 + e)"+' times an upper sum of / o ^| det D^ |, and thus by
(l+e)"+' /"/o*(>')|detD*(>')|rf>'.
Jv
Sending e to 0 leads to (6.58) below.
Obtaining a lower bound for vol,, (^ (C)) is a delicate matter. For example,
suppose n = 2 and let ^(C) be a long rectangle [ 0, ^~M x [ 0, ^ ], with volume 1.
Then by moving each point in ^(C) over a distance of only 6, by sending x e R^
to (xi, 0), the rectangle collapses to the interval [0, ^~^ ] x {0} along the Xi-axis,
which has volume 0. In Step III this difficulty is avoided by applying the preceding
arguments to the inverse of ^.
Step I (Local inequality). We begin with the preparations needed to establish
the estimate (6.56) below. In view of Theorem 1.8.3,
K := supp(/ o *) = *-'(supp/) C V
is a compact subset of V. Below we shall work with cubes covering K; these are not
necessarily contained in K and therefore we enlarge if in a suitable way. According
to Lemma 6.8.1 we can find ^ > 0 such that
Ks = {y €"&!' \ there exists y' € K with \\y-y'\\<S}c V. (6.51)

6.13. Appendix: proofs of Change of Variables Theorem 481
In particular, every cube C C R" centered at a point of K and with diameter less
than S is contained in the compact set Kg.
From the definition of differentiability of ^ at 3?° e V, there exists for every
?j > 0 a ^ = S(y°, ?j) > 0 such that for every y e V with Hy — y^H < S,
\\^(y) - TyO^(y)\\ = \\^(y) - ^(y°) - D^(y°)(y - /)|| < 7? ||>- - >-°||.
Actually this estimate is valid uniformly for y° e Ks; more precisely, for every
?j > 0 there exists ^ > 0 such that every y and y° e Ks for which the line segment
between y and y° lies entirely in Ks satisfy
II3; - y°\\ < S =^ \\^(y) - TyO^(y)\\ < rj\\y - /||. (6.52)
Indeed, use Formula (2.25) and the fact that D^\i^ : Ks ^ Aut(R") is continuous,
and therefore uniformly continuous, on the compact set Ks.
From (6.50) and (2.5) we obtain
||(r,,ovi/)-'(x) - (r,ovi/)-'(x')|| < I|£>*(/)-'IIeuc.II^ -^'11 (^.^' ^ R")-
(6.53)
Once more using that D^ is continuous on K and that K is compact, we find
0 < m = max [[^^(y)"'!! , < 00. (6.54)
We now claim that for arbitrary ?j > 0 we can find ^ > 0 with the following
properties. Condition (6.51) is satisfied and for all y° ^ K and y ^ Ks with
II3? — 3?° II < ^ we have, in view of (6.53), (6.54) and (6.52),
ll(r,ovi/)-'(vi/(3,))-3,|| = ||(r,ovi/)-'(vi/(y))- (r,ovi/)-i(r,ovi/(j-))||
< ||DvI/()-0)-' \\^^J\^{y) - Ty.^{y)\\ < mrjWy - y%
€
Now let e > 0 be arbitrary, select r) = — and a ^ > 0 corresponding to this r), and
m
deduce
\\(Ty.^r\^{y))-y\\<€S for >-e C,
where
C is a cube centered at an arbitrary y° ^ K with diameter less than 6. (6.55)
Hence we get, writing C^ for the cube concentric with C whose sides are multiplied
by 1 + e,
iTy,^r\^iy)) e C,s C C% and so ^{C) C Ty^^iC).
It follows that the following upper Riemann integral satisfies the inequality
^(vI/(C)) := / l*(C)(x)rfx < / It^^ic^){x) dx = vol,, {T,o^(C')).
J R" J R" ■*

482 Chapter 6. Integration
Applying Proposition 6.13.4 with ^ replaced by the bijective affine mapping r^o^,
we find
vol,, (TyO^iO) = I det D^iy°)\vol„iC') = (1 + e)"| det DvI/(>-°)| vol„(C).
Thus, for any cube C as in (6.55),
vol„(vI/(C)) < (1 +e)"|detDvI/(>-°)| vol„(C). (6.56)
Step n (Global inequality). Let <S = {C,- \ i e / } be a finite set of nonoverlap-
ping cubes C; centered at y9 e K with diameters less than S as above, satisfying
Kc[JQcKsCV.
Since / o ^|^. : C,- —> R is continuous, there exist }?,■ e C,- such that we have
maxygc, / o ^(y) = / o *(}'/) for i € I. Define _/ : V ^^ R by jiy) =
I det D'^{y)\. Then y > 0 and, for y, y° e V,
j(yr'j(y°) = \j(yr\j(y°) - j(y)) +1| < j(yr'\j(y°) - j(y)\ +1-
Furthermore, j is uniformly continuous on Ks. Therefore we have, by shrinking
^ > 0 if necessary,
|detDvI/(3;°)| = jiydjiyir'jiy^) < (1 +e)| detDvI/(>-,)| (i e /).
We now combine this estimate with (6.56) for C = C,- in order to get
^i^(C,)) < (1 + e)"+' I det D^(y,)\ vol„(C,) (i e /). (6.57)
After these preparations we are able to treat the global problem. We have
/ = /+ — /_ with /± > 0 as in Theorem 6.2.8.(iii) and f± continuous. By going
over to the f± and using the linearity of integration we may assume 0 < /. In view
of
we obtain from (6.57)
f fix) dx <Tfo ^(y,) I l*(c,)(x) dx = Tfo vI/(>-,)^(vl/(C,))
< (1 + e)"+' Y. f ° *(>''-)l ^^t D^(yi)\ vol„(C,-)
16/
< (1 + e)"+' V max(/ o *| det D*!)(>>) vol„(C,-).

6.13. Appendix: proofs of Change of Variables Theorem 483
The sum at the right-hand side is an upper sum of / o ^| det D^\ determined by
the partition <S that covers K = supp(/ o ^); thus
f fix) dx < (1 + e)"+' f fo <iJ(y)\ det D<i/(y)\ dy.
Ju Jv
Since the estimate is valid for every e > 0 it implies
f fix)dx < /" /ovI/(y)|detDvI/(>-)|^>-. (6.58)
Ju Jv
Step in (Reverse inequality). Now apply this inequality with ^ : V —> f/
replaced by *"' : f/ ^^ V, and / : f/ ^^ R by / o *| det D*| : V ^^ R,
respectively; the conditions are satisfied since supp(/ o ^| det Z)^|) = AT is compact in
V. As|detD*(*-'(x))||detD*-'(x)| = 1 we find
\ fo<\i{y)\dQiD<\'{y)\dy< f f(x)dx.
Jv Ju
The desired equality from the Change of Variables Theorem follows by combining
these two inequalities.
Remarl^. Note that the proof makes repeated use of the fact that ^ is a C' dif-
feomorphism, fully exploiting all implications thereof.
Third proof. In the remainder of this section we give a third proof of the Change
of Variables Theorem 6.6.1. Again the main ingredient is reduction to the case of
a bijective affine transformation as treated in Proposition 6.13.4. In Step IV below
this reduction is effectuated by showing that
-J- I {fo<\>'){y)d^iD<\'\y)dy = 0,
at Jv
dt
if (^Or6[o, 1 ] is a one-parameter C' family of C' diffeomorphisms with the property
that supp(/ o ^') n 9 V = 0. In particular, we construct such a family (^') that
transforms ^ into a local best affine approximation to ^. The other technical tool
is the Global Inverse Function Theorem 3.2.8. This approach might be the least
intuitive of the three. The circle of ideas involved here originates from homotopy
theory, a subject in algebraic topology.
Step I (Localization). This is the same as Step in on localization in Section 6.9.
The precise nature of the bounded open neighborhoods Ox oi x e U that are used
to cover the compact set K = supp(/) C U is specified at the end of the following
step.

484 Chapter 6. Integration
Step II (Deformation). We consider a C' diffeomorphism ^ : V —> f/.
According to Step I it is sufficient to study the diffeomorphism in suitable neighborhoods
of an arbitrary but fixed point ^~^(x) = y e V. We will show that locally, near y,
the diffeomorphism ^ can be deformed through a one-parameter C' family of C
diffeomorphisms into its best affine approximation at y.
From the definition of differentiability of ^ at y (compare with (2.10)) we see,
for y + h e V,
<i/(y + h) = <i/(y) + D<i/(y)h + ey(h) =: Ah + ey(h),
||e,(/2)||=a(||/2||), h^O.
Here A is the best (bijective) affine approximation to ^ at y. Next define, for
f €[0, 1],
vl/(f, y + h) = ^(y + h) - t€y(h) = Ah + (1 - t)€y(h). (6.59)
Then \i/(0, y + h) = <i/(y + h) and *(1, y + h) = Ah. Further introduce
3^:[0, 1] X V^[0, l]xR" given by ^(t,y) = (t,^(t, y)).
In order to prove that 'I' is a C' diffeomorphism onto its image we now verify that
the conditions of the Global Inverse Function Theorem 3.2.8 are satisfied. In fact,
Dy<i/(t,y) = D<i/(y) e Aut(R") immediately gives D3'(f, y) e Aut(R"+'), for
(r, 3?) e [0, 1 ] X V. Moreover, ^(t,y+h) = ^(t',y + h') implies t = t' and also
Ah + (1 - t)€y(h) = Ah' + (1 - t)€y(h').
so
h-h' = (l- t)D^(yr\,y(h') - ,y(h)).
Next we apply the result of Example 2.5.4 with the mapping Sy and e equal to
(2\\D^(y)~^ IIeucI^' ^"^ order to find ^ > 0 as in the example. Thus we obtain, for
h,h' eB(0;S) and t e [0,1],
\\h-h'\\ < (l-t)\\D^(yr'\\^^J,y(h')-,y(h)\\ < -\\h-h'\\.
Consequently h = h', which proves the injectivity of ^. On account of the Global
Inverse Function Theorem, ^ restricted to [ 0, 1 ] x ^(y; ^) is a C' diffeomorphism
onto its image. Finally, take Ox equal to the open neighborhood ^{B(y; S)) of
X = ^(y); note that S depends on the choice of 3? e V.
Step ni (Supports). Write *' = *(r, •)• We need control over supp(/ o *') c
V. Recall that if is compacting. According to Lemma 6.8.1 we can find e > Osuch
that K C Kg C U, with K^ compact too. And as a consequence of Theorem 1.8.3
the image set L := ^~^(Ke) is compact in V. In view of Steps I and II it is sufficient
to consider the subsets K H Ox- According to (6.59),
vl''(>' + h)eKnOx =;> y + he^-\K + t€y(h)).

6.13. Appendix: proofs of Change of Variables Theorem
485
By shrinking^ > OasinStepIIifnecessary, we may arrange that AT+fe3,(/2) C K^,
for all (t, /z) e [0, 1 ] X B(0', S). Consequently we obtain
supp(/o vl/') c L C V (fe[0, 1]).
(6.60)
Step IV (Differentiation). Suppose ^ restricted to [0, 1 ] x B(y°; ^) is a C
diffeomorphism as in Step 11. Then (^Or6[o,i ] is a C' family of C' diffeomorphisms
defined on a subset of V. In the formulae below we write D for the total derivative
with respect to the variable y e V,m particular, D^'(y) for Dy^(t, y). Further
assume that / e C^(O^o) with ^(y^) = x°. Then we introduce
I(t)= f(fo^')(y)dctD^'(y)dy (fe[0,l]).
Jv
Under these assumptions we will prove that / (t) actually is independent off, which
implies
/"(/° *)(}') detD*(3;)^>' = detD*(>'°) f f(<i/(y°) + D<i/(y°)y)dy.
Jv Jv
Given t and t + u e [ 0, 1 ] we define
S"eC'(V,R") by 3" = (*0~'*'+"-
Observe that S° = /. The chain rule now implies the following identity of functions
on V:
/ o *'+" det D*'+" = (/ o *0 o 3" (det D*0 o 3" det D3" =goa" det D3",
(6.61)
where g = / o *' det D*' e C'(V). Next define the C vector field
? : V ^ R" by ?(>-) = —
du
s"(y).
H=0
Note that the mapping t h> D^'(y) is continuously differentiable, while
y^^yij'(y) = -eyo(y - /) = ^(y°) + D^(y°)(y - y°) - ^(y)
dt
is continuously differentiable near y° too. On account of Theorem 2.7.2 on the
equality of mixed partial derivatives and Exercise 2.44.(i) we therefore obtain
d_
du
detD3"(>')=trD
H=0
du
E"{y)=^D^{y) = Avj^{y),
«=o

486 Chapter 6. Integration
where we recall the definition of the divergence of § from Formula (5.30). Then
we have, by the Differentiation Theorem 2.10.13 or 6.12.4 and Formula (6.61),
/'(O =-f| Iit + u)=f-f\ igoS")iy)detDS"iy)dy
= f (Dg(ymy) + g(y)diY^(y))dy= f diY(g^)(y)dy
Jv Jv
= f Yl Djig^j)iy)dy = 0.
The last equality is obtained by the Fundamental Theorem of Integral Calculus on
R, see Case I in the proof of Theorem 7.6.1 on integration of a total derivative, and
by the inclusion (6.60). In Example 8.9.3 one can find an alternative computation.
Step V (Case of affine transformation). The previous steps enable a reduction
of the proof of the theorem to the case treated in Proposition 6.13.4. Two details
still have to be settled. Note that det D<if(y) ^ 0, for all y e V. Hence it is a
consequence of the Intermediate Value Theorem 1.9.5 that det D^(y) has constant
sign on the connected components of V, see Definition 1.9.7. Considering V as the
union of its connected components and splitting the integral accordingly, we can
take the sign of det D^(y) outside the integrals, which justifies the omission of the
absolute value signs in Step IV. Furthermore, we may use approximation arguments
from Section 7.7 to replace the condition of / e C (f/) by that of / e C(U).
Remark. Actually, in the notation of the Steps 11 and IV above one can find
a C' family (*0r6[o,i] of C' diffeomorphisms satisfying ^° = ^ and *' =
sgn{detD^(y))I, where / is the identity mapping. This is a consequence of
the connectedness of the subset Aut°(n, R) = [A € Aut(R") | det A > 0} of
End(R"), which can be proved using Lemma 6.13.2. Arguing in this way one may
bypass Proposition 6.13.4.

Chapter 7
Integration over Submanifolds
The knowledge acquired about manifolds and integration will now be used to
develop the theory of integration over a submanifold in R" of dimension d < n. In
particular the d-dimensional volume (length, (hyper)area, etc.) of bounded sub-
manifolds will be defined. By way of application we study the generalization to
R" of the Fundamental Theorem of Integral Calculus on R. This is a problem with
two aspects: finding correct formulae on the one hand, and antidifferentiation of a
function of several variables on the other. The first aspect culminates in the theorem
which asserts equality between the integral of the total derivative of a function over
an open set, and an integral of the function itself over the boundary of that open set.
Gauss' Divergence Theorem then is a direct corollary.
7.1 Densities and integration with respect to density
In Chapter 6 we introduced in particular the integral of (absolutely) Riemann inte-
grable functions defined on open subsets of R", which we shall regard here as C*
submanifolds in R" of dimension n. We now wish to develop a theory of integration
over C^ submanifolds, for k > 1, in R" of dimension d < n. Note that according
to Corollary 6.3.8 such submanifolds are negligible in R", if they are compact. We
shall therefore have to be somewhat careful to take due account with respect to
integration of the manifolds V in R" of dimension d <n.
First we consider this problem locally, that is, in a neighborhood U in R" of a
point X e V. According to Theorem 4.7.1 there exist an open subset D C R'' and
a C'' embedding (j) : D ^ W such that 0(D) = V n f/. Let / : V ^^ R be a
bounded function for which
supp(/) C <PiD) is a compact set. (7-1)
487

488 Chapter 7. Integration over submanifolds
Then / o 0 : D —> R has a compact support in R''; and it is tempting to call /
Riemann integrable over V \f f o <p \s Riemann integrable over D C R''. And
further, in that case, to define the integral of / over V, with the notation I^(f), as
the integral of f o<p over D,
-L
hif)=\ifo4>){y)dy. (7.2)
Jd
Now let D C R'' be open, let 0 : D —> V be another C* embedding, and
assume
supp(/)C0(£>)n0(D). (7.3)
Then
supp(/ o 4>) c £>^.^ := 4>-\HD)) C D.
It follows from Lemma 4.3.3.(iii) that the mapping 0~' o 0 : D^ ^ —> D^^ is a C'^
diffeomorphism of open subsets in R''. Furthermore,
(/ ° 0) ° (0~' ° 0) = / ° 0 on D^ J.
On account of the Change of Variables Theorem 6.6.1, therefore, f o<p\s Riemann
integrable over D if and only if
(/ o 0) o {4>-' o $) I det Di,p-' o $)\ = ifo$)\ det Di,p-' o $)\
is Riemann integrable over D; the latter applies if and only if / o 0 is Riemann
integrable over D. Subject to this assumption we then have
f (f o ct>)(y) dy= fifo ^)(y) \ det D(0-' o $)(y)\ dy.
Jd Jd
In general, therefore,
hif) = f (fo ^)(y) dyy^ fifo ^)(y) dy = If(f).
Jd Jd
This result shows that the Riemann integrability of / over V is independent of the
choice of the parametrization 0 of V, but that the value /,^(/) of the integral of /
over V depends on 0, unless by chance | det D((j)~^ o(j))\ = 1, that is, unless 0~' o0
is a volume-preserving coordinate transformation in R''.
We are thus confronted with the presence of extra factors | det D{<p~^ o 0)| in
the integrand. It is natural, therefore, to incorporate these from the start in the
definition of the integral /,^(/) of / over V, in such a way that this integral is
independent of the chosen parametrization (p. Thus, assume there is a continuous
function p^ : D ^ VL associated with the embedding <p : D ^ W and let, unlike
(7.2), the integral I,j,{f) of / over V be defined by
-L
hif)= / ifo4>){y)P4>iy)dy.

7.1. Integration with respect to density 489
We then require
hif) = lif°<i>)iy)p<i>iy)dy
Jd
= f (f ocP) o(ct>-' o^)(y) p^(ct>-' o^(y))\dctD(ct>-' o^)(y)\dy
Jd
= f(fo^)(y)p^(y)dy = If(f).
Jd
Hence the following:
Definition 7.1.1. A continuous d-dimensional density p on a C* submanifold V in
R" of dimension disa mapping which assigns to every C^ embedding 0 : D^ —> V,
where D^ C R'' is open, a continuous function p^ : D^ —> R in such a way that
P$(y) = P4>ir' o ^(y)) I det D(0-' o ^)(y)\, (7.4)
for any C^ embedding (p : D^ —> V and every y e (j)~^{(j)(D^)). O
Remark. A continuous density p on V is uniquely determined by a collection
of continuous functions p^ : D^ ^ VL satisfying (7.4), where <p £ <1>, if <1> is a
collection of C'' embeddings such that
Vc|J<^(^*)- (7-5)
To see this, consider an arbitrary C'' embedding <p : D ^ V. For every y e D we
can find a 0 e <1> with 0(y) £ <p{D^). We now define
P^(y) = P^{4>~' o 0(>O) I det D(ct>-' o ^)(y)\.
It is easy to verify that this definition does not depend on the choice of 0 e <1> for
which (j)(y) e <p{D^), and that the collection {p'^\<p arbitrary embedding} defined
in this way satisfies requirement (7.4). Hence all that is required for the introduction
of a continuous density on V is a minimal collection <1> of embeddings that satisfy
(7.5); thus for a sphere in R^ two embeddings suffice.
Remark. It is also possible to formulate the theory above in terms of coordinati-
zations or charts (see Definition 4.2.4),
K := <p'^ : <p(D) ^ D =: U^; hence k : k~\U^) ^ U^,

490 Chapter 7. Integration over submanifolds
instead of the embeddings <p. A density p then assigns to every chart k a continuous
function Pi^ : U^ ^>- R such that
Pi^iy) = P.iK olc-\y))\detD(K oK-')(y)\ (y € Ui; n (jc o k-')(U,)).
Note that k ok~^ is the transition mapping from the last Remark in Section 4.3.
We now want to free ourselves from the requirements in (7.1) and (7.3) that the
support of / be contained in (p (D), or even in the intersection of several such image
sets. Therefore we have:
Definition 7.1.2. - Theorem. Let V be a C*" submanifold, with k > 1, in R"
of dimension d. Let / : V —> R be a bounded function with compact support
supp(/) =: K. Let <!>' be a collection of C^ embeddings 0 : D^ —> V with
D^ C R'' open and with K C [J{(piD^) | 0 £ *'}. Let {x,^ I 0 € *} be a
continuous partition of unity on K subordinate to the open covering {f/^ | 0 e <!>'}
of K (see Theorem 6.7.3); here <p{D^) = V (1 f/^, with U^ open in R". Then / is
said to be Riemann integrable over V if for every 0 e <1> the function
is Riemann integrable over D^. If this is the case, the integral of f over V with
respect to the density p, notation J^ f(x)p(x) dx, is defined by
f f(x)p(x)dx = Tf (X* /) ° <P(y) pfiy)dy- (7.6)
Here it is of course essential that the left-hand side in (7.6) is in fact independent
of the choice of the collection <1> and of the partition {x<t> \ <P ^ ^}- Indeed, let <5
and {X^ I 0 ^ *^ }> respectively, be another such choice. Then
XI / (x<t> f) ° <P(y) p<t> (y)dy
<t>e^ ■'^^ ^6$
= XI / X$ o $(y) x^ o $(y) f o $(y) p^{(j)~^ o $(y))
.\dctD(ct>-'o^)(y)\dy
= X / X ■'^'^ ° ^^y^ ^^ ° ^^y^ f ° ^^y^ p^^y^ ^y
= E/' ix^f)°^(y)p?(y)dy-

7.1. Absolute Riemann integrability w.r.t. density 491
Here we have successively used X^^gj XJ — 1 on iir; the substitution y = <p~^ o<p(y)
and the Change of Variables Theorem 6.6.1; Formula (7.4); and, finally, X^j,^^ X(t> —
lonK. n
The continuous density p is said to he positive if p^ {y) > 0, for every embedding
<p and all y e D^. In this case p may be regarded as a continuous "ubiquitous"
mass density on V. In cases where a p^ also takes values < 0, the physical analog
is a continuous charge density. O
The following lemma gives a description of all continuous (i-dimensional
densities on V in terms of one positive density.
Lemma 7.1.3. Let p be a fixed positive continuous d-dimensional density on V.
(i) For every continuous function f on V the mapping f p with f p : <p \-^
f o<p p^ defines a continuous d-dimensional density on V.
(ii) The mapping f h> f p is a bijection from the space of all continuous
functions on V to the collection of all continuous d-dimensional densities on
V.
Proof, (i). The function f o<p p^ satisfies (7.4), because
(/ 0 ^P^)(y) = f o 4>{4>~' o Hy))P4>ir' ° Hy)) I det Dir' ° Hy))\-
(ii). This assertion follows if we can prove that / i-^- / p is a surjection. Thus,
given a continuous density p we have to determine a function / such that p = f p.
We define, for every embedding <p, the continuous function f^ : <p{D^) —> R by
f,{x) = ^"f'^fl {X € 4>{D,)). (7.7)
Note that positivity of yo is essential here. Ifx e 0(D^) (10 (D^), one has, by (7.4),
_ P^(^"'(^)) _ p;(0-'°0(0-'(^))) _ ,.
* P^G-\X)) p^(0-la 0(0-1 (X))) •'*^''^'
Consequently there exists a unique function / on V such that f{x) = f,j,{x), for all
X in <p{D^). Because the function f\^(j) \ = f<i> is always continuous, it follows
that / is continuous on V. Furthermore, (7.7) implies that 'p = f p. □

492 Chapter 7. Integration over submanifolds
7.2 Absolute Riemann integrability with respect to density
In Definition 7.1.2- Theorem a restriction was made to functions / : V —> R with
compact support. As in Chapter 6, this is not without its drawbacks. Therefore we
imitate Lemma 6.10.2 and Definition 6.10.5 in the following:
Definition 7.2.1. Let the notation be as in Section 7.1. Let p be a positive continuous
density on V, and let / : V —> R. The function / is said to be absolutely Riemann
integrable over V with respect to p if / is locally Riemann integrable in V, and if
sup / \f{x)\p{x)dx < oo.
Ke${V) Jk
For such an / one sees, as in Definition 6.10.5, the existence of the integral
fy f(x)p(x) dx of f over V with respect to p, with the following property:
/ f(x)p(x)dx= sup / f+(x)p(x)dx — sup / f^(x)p(x)dx.
Jv k^${V)Jk' k^${V)Jk
A subset i4 of V is said to be Jordan measurable with respect to p if the
characteristic function 1a is absolutely Riemann integrable over V with respect to
p\ in that case
volrf(i4) := / p{x)dx := / \Aix)p{x)dx
is said to be the d-dimensional volume or the d-dimensional Jordan measure of A
with respect top.
The set A is said to be negligible in V if vo\d{A) = 0. Note that the question
of A being negligible or not is independent of the choice of the positive continuous
density p on V; this follows from Lemma 7.1.3. O
By means of the techniques from Sections 6.8 and 6.7 one easily proves the
following:
Lemma 7.2.2. Let K C V be a compact set which is negligible in V. Then there
exist, for every e > 0, a continuous function Xa: : V —> R with compact support,
and an open set U^ in V such that
0<Xa:<1, KcUk, Xk = i onUK, / XK(-x)p(x)dx < e
/ Xk(-x)p(x)
Proposition 7.2.3. Let U C V be an open subset in V and assume that dyU is
negligible in V. Let f : V —> R Z>e a locally bounded function. Then I u /
is absolutely Riemann integrable over V with respect to p if and only if f\rj is
absolutely Riemann integrable over U with respect to p\jj. If this is the case, then
f (lu f)(x)p(x)dx = f (f\u)(y)(p\u)(y)dy.
Jv Ju

7.2. Absolute Riemann integrability w.r.t. density 493
Proof. In conformity with the definition of absolute Riemann integrability over V,
consider arbitrary K e JC(V). One has that (see Definition 1.2.16)
L ■.= KndvU
is compact in V; to see this, use the fact that the intersection of two closed sets
is also closed. Furthermore, L is negligible in V, because L is a subset of the
negligible dyU. Let e > 0 and let Xl ^iid Ui be as in the preceding lemma. Set
X = Ijf (1 — Xl)- Because I — Xl vanishes in the neighborhood Ul of dyU (1K, it
follows that X vanishes in a neighborhood of dyU. Because Ik — X = ^k Xl, one
has, as / is locally bounded,
/ (IkIu f)(x)p(x)dx - (x^u f)(x)p(x)dx
\Jv Jv
<sup|/(x)| / XL(x)p(x)dx <esup\f\.
xeK Jv K
A similar estimate holds if / is replaced by | /1. But this shows that, in examining the
absolute Riemann integrability of Ik^u f over V with respect to p, it is immaterial
whether or not one imposes on the s&X K e X{V) the extra condition that K C
U. And this last point corresponds to the examination of the absolute Riemann
integrability of /| jy over U with respect to p\jj. □
Definition 7.1.2 - Theorem suffers from the complication that jy f(x)p(x) dx
is defined in terms of bump functions x<t>-: if the manifold V has to be described by
more than one parametrization <p. Indeed, the functions x<t> ensure that overlapping
subsets <p (Dif,) of V give proper contributions to the integral. The following theorem
demonstrates that when the case arises, the integral can be calculated without these
bump functions.
Theorem 7.2.4. Let p be a positive continuous density on V and let <1> be a finite
collection ofC^ embeddings <p : D^ ^ V such that
(i) <P(D^)n$(D^) = Q,if<p^$;
(ii) N := V \ [Jj,^^(t>(D^) is a negligible set in V.
Let f : V —> R Z>e fl locally bounded/unction. Then f is absolutely Riemann
integrable over V with respect to p if and only if f o<p p^ is absolutely Riemann
integrable over D^,for every 0 e <1>. If this is the case, it follows that
f f(x)p(x) dx = Tf (/ o ct>)(y)p^(y)dy. (7.8)
Jv j.^^Jdj,
tt>e^

494 Chapter 7. Integration over submanifolds
Proof. One has
Because N is negligible in V, it follows that / is absolutely Riemann integrable
over V with respect to p if and only if this holds for
/ ■= zJ ^'^(^*) f-
When this holds, / and / have the same integral over V with respect to p. For /
absolutely Riemann integrable over V with respect to p we then see, multiplying /
only by continuous functions X : V —> R with compact support supp(x) C <p{D^),
that f\(h(D ) is absolutely Riemann integrable over 0(D^) with respect to p. But
that is equivalent to the absolute Riemann integrability of f o<p p^ over D^.
Now assume, conversely,
f o<p p^ absolutely Riemann integrable over D^, for all 0 e <1>. (7.9)
Then
4>{D^)cW:=V\\Jhd^);
and so IV, being a complement of open sets in V, is closed in V. Because 0(£>^)
is the smallest closed set in V containing 4>{D^), it follows that <p(D^) C W.
Because <p{D^) is open in V, we have the disjoint union
Consequently
MHD^)) C W \ <j>{D^) = V\[j HD^) = N.
Hence dv{<j)(D^)) is negligible in V. On account of the preceding proposition,
it follows from (7.9) that 1,^(d^) / is absolutely Riemann integrable over V with
respect to p, and that
/ (l,^(D^)/)(-^)p(-^)^-^ = / f(x)p(x)dx= / (f o(j))(y)p^(y)dy.
Jv J(I>{D^) JD^
Summation over 0 e <1> now leads to (7.8). □
Remark. In practice, <1> often consists of a single element <p, that is, in such a
case there exists a C' embedding 0 : D —> V such that V \ <j)(D) is negligible in
V. Then / is absolutely Riemann integrable over V with respect to p if and only if
f o(j) p^is absolutely Riemann integrable over D. If this is the case, then
/ f(x)p(x)dx= / (f o(j))(y)p^(y)dy.
Jv Jd

7.3. Euclidean A-dimensional density 495
7.3 Euclidean ^-dimensional density
Let V be a C'^ submanifold in R" of dimension d. Then V, by virtue of its being a
submanifold of R", possesses a "natural" positive rf-dimensional density do; this ui
is said to be the Euclidean d-dimensional density on V, to be introduced below. It
will turn out that, if d = n, integration with respect to this co is identical with the
n-dimensional integration from Chapter 6.
From Example 2.9.6 it follows, for a mapping A e Lin(R'', R"), that A'A e
End(R'') satisfies det(A'A) > 0. Accordingly we can now define
CD : Lin(R'', R") ^R by cd(A) = ^det(A'A) = s/det({a~aj}).
We then have the following properties for co:
ca(AB) = \detB\ca(A) (B e End(R'')); (7.10)
cd(CA) = cd(A) (C e 0(R")). (7.11)
Indeed, (7.10) follows from
det(ABy (AB) = detB'(A'A)B = det B' det(A'A) del B = (det Bfco(Af.
According to Definition 2.9.4 any C e 0(R") satisfies C'C = /, and (7.11) is
found from
det(Cyl)' (CA) = detA'(C'C)A = cd(A)^.
Definition 7.3.1. - Theorem. Define, for every C'^ embedding cp : D^ ^ V, with
k > \, the function cd^ : D^ —> R by
CD^iy) = CD(Dct>(y)) = ^dct(DcPiy)' o Dct>(y)) (y e D^).
Then co : cp h> cOif, is a positive rf-dimensional C^ density on V, the Euclidean
d-dimensional density on V. Indeed, let (p be another embedding; one then has, by
the chain rule,
D$(y) = D(ct> o (ct>-' o my) = DcPdcp-' o ^)(y)) o D(ct>-' o ^)(y),
where D(0-' o $)(y) e End(R''). Using (7.10), one then finds
cofiy) = co(D^(y)) = \ det D(0-' o ^)(y)\co(Dct>((ct>-' o $)(y)))
= co^(ct>-' o ^(y)) I det D(ct>-' o ^)(y)\;
that is, requirement (7.4) for a density is met.
If / is Riemann integrable over V, then by Definition 7.1.2 we have the integral
of f over V with respect to the Euclidean density, with notation
/ fix)dax.
The subscript d in ddX emphasizes that we are dealing with integration with respect
to the Euclidean density on a rf-dimensional manifold. O

496 Chapter 7. Integration over submanifolds
Motivation for the preceding definition. In the special case of d = n one has,
on account of the Global Inverse Function Theorem 3.2.8, that 0 : D —> (j)(D)
is a C^ diffeomorphism of open sets in R". Additionally, because the matrix of
D(j) (y) : R" —> R" is square, one has in this case
co^iy) = ^dct(Dct>(yyoDct>(y)) = ^dct DcPiy)' dct Dct>(y) = \ det Dct>(y)\.
In view of the Change of Variables Theorem 6.6.1 it now follows for L,d-. f{x) dnX,
the integral of / over 0(D) with respect to the Euclidean density co on <p{D), that
I fix) d„x =f(fo <P)(y) I det D<P(y)\ dy = I f(x) dx.
J(t>{D) Jd J<t>{D)
In other words, integration with respect to the Euclidean density on (p (D) on the
one hand and n-dimensional integration over 0(D) on the other coincide.
(The case d < n). If cp : D ^ V with D C R'' open and d < n,we should like
to imitate the foregoing and define Lf^-. f(x) d^x, in accordance with the Change
of Variables Theorem 6.6.1, as
f fix) dax =f(fo (j>) iy) I det D(j> {y) | dy. (7.12)
J(I>(D) Jd
However, there is a problem in that D<p{y) : R'' —> R", and that as a result
det D(j)(y) is undefined. But we do know, from Theorem 5.1.2, if (j)(y) = x, that
D<Piy):R'' ^ r,VcR"
is a linear isomorphism onto the tangent space T^V to V at x. But there exists
C e 0(R") by which T^V is "laid flat", that is, for which
C(r,V) = R''x{ORw}.
Therefore C o D<p{y) is a bijective linear transformation from R'' to R'' x {Or-i-^/}.
On its image, the projection Pd e Lin(R", R'') onto the first d coordinates is a linear
isomorphism, that is
D{y) := PdoCo D<piy) e Aut(R'').
The correct version of (7.12) therefore is
f fix) ddx = fifo ct>)(y) I det D(y)\dy. (7.13)
J(t>{D) Jd
Since Pj o P^ e End(R") equals the identity on R'' x {Or^-^/} = im(C), we find
( det D(y))'^ = det D(y)' det D(y) = det (D(y)' o D(y))
= det (Dipiy)' o C'^ o P^ o Pd o C o D<piy))
= det{D(j>{y)'oD(j>{y)) = co^{y)^.

7.3. Euclidean A-dimensional density
497
And this in fact yields, by (7.13),
/ fix) ddX = (f o (j))(y) co^iy) dy.
J(t>{D) Jd
The Euclidean rf-dimensional density co on V can therefore be characterized as
follows: CO equals the standard volume factor in R'' ~ R'' x {Or-i-^/ } — T-gV,if T^ V
"lies flat", and in addition co is invariant under elements in 0(R")-
(Area equals volume divided by length). It is possible to formulate the
foregoing in a somewhat different manner. The column vectors Di<p{y),..., Dd<j)(y)
in R" of the matrix D<p{y) span T^V. Now dim(rf V)-*" = n — d, and we can
therefore choose vectors v^+i (y) through v„ (y) in R" such that together they form
an orthonormal basis for (TxV)-^. Define D<p{y) e GL(n, R) as the matrix having
in the first d columns the column vectors of D<p{y), and in the last n — d columns
the vectors Vd+\ (>'), • • •. v„{y).
D<j)(y)= n ( D<j)(y) \ Vd+i(y)
n-d
Vniy) )
(7.14)
One then has
{AQlD(j){y)f = det(D(p(yy o^(p(y))
(Dj<j>(y), D,<j>(y))
{Dj<P(y), Dd<P(y))
n-d
(Dj<j)(y), Vd+i(y)]
{Dj<P(y), v„(y))
ISjsd
n-d
(vd+j(y), Di<j)(y))
(vd+jiy), Dd<p(y))
(vd+jiy), Vd+i(y)}
(vd+j(y), v„(y))
l<j<n—d
n-d
d
n-d
(Dj<j>(y), Di<j>(y)}
0
'n-d
= detiD<p(yy oD<p(y)).
Therefore
co^(y) = \dctDct>(y)U (7.15)
in other words, C0ii,(y) is the n-dimensional volume of the parallelepiped in R" (see
Example6.6.3)spannedby the vectors Di0(}'), ..., Dd<j)(y), Vd+i(y), ..., v„(}')
inR". Thisvolumeisindependentof the choice of the vectors Vrf_|_i()'), ..., v„(y),
provided only that they form an orthonormal basis for (TxV)-^.

498 Chapter 7. Integration over submanifolds
lA Examples of Euclidean densities
I. Arc length. Let (i = 1. If D C R is open and (p : D ^-BJ' a. C'^ embedding,
for A; > 1, then V = im(0) is a C* curve in R", while, if we identify Hnear operators
and matrices.
D<j>(y) = <j>\y) = I
' €(y) J
e Lin(R, R") (y e D).
We have D<p(yy o D<p(y) = ((D<p(y), D<p(y))), and therefore
CD^iy) = co{D(j>{y)) = \\D(j>{y)\\ {y e D). (7.16)
The Euclidean density co in this case is said to be the arc length, and accordingly the
integration with respect to co is said to be the integration with respect to arc length.
Thus we have, for / Riemann integrable over V,
f f(x)d,x = fifo ct>)(y)\\Dct>(y)\\ dy.
Jv Jd
In particular, the arc length of V is defined as
d\X,
I
if the integral converges. Note that the arc length of V has now been defined
independently of the parametrization of V.
Example 7.4.1 (Circle, ellipse and cycloid). The segment of the unit circle {x e
r2 I ||x|| = 1} lying between (1,0) and (cosx, sinx), for 0 < x < 2n, is
parametrized by
<p(t) = (cos^, sin^) (0 < ^ < x).
Therefore the arc length of this segment is
V (— sin f)2 + cos^ t dt = x.
f
Jq
We now define the angle between two intersecting lines in R^ as the length of the
shortest segment, lying between those lines, of the unit circle about the point of
intersection. The result above implies that the angle between the positive direction
of the Xi-axis and the line through the points (0, 0) and (cosx, sinx), (indeed)
equals X.

7.4. Examples of Euclidean densities
499
e
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0
5708
5308
4890
4454
3994
3506
2984
2417
1785
1048
1
5669
5267
4848
4409
3947
3456
2930
2357
1717
0965
2
5629
5226
4805
4364
3899
3405
2875
2296
1648
0879
3
5589
5184
4762
4318
3851
3354
2819
2235
1578
0791
4
5550
5143
4718
4273
3803
3302
2763
2173
1507
0700
5
5510
5101
4675
4227
3754
3250
2707
2111
1434
0605
6
5470
5059
4631
4181
3705
3198
2650
2047
1360
0505
7
5429
5017
4587
4134
3656
3145
2593
1983
1285
0399
8
5389
4975
4543
4088
3606
3092
2534
1918
1207
0286
9
5348
4933
4498
4041
3557
3038
2476
1852
1129
0160
Illustration for Example 7.4.1
E{k) = l.a, table giving first four decimals of a (rounded off) with k^
increasing in steps of 0.01
The length of the ellipse {x eR^ | -2+72= 1}, with a > Z> > 0, is given by
4aE(e). Here
e =
Vfl2 - Z>2
is said to be the eccentricity of the ellipse, and
/.f
= / y/l - k?- sin^ t dt
Jo
E(k)
(0<k<l)
is Legendre 'sform of the complete elliptic integral of the second kind with modulus
k. Indeed, the ellipse is the image under the embedding t \-^ {a cos t, b sin t).
The arc length of the segment of the cycloid (p : t h> (^ — sin f, 1 — cos t) from
Example 5.3.6 lying between 0(0) and <p(27i) is given by
/ J(l-cost)^ + sm'^tdt ='/i\ VT
= 2 / sin - rff =
h 2
cos t dt
-4 cos
2jr
= 8.
1^
Special case. Let n = 2 and assume 0 : R —> R^ has the form 0(r) = (f, fit))
of the graph of a C* function / : R —> R, for A; > 1. Then the arc length of the
segment of the graph of / lying between {a, f(a)) and {b, f(b)) is given by
/'
wihfitmdt
Ja
\ + f'{tYdt.
{lAl)

500
Chapter 7. Integration over submanifolds
f(t + dt) = f(t) + f'(t)dt + ---
f'(t)dt + ---
t+dt
Illustration for Example 7.4.1: Special case
Example 7.4.2. The length / of a circle in R^ of radius R islnR. Indeed, let
fit) = -s/R^ - f2. Then
(—t \^ t^ R^
And therefore
, dt = 1R I , dt = 1r\ arcsinr = InR. i^
Special case. Let n = 2. Many curves in R^ are described in polar coordinates
(r, a) for R^ by an equation of the form r = f((x), for example
r = R (circle); r = 2(1 +cosa;) (cardioid);
r = cos 2a; (rose with four petals).
Such curves therefore occur as images under mappings of the form (p : a h>
f (a) (cos a, sin a). If this is a C^ mapping, for A; > 1, then
r.j.^ s / f ((X)COSa - f(a)sma \ nr.,/ Ml /^. ., , J,,,
^'^("> = ( /'(«)sina + /(a)cosa)' mm = VfWTf^
r

7.4. Examples of Euclidean densities
501
Example 7.4.3. The length of a circle in R^ of radius i? is /^ Rda = 2nR; and
of the cardioid
2/ yj{I + cos a)'^-trsm^ a da =2^/2 I Vl + cos a da
J —JT J —TT
r a
= 41 cos —da = 16.
1^
Example 7.4.4 (Parametrization by arc length). Assume the C' embedding 0 :
] fl, Z> [ —> R" has an arc length of /. We then have the corresponding arc-length
function (compare (7.16))
X:]a,b[^]Q,l[ with X{t)
= f W
Ja
(y)\\dy.
Because X'(t) = \\<j)'(t)\\ > 0, the function A. is strictly monotonically increasing
on]a,b [; therefore A. has a differentiable inverse function A."' : ] 0, / [ —> ] a, Z> [.
If A.(f) = s, then by the chain rule
(x-'y(x(t))x'(t) = 1, that is, (x-'Yis) = 110'(Oir'.
Consequently, we find for the curve
f:]0,l[^R" with f(s) = (t>(X-'(s)) = (j>(t),
by means of the chain rule
\\r(s)\\ = \\ct>'(x-\sm \(x-'y(s)\ = wm wm-' = i.
That is, the curve i/f : ] 0, / [ —> R" is a C' parametrization ofV = (j)(]a,b[) with
the arc length as a parameter, and the tangent vector i/'(s) to V at i/(s) is always
of unit length. ik
n. Area. If rf = 2, then V is a C'^ surface in R", while, if we identify linear
operators and matrices, for y e D,
D<j>(y) = ( D,<j>(y) D2<j>(y) ) =
/ 901, , 9<^V X \
ayi dy2
e Lin(R^ R").
We obtain
DcPiy)' o Dct>(y) =
WDMHy) (D,,p, D2<P)(y)
{D2<P, Di<P)(y) \\D2MHy)

502
Chapter 7. Integration over submanifolds
and consequently
c^iy) = ^\\D,ct>f\\D2ct>f-{D,ct>, D2ct>)Hy) (y e D). (7.18)
If a is equal to the angle between the vectors Di<p{y) and D2<j)(y), we find
{Di<piy), D2(piy))
II n ^r Ml II n ^r mi = ^°^^' ^° ^-P^y^ = \\Di<p(y)\\ \\D2<p(y)\\ sma.
\\Di<j)(y)\\ \\D2<j)(y)\\
(7.19)
In this case the Euclidean density co is said to be the Euclidean area; accordingly, the
integration with respect to co is said to be the integration with respect to Euclidean
area. Thus, for / Riemann integrable over V,
f fix) d2X = fifo ct>)(yy\\DM^\\D2ct>\\^-{D,ct>, Dz^)^ (y) dy.
Jv Jd
In particular, the Euclidean area of V is defined as
d2X,
I
if the integral converges. Note that the area of V is now defined independently of
the parametrization of V.
<^{y) + dyi dy2 £>i0(>') x D24>{y)
(P(y)+dyiDicP(y)
(p(y) + dy2D2(p(y)
y2
y + (0, dy2)
y + (dy„0)
Illustration for 7.4.11. Area
Area of parallelogram spanned by dy\ Di<p{y) and dy2 D2<j)(y) equals
||(D,0 X D2ct>)(y)\\ dy, dy2 = ||(£>,0 x D2ct>)(y)\\ dy
Special case. In the case where n = 3, Formula (7.19) takes the form, see the
remark on linear algebra in Section 5.3,
My) = UDi,pxD2<P)(y)\\.
(7.20)

7.4. Examples of Euclidean densities
503
Example 7.4.5 (Torus). According to Example 5.7.2, the toroidal surface T C R^
has the parametrization <p '■ ] —^, ^ [ x ] ~^. ^[^ T, with
while
<p : (a, 9) i-> ((2 + cos 9) cos a, (2 + cos 9) sin a, sin 9),
90 90 / ^osacos9
— (a, 9) X —(a, 0) = (2 + COS0) I sin a cos 0
da 69 I • n
siny
Therefore, from (7.20),
co^{oi, 9) = 2 + COS0.
Consequently the Euclidean area of the torus equals
/
(2 + cos9)dad9
= f da f
J ~7T J — JT
2d9 = 271 -An = 871^
This result leads to the following remark. Intersect the toroidal surface with a
plane through the xs-axis. We slit the toroidal surface open along one of the circles
thus created. Next, we straighten the toroidal surface, in such a way that the central
circle retains its length, and we obtain the form of a right cylinder. This cylinder
then has a ruling of length Atz , and a perimeter of length 27r; therefore the Euclidean
area of the cylinder equals Stt^ (check this). Evidently, the total area is not altered
by this deformation from torus to straight cylinder.
Furthermore, the average of the Gaussian curvature K from Example 5.7.2 over
the torus V equals
If If cos9
/ K(x)dx = —- / (2 + cos9)dad9 = 0. i^
area
Illustration for Example 7.4.5: Torus
Example 7.4.6 (Cap of sphere and kissing number). Let V be that part of the
sphere {x e R^ | ||x|| = i? } lying inside the lateral surface of the cone given by
the equation X3 = Jx^ +^2 ^^^ ^' ^^^ 0 < i/f < |-. In spherical coordinates for
R3
X = r(cosacos9, sinacos0, sm9) (0 < r, —n <a<n, —— < ^ < —-),

504
Chapter 7. Integration over submanifolds
the set V is described as the image under the mapping
<p : (a, 6) i-> R{cosoicos6, sin a cos 0, sin0),
where —n <oi<7i and t/^ < ^ < f. Now
= i?(—sinacos^, cos a cos 0, 0),
= i?(—cosa sin0, — sin a sin 0, cos6),
= R^(cosacos^ 6, sinacos^^, cos0 sin0),
= i?^cos6l.
■
^{OL, 6)
da
— (a, 6)
d(p d(p
3<P 3<p
— X —(a, 6)
3a de
And so
are
a(V) = /
R^ cose da de
= R^ f da f'
J ~7T J ^
cos e de
= 2nR^i\ -sinf).
In particular, for i/f = 0, we find that the Euclidean area of a hemisphere equals
271 R^\ therefore that of the entire sphere equals An R^.
We now give an application of the preceding calculation. Consider two spheres,
say Si and S2, in R^ of radius 1 that have one point in common. The lines through
the center of Si that are tangent to S2 form a conical surface, and the maximum
angle between two of these tangent lines equals j. The solid angle subtended by
S2 from the center of Si therefore equals 2n{\ — ^v3). This implies that in R^ a
maximum of
An
14 <
n{2 - v^)
= 8 + 4V3 < 15
unit spheres, pairwise having at most one point in common, can be tangent to a unit
sphere.
The actual maximum number, the so-called kissing number, equals 12; a proof
of this is not altogether simple to give.' A configuration in which this number is
realized is obtained by taking the 12 vertices of a regular icosahedron (eI'xoolv =
twenty) of diameter 4 as the centers of spheres of radius 1. Then all of these 12
spheres are tangent to the sphere of radius 1 about the center of the icosahedron. Note
r20-3
that the (^ = 30) edges of this icosahedron are of length /8 - ^ = 2.1029 ■ • •.
In this configuration, therefore, the 12 spheres are not tangent to each other. ik
' See Chapter 12 in Aigner, M., Ziegler, G. M.: Proofs from THE BOOK. Springer-Verlag, Beriin
and Heidelberg 1998.

7.4. Examples of Euclidean densities
505
Example 7.4.7. Let V be the surface in R"* that occurs as the intersection of the
ellipsoid {x e R"* \x1-\-x\+xl + Sxj = 1},
with the
cone [x e R"* \ x^ + x^ + x^ — xj = 0}.
Subtracting the equations we see that x e V satisfies
1
X4 = ±-,
hence
2 . 2 I 2 _
x^ ~r X2 ~r -^3 —
1
It thus emerges that V is the disjoint union of two spheres (of dimension two). We
now have parametrizations
0± : ] -77:, 77: [ X J --, j[^ V'
1
(p±(oi, 6) = -(cosacosB, sinacos^, sin0, ±1).
Then, for <p = <p±.
d(j) 1
— (a, 6) = -(—sinacos^, cosacos^, 0, 0),
da 2
d<p 1
— (a, 6) = -(—cosasin^, — sin a sin 0, cos 6, 0),
d6 2
da
cos^9
dd) 2 1
— (a, 6)
de
(^(«,.),^(«,.)) =0.
Consequently, one finds from (7.18)
d(p
da
(ct,e)
d(j)
de
co^.^{a, 6) =
Therefore it follows that V has Euclidean area
cos 9
(ct,e)
cos 6
f cos6i, ,^ i r , f^
', I dade = - I da I
y[-jr,jr]x[-f,f ] 4 2 y_;r J^z.
COS ede = 2n.
1^
Example 7.4.8 (Newton's potential of a sphere). We employ the notation of the
preceding Example 7.4.6 and that of Example 6.6.7. The Newton potential <pA of
the sphere A = {x e R^ | ||x || = i? } is the zero function, because A is a negligible
set in R^. However, if we now define the Newton potential <pA as an integral with
respect to the Euclidean density coon A, then, for x ^ A,
1 /• 1
4-^ J A \\x-y\\

506 Chapter 7. Integration over submanifolds
In view of the rotational symmetry of A we may assume that x = (0, 0, a), with
a = \\x\\ ^ R. Upon introducing spherical coordinates we obtain, for y S A,
y = R(cosacos9, sin a cos 0, sin0).
Hence
\\y - x\\ = v/i?2cos2 6l + (i?sin6l-fl)2 = s/R^+ a^-laRsinO.
Therefore
R'^cosB
*« =i/!(/-!
_ ^R^ + a^-2aRsm0
R~
d0 da
2a R
-y/R^ + a^-2aRsme \' =—(-\R-a\ + R + a)
J-f 2a
r R
—{-(R-a) + R+a) = R (a < R);
2a
R /?^
—(-(a-R) + R+a)=— (a > R)-
2a a
with the result that
—R, if X inside A;
(Pa(x) = p2
, if X outside A.
Note that <pA can be continuously extended over A. ik
Example 7.4.9 (Geometrical interpretation of Gauss curvature). The notation
employed is that of Section 5.7. Let V be a C^ surface in R^. Let x e V be fixed
and let B,. be the closed ball in R^ about x and of radius r. Then, with K{x) the
Gauss curvature of V at x and n : V —> S^ the Gauss mapping, we have
area«(Vng,)
ATtx) = lim . (7.21)
rio area(VnBr)
Indeed, let <p : D,. ^ V r\ B,- be a parametrization of V (1 B,-. Note that, by
restriction, this gives a parametrization of V (1 Br', for r' < r. Then
no(p : D,. -^ niVnB,.)
is a parametrization of the image set n(V (1 B,.). The tangent space to V (1 B,. at
X = (j)(y) is spanned by the tangent vectors Di<p{y) and D2<j)(y), while the tangent
space to n(V n B,.) at n(x) = n o (j)(y) is spanned by (see Formula (5.12))
Dj(no<P)(y) = Dn(x)Dj<P(y) (1 <; < 2). (7.22)

7.4. Examples of Euclidean densities 507
If i4 e Aut(R^), one has for u,v,w € R^,
(Au X Av, Aw) = det(Au Av Aw) = det A det(u vw) = det A {u x v, w)
= det A{u X V, A~^Aw )=detA{ (A~^y(u x v). Aw).
That is
Aux Av = det A (A-y(u XV). (7.23)
Because we regard Dn (x) as element of End (T^ V) and because K (x) = det Dn (x),
>3 , T>3
it follows from (7.22) and (7.23), with A : R^ ^^ R^ defined hy A\j. y = Dn(x)
.± = I, that
Di(« o <p)iy) X D2in o <p)iy) = K{x) {D,4> x D2<p)iy).
and A\fj, y.± = I, that
We obtain
area«(VnB,) = f \K o<P(y)\\\(Di<p x D2<P)(y)\\dy, (7.24)
JDr
area(VnB,) = f \\iD„p x D2<p)iy)\\dy.
JDr
The assertion (7.21) now follows by application of an argument analogous to that
of the Mean Value Theorem of integral calculus to the integral in (7.24). ik
in. Hyperarea. (See also Example 5.3.11.) Let d = n — \, then V = im(0) is a
C'^ hypersurface in R". According to Formula (5.3) one has
detiD<P(yy o D<P(y)) = \\{D,(j> x • • • x D„^,(j>){y)f.
Therefore
o^4>iy) = WiD^cP X •. • X D„_i0)(>-)|| (>- e D). (7.25)
In the case where rf = 2 (and son = 3), note the agreement between Formulae (7.20)
and (7.25).
In the case d = n — \the Euclidean density co is said to be the Euclidean
hyperarea, and accordingly the integration with respect to co is said to be the integration
with respect to Euclidean hyperarea. Therefore, for / Riemann integrable over V,
I f(x)d„_,x = fifo <P)(y)UD,<p X ... X D„_,<P)(y)\\ dy.
Jv Jd
In particular, the Euclidean hyperarea of V is defined as
hyperarea(V) = / rf„_ix,
Jv
if the integral converges. Note that the hyperarea of V is now defined independently
of the parametrization of V.

508
Chapter 7. Integration over submanifolds
Special case. Let D C R" ' be an open set and let 4>{y) = (y, h(y)), for a C''
function /?, : D —> R. One then has, on account of Formula (5.7), for every y € D,
co^y) = ^l + \\Dh(y)p.
Note that the special case in (7.17) follows from this equation.
1
We have cos f = — „^, , .,,T,,n-
(l + ||D/7,(>')|p)V2
<P(y) + (0,l)
ct>(y) + (-Dh(y)A)
yn
yi
For the hyperarea cd^ (y)dyi ■ ■ ■ dy„_, of an element of the hypersurface
y,, — h(y) =0 over a rectangle in R"~' with vertex y and edges of lengths
dyu ..., dyn-i, one has approximately
dyx--- dy„^
cos i/f; hence co^(y) =
oi,l,iy) dyi ■ ■ ■ dy„^i ^ cos^
Illustration for 7.4.111. Hyperarea
= {\^\\Dh(y)tyl^

7.4. Examples of Euclidean densities 509
Example 7.4.10 (Area of two-sphere). The unit sphere S^ in R^ is the union of
two surfaces, determined as indicated above by the functions /2± : B^ i-> R, with
B^ the closed unit disk in R^ and
h±{y) = ±.yi - \\y\\\
We have
Dh^{y) = --\-y, ^l + \\Dh^{yW= ^
h±(yy
Vi - \\yP
The two surfaces have the one-dimensional equator in common; accordingly, for
the two-dimensional calculation of the area this is negligible. Consequently, by
Example 6.6.4,
area(S^) = 2 / , ^V = 2 / dcp f
Jb^ Jl - llylp J-n Jo
VF^
dr = An.
1^
r cos B da
Illustration for Example 7.4.11
The shaded volume approximately equals r cos 9 da r d9 dr = r^ cos 6 dr da d6

510 Chapter?. Integration over submanifolds
Example 7.4.11 (Hyperarea of three-sphere). The unit sphere S^ in R'' is the union
oftwohypersurfaces, determined as indicated above by the functions/2± : B^ —> R,
with B^ the closed unit ball in R^ and
h±{y) = ±Vl - llylP.
The two surfaces have the two-dimensional equator in common; for the three-
dimensional calculation of the hyperarea this is negligible. Therefore we now have,
as in Example 7.4.10,
hyperarea(S^) = 2 / , dy.
Parametrize B^ with
^ : (r,a,9) i-> r (cos a cos 0, sin a cos 0, sin0),
71 71
(0 < r < 1, —TT < a < 71, —— < 6 < —).
Then
(cos a cos 9 — sin a — cos a sin 0 \ / 1 0 0
sin a cos 6 cos a — sin a sin 0 I I 0 r cos 6 0
sine 0 cos^ / \ 0 0 r
Hence det D^(r, a, 9) = r^cosB. By the Change of Variables Theorem 6.6.1,
therefore
hyperarea(S^) =2/ da \ ' cos6d6 \ dr
J-. 7-1 Jo VT^
= 477: [sin61]^, / ( , - ^1 - rZ ) dr
= 871-
arc sin r r
yfl-r^
_ 2
2^' ■
1
= A7Z arcsin 1 = 27z^,
where use has been made of the computation of the antiderivative as in
Example 6.5.2. i^
Example 7.4.12 (Spherical coordinates in R"). Let S""' = {x e R" | ||x|| = 1}
be the (n — 1)-dimensional unit sphere in R". Assume that 0 : D —> S"~' with
D C R"~' open is a C' parametrization of an open part of S"~' having negligible
complement (see Exercise 7.21.(vii) for explicit formulae). Then the mapping
* : R+ X D ^^ R" given by *(r, y) = r(p(y)

7.5. Open sets at one side of their boundary 511
is a C' diffeomorphism onto an open dense subset in R", with
D<^{r,y) = {(j>{y)\rD(j>{y)).
Therefore we find
|detDvI/(r,3;)| =r"-'|det(0W | D(j>{y))\= r"-'\AQ\~D(j>{y)l
in the notation of Formula (7.14), because <p{y) is a unit vector perpendicular to
I>0.)5"'~'- But then, by Formula (7.15)
\A^\.D^{r,y)\=r"-'oi^{y).
Hence it follows, from the Change of Variables Theorem 6.6.1 and Corollary 6.4.3,
that for every / e Cc(R") (compare with Example 6.6.4),
I f(x)dx = f r"-' I f(ry)d„^,ydr
(7.26)
= f f r"-'f{ry)drd„^,y. ^
7.5 Open sets at one side of their boundary
The Fundamental Theorem of Integral Calculus 2.10.1 on R asserts that, for C'
functions / on [a, Z>] C R,
/
'' df
a dX
This establishes a relation between the integral over the open set ]a,b[ of the
derivative of a function / on the one hand, and the values of that / at the boundary
points b and a on the other. It is a characteristic feature of this relation that ]a,b[
is bounded by its boundary points a and b, and that the boundary points b and a
are counted as positive and negative, respectively. In Section 7.6 we formulate an
analog for R" of the Fundamental Theorem, applicable for suitable open sets Q in
R". Admissible sets in that respect are bounded open sets Q which are bounded by
their boundary dQ (see Definition 7.5.2), and whose boundary 9S^ is a C' manifold.
At this point, recall that according to Formula (1.4) one has dQ = S^\S^, ifS^is
open. Undesirable situations are outlined below. Note that in situation IV a point
x° e dQ possesses an open neighborhood U in R" with
UnQ = U\dQ. (7.27)

512
Chapter 7. Integration over submanifolds
0<0
Q
Q
d£i
Q
I: 9 fi note' II: 9 S^ image under HI: S^ unboundedIV: S^ not bounded
immersion, not by 9S^
embedding
Notation. Let Q. C R" be an open set that satisfies, in the terminology of
Example 5.3.11, the requirement
9fi is a C* hypersurface in R" (Jfc > 1).
According to the theory in that example, ^Q. can locally be described as follows.
For every x° e 9S^ there exist an open neighborhood U of x° in R", an open subset
D C R"~', and a C*" embedding 4> : D ^ W with
d^r\U = imi^) = {^iy) \y€D}.
(7.28)
For X = <p{y) e ^Q.r\U the column vectors Dj<p{y) in the matrix D<p{y), for 1 <
j <n — l,form a basis for rv(9^), the tangent space to 9^ atx. Letx° = (j)(y°),
and let v e R" with v ^ T^o(dQ) be chosen arbitrarily, then det {v \ D<p{y°)) ^ 0
because of the linear independence of the occurring vectors. Now define
One then has
and so
*:RxD^^R" by <if{t,y) = tv+ (j>{y). (7.29)
Dvl/(f, y) = {v\ D(j>{y)) ((t, y) e R x D), (7.30)
det D*(f, y°) ^0 (re R). (7.31)
The Local Inverse Function Theorem 3.2.4 asserts that we can find ^ > 0 and that
we can shrink D and U if necessary, in such a way that
^ : ] —^, ^ [ X D —> f/ is a C* diffeomorphism of open sets.
(7.32)
Note that dQr\U C *({0} x D); on the other hand tv + (p(y) = (p(y') edQr\U,
with 1^1 < ^ and y, y' e D, implies f = 0 and y = y'. As a result we have
dQnU = <i/({0} X D).
(7.33)
That is, dQ has now been flattened locally. Indeed, (7.33) asserts that in (t, y)-
coordinates dQnU is given by the condition t = 0.

7.5. Open sets at one side of their boundary 513
Also, according to Example 5.3.11, there exists a C* function g : f/ —> R with
Dg 7^ 0 on f/ and
^^f^U = Nig) = {x€U \gix) = 0}. (7.34)
Theorem 7.5.1. For an open subset Q. C R" the following three assertions are
equivalent.
(i) ^Q.is a C^ hypersurface and for every x° e dQ there does not exist an open
neighborhood U of x° in R" such that U H Q = U \ dQ (compare with
(7.27)).
(ii) For every x° e dQ and for every v e R" with v ^ 7^,.o(9S^), there exist U, ^,
S and D as in notations (7.28), (7.29) and (7.32) with
QnU = <ifQ-S, 0[x D) = {tv + <p(y) \-S <t <0, y € D} or
QnU = <ifQO, S[x D).
(iii) For every x° e dQ there exist U and g as in notation (7.34), for which
QnU = {x eU \g(x)<0}.
Proof, (i) ^ (ii). Let U be as in (7.32) and write
Uo = {^(0,y)\y€D}, U± = {^(t,y)\\t\ <S,t^O,y€D}.
From (7.33) we know already that dQ HU = Uq; and because Q (1 dQ = 0, it now
follows that ^nU = i^nU+)Ui^nU^). We will prove that either ^(MJ = U+
or S^ n f/ = f/_.
We may assume that D is a convex set in R"~' (this may require U to be
shrunk). Because x° e Q one has QHU ^&; and this leads to either QnU+ ^ &
or S^ n f/_ ^ 0. We now prove
QnU±^& =;> QnU± = U±. (7.35)
For the proof, let x = *(r, 3?) e S^ (1 U+, and assume x' = *(f', y') e U+ is
chosen arbitrarily; then f > 0 and t' > 0. Next define, for s e [ 0, 1 ], the vector
x(s) e f/ by
x(s) = *((1 - s)t + St', (1 - ^)>' + sy').

514 Chapter?. Integration over submanifolds
In particular, then, x(0) = x and x{\) = x'. Because t > 0 and t' > 0, it follows
that
{\-s)t + st'>0 (^£[0,1]); (7.36)
which implies x{s) e U+, for ^ e [ 0, 1 ]. If x' ^ ^, the point xCcr) e U+ with
a = swp{s e [0, 1] I x(s) e Q}
would be contained in dQ (1 U. But in view of (7.33) and (7.36) this forms a
contradiction, and so (7.35) does indeed follow. If the two sets Qr\U_^ and Qr\U^
are nonempty, we obtain
ur\Q = u+uu^ = u\d^-
but this was ruled out in (i). As a result, one has either Qr\ U = f/_|_ or f/_; and
this is precisely (ii).
(ii) ^ (iii). Define g = p^ o ^~\ where pi : x i-> Xi : R" —> R is the projection
onto the first coordinate. Then g(x) = t if x = ^(f, y), and by Formula (7.31) it
follows that
Dg(x°) = p,o(D^(t,y°)r'y^O.
Therefore g is a submersion in x°.
(m)^(i). Lett/ C R" be as in (7.34). Consider the curve/ : t i-> x°+r gradg(x°)
in R". We then have in view of Formula (2.18)
g o y(0) = 0, (go yYiO) = II gradg(x°)f > 0
Hence )/(f) e U and g(y(t)) > 0, forO < ^ < ^, where ^ > 0 should be sufficiently
small; and consequently y(t) ^ QU dQ; but this implies
y(t)eU\dQ and yit)^Uri^,
thatis, f/\aS^ y^U n^.

7.5. Open sets at one side of their boundary
515
Definition 7.5.2. We say that the open set ^ C R" at the point x° e 9S^ has a C''
boundary dQ and lies at one side of dQ, if any one of the equivalent properties (i)
- (iii) in the preceding theorem is satisfied. If this is the case for every x° e dQ,
we say that Q has a C'^ boundary ^Q. and lies at one side of dQ. O
Now let X e dQhe such a boundary point, let v e R" with v ^ Tx(dQ). In
terms of the function g in Theorem 7.5.1.(iii), there are two possibilities:
(i) (gradgix), v) > 0 or (ii) (giadgix), v) < 0.
As in part (iii) ^ (i) in the proof of the theorem, we see that in case (i) there exists,
for every differentiable curve y in R" with y(0) = x and Dy(0) = v,aS > 0 such
that
(i)' y(t) eQ (S <t < 0), y(t) iQ. (0 < t < S);
in other words, y leaves Q via the boundary. In case (ii) we have instead of (i)'
(ii)' y(t) iQ. (S <t < 0), y(t) eQ (0 < t < S);
that is, y enters the set Q via the boundary. This makes conditions (i) and (ii)
independent of the choice of the function g in Theorem 7.5.1 .(iii).
We recall the definition from Example 5.3.11 of a normal n(x) to dQ at x
n(x) _L T,(dQ) and ||n(x)|| = 1.
Illustration for Definition 7.5.3: Outer and inner normals
Definition 7.5.3. Assume the open set Q has a C^ boundary dQ and that Q lies at
one side of dQ. If x e dQ, v e R", v ^ Tx(dQ), we say that v points outward
from Q if condition (i) or (i)' above is met; or that it points inward in Q if (ii) or
(ii)' above holds. In particular, the outward-pointing normal to 9 S^ at x, notation
v(x), is said to be the outer normal to dQ at x; the other normal, —v(x), is said to
be the inner normal to 9S^ at x. Thus, these normals are defined independently of
the choice of the function g in Theorem 7.5.1 .(iii). O

516 Chapter?. Integration over submanifolds
Lemma 7.5.4. Let Q. C R" be an open set having a C^ boundary 9^ and lying at
one side ofd^. Let x e 9^ and let v e R" with v ^ r,.(9fi) be pointing outward
from Q.. Condition (ii) in Theorem 7.5.1 then takes the following form:
QnU = <ifQ-S,0[x D) = {tv + <p(y) \-S < t <0, y € D}.
The outer normal v(x) to d^ at <j)(y) = x, with the substitution of —yi for the
variable yi when applicable, is given by
KHy)) = \\{D,(j> X .. • X D,_,(j>){y)r\D,(j> X .. • X D,_,(j>){y) e R". (7.37)
In what follows it will invariably be assumed that the outer normal is given by
Formula (7.37). Furthermore, then
det D^{t, y) = {v, {D,4> x • • • x D„_,4>){y)) > 0 {{t, 3;) e R x D). (7.38)
In the special case ofx = <p{y) = {y, h (y)), for a C^ function /z : D —> R, one has
v(x) = (-1)"-'(1 + \\Dh(y)fr'^\-Dh(y), 1) e R". (7.39)
Ifd^ is locally described as in Theorem 7.5.1.(Hi), then
v(x) = II grad g(x) r' grad g(x) e R". (7.40)
Proof. By Formula (5.5), Equation (7.37) is correct to within its sign. When this
cross product yields the inner normal, we replace the variable yi in the embedding
<P by —y\, as a result of which the cross product changes its sign. Formula (7.38)
follows from (7.30). Further, (7.39) is derived from (5.6). And finally, (7.40) follows
in a straightforward manner from Example 5.3.11, plus the fact that the direction of
grad g(x) is the direction in which, starting at x, the function g increases the most
rapidly, see Theorem 2.6.3.(i). □
Remark. In the considerations above, gradg(x), for x e dQ (1 U, plays an
important role. Note that Exercise 4.32 implies that, for all x e dQ (1U, the vector
gTadg(x), to within a scalar factor f(x) ^ 0, is uniquely determined by the set
9fi n f/ = {x e f/ I g(x) = 0}. Since the function / is continuous on dQ (1 U,
it further follows that the sign of / is constant on the connected components of
dQnu.
Remark on the description of a boundary. Given a point x° in an (n — 1)-
dimensional C* hypersurface 9 S^ in R", there are various customary ways to describe
a full neighborhood of x° in R".

7.5. Open sets at one side of their boundary 517
(i) Describe ^Q. near x° as the image under a C'^ embedding <p, and move away
from ^Q. following a fixed direction v e R" which at x° is transverse to 9S^
(this is the situation described in Lemma 7.5.4), that is
X = *(f, y) = tv +(p(y) with V ^ T^o(dQ).
(ii) Describe dQ near x° as the image under a C^ embedding (p, and move away
orthogonally with respect to dQ, that is (see Exercise 5.11)
X = ^(t, y) = tv{<j)(y)) + <j)(y).
(iii) Describe dQ near x° as the image under a C^ embedding (p, and move away
from 9S^ by following a transverse C^ flow, that is (see Section 5.9)
9<1>
X = ^(t, y) = ^'(,p(y)) with ^(0,x°) ^ r,o(9S^).
at
The situation in (i) is a special case of this.
(iv) Describe dQ near x° as zero-set of a C^ submersion g, and apply the
Submersion Theorem 4.5.2.(iii); this gives
g(x)=t ^:^ x = ^(t,y).
In all of these cases ^ : R x R"~' D-> R" is a C^ diffeomorphism (except in (ii),
where one merely obtains C^~^). In case (iii) the proof is obtained, as for (i), by
using the Local Inverse Function Theorem 3.2.4.
Example 7.5.5. Assume a, b and c > 0, and let V be the ellipsoid in R^
2 2 2
V = {X e r3 I g(x) = ^ + ^ + ^ - 1 = 0}.
a/^ b^ c^
Let (i(x) be the distance in R^ from 0 to the geometric tangent plane x + TxV io V
atx. Then
J- = ^(gradg(x), v{x)) {X e V). (7.41)
d{x) 2
Indeed, Dg{x) = 2{^, %, ^), and so
1 X,h\ Xnhn X-ih-i
hsT^V ^-^ -(gradg(x), /7,) = -^ + ^ + ^ = 0.
2 a/^ b^ c^
Because k e x + Tj. V if and only \fk — x e Tv V, it follows that
„, x\ki xiki xsks
x + T,V = {keR'\^ + ^ + ^ = l}. (7.42)
a/^ b^ c^

518 Chapter?. Integration over submanifolds
Since v{x) is perpendicular to Tx V and has unit length, d{x) is determined by the
requirement
dix)v{x) €x + T^V.
One has
, ^ 1 /X\ X2 xi\ . , , ^ Ixl; Xo xl
Therefore, according to (7.42),
,, ^ , ^ d(x) /xf Xy x^\
d(x)c(x) = ^-^(4 + 4 + ^) = 1-
c(x)\fl4 M cV
Consequently, (7.41) results from
_i . ._ . s2_!i //fi ^ ^\ J_/fi :^ i^U ^v
d(x) -'^W-C(X) ^^^^ -Ua2' ^2' c2J' c(x)lfl2' ^2' c2J/-
7.6 Integration of a total derivative
Now we are prepared enough to prove the following:
Theorem 7.6.1 (Integration of a total derivative). Let Q C R" be a bounded
open subset having a C' boundary dQ and lying at one side ofdQ. Let v(yy be
the outer normal to dQ at y e dQ considered as a row vector, and let d„^iy be
integration with respect to the Euclidean (n — \)-dimensional density on the C'
hypersurface dQ. Let f : Q —> R be a C^ function such that f and its total
derivative Df : Q —> Lin(R",R) ~ R" can both be extended to continuous
mappings on Q. Then the following identity of row vectors holds in R", where the
integration is performed by components:
f Df(x) dx= f f(y) v(yy d„^,y. (7.43)
Jn Jan
Remarks. According to Corollary 6.3.8, the boundary dQ is an n-dimensional
negligible set in R", and therefore, by Theorem 6.3.2, Q is Jordan measurable.
Hence, the integral on the left-hand side in Formula (7.43) is well-defined.
By taking the j-th component in (7.43), for 1 < J < n, we find
f Djfix) dx= f f(y) vj (y) d,,^,y. (7.44)
Jn Jan
Further, Formula (7.43) is equivalent to the assertion
f Df(x)v dx= f f(y) { v(y), v ) d„^,y {v e R"), (7.45)
Jn Jan

7.6. Integration of a total derivative 519
and also to
giad f(x)dx= f(y)v(y)d„-iy.
Jn Jan
'S2 Jd^
Let n = 1 and Q = ]a,b[. Then 3^^ = {a,b}, v(a) = -1, v(b) = +1.
Integrating a function over a point (a zero-dimensional manifold) with respect to
the Euclidean zero-dimensional density is, by definition, tantamount to evaluating
the function at that point. Therefore the Fundamental Theorem of Integral Calculus
on R is a special case of (7.43),
f ^(x)dx=f f(y)(-l)day+ f f(y)(+l)doy = f(b)-f(a).
Ja "-^ J{a] J{b]
Proof. We first demonstrate that, for every point x e Q = S^U9S^, we can find
an open neighborhood Ux of x in R" such that Formula (7.45) holds for functions
/ as specified in the theorem, if moreover these functions satisfy supp(/) C f/.v
Since both sides of Equation (7.45) depend linearly on i; e R", it suffices to prove
that formula for vectors v belonging to a basis for R". Note that the choice of that
basis may depend on the point x considered.
Case I. Assume x e Q. Because Q is an open set in R", we can find vectors a
and b € Q such that x € Ux := [y € R" \ aj < yj < bj (1 < 7 < n)} C fi;
and therefore Ux H dQ = 0. We now successively choose v = ej, the standard
j-th basis vector in R", for 1 < j < n. For a function / as in the theorem, also
satisfying
supp(/) C Ux C fi, (7.46)
Corollary 6.4.3 gives
/ Djf(x)dx= / Djf(x)dx
J^ Jr"
= I ■ ■ ■ I I Djf(x)dxj dxi ■ ■ ■ dxj_i dxj_^_i ■ ■ ■ dx„.
On account of (7.46) one has, for fixed (xi,..., Xj_i, Xj+i,..., x„),
J {Xi,..., Xj_i, Xj, Xj_^_l,...,x„)^{J ^^ aj < Xj < bj.
(7.47)
Therefore, application of the Fundamental Theorem of Integral Calculus on R to
the function Xj i-> /(x,,..., Xj_i, Xj, Xj_^_l,..., x„), defined on [a^, bj], gives
/
j J V 15 ■ ■ ■ 5 J' ■ ■ ■ J -^n) "■■^j
= f(xi,..., bj,..., x„) - f(xi,..., aj,..., x„) =0-0 = 0.

520 Chapter 7. Integration over submanifolds
Consequently, (7.47) implies
Djf(x)dx =0.
/
On the other hand, / = 0 on 9^ under the assumption (7.46), and therefore one
also has
Jan
which proves (7.44) in this case.
Case n. Assume x e dQ. In this case we select each of the basis vectors v
such that V is not included in Tx(d^) and points outward from Q. This we do by
choosing a basis (hi,..., /7„_i) for the linear subspace Tx(dQ) of dimension n — 1,
and vq ^ Tx(dQ). The vectors vi = hi + vq, ..., i;„_i = /7,„_i + vq, v„ = vq then
form a basis of R". They are not contained in T^(dQ), and after changing over to
their opposites if necessary, we may assume that they point outwards from Q. Now
successively consider v = Vj, for I < j < n. By Lemma 7.5.4, the pair (x, v)
defines an open neighborhood U = U(v) ofx;let^ ■]—S, 0[xD—> QHU with
\i/(t,y) = fi;+0()') be the corresponding C' substitution of variables. Application
of the Change of Variables Theorem 6.6.1 to a function / as in the theorem, with
/ moreover satisfying supp(/) C ^ n f/, gives
I Df(x)vdx= I Df{<i/(t,y))v\detD<i/(t,y)\dtdy. (7.48)
Jn J]-5,0[xD
Now, by the chain rule and Formula (7.29),
9* 9(/o*)
D/(vI/(r, y))v = Df(^(t, y))—(t, y) = ''. \t, y). (7.49)
dt at
And by means of (7.38) we find
I det D^it, y)\ = (V, {Di4> x • • • x D„^,4>){y)) {{t, 3;) e R x D); (7.50)
note that the expression on the right-hand side is independent of ^. Then use
Corollary 6.4.3, (7.49) and (7.50), subsequently the Fundamental Theorem of Integral
Calculus on [ —^, 0 ], plus the fact that (/ o ^) (—^, y) = 0, to write the right-hand
side in (7.48) as
LL
° 9(/°*)
-{t, y) dt { V, {D,4> X ... X A,-i0)(>')) dy
at
if o vl')(0, y) {{D,4> X ... X D„^,4>){y), v) dy
-L
= if o <j))(y){ (v o (j))(y), V ) co^iy) dy = f(y) { v(y), v ) d„^iy.
Jd Jan
(7.51)

7.6. Integration of a total derivative 521
The penultimate identity follows because of Formulae (7.37) and (7.25).
Combination of (7.48) and (7.51) now gives (7.45). By intersecting the neighborhoods
U(Vj) of X, for 1 < 7 < n, we find the desired open neighborhood U^ of x in R".
End of Proof. Q is compact, according to the Heine-Borel Theorem 1.8.17. By
virtue of Theorem 6.7.4 there exists a C' partition {xi I i € / } of unity on Q
subordinate to the open covering {Ux \ x e S^} of Q. Because Formula (7.43)
has been proved above, for / replaced by Xi f, for every i e I, summation over
i e / gives Formula (7.43) for /, since X!,g/ Xi = 1 on S^ and therefore also
D(j:,.^jXif) = DfonQ. ' □
Corollary 7.6.2 (Integration by parts in R"). Assume Q and f to be as in the
theorem above, and assume g satisfies the same conditions as f. Then, for I < j <
n,
f Djf(x)g(x)dx= I (fg)(y)vj(y)d„^,y- f f(x)Djg(x)dx.
Jn Jan Jn
Proof. Replace / by fg in Formula (7.44) and apply Leibniz' formula Dj(f g) =
gDjf + fDjg. ' □
■I- i
*
Illustration for Remark in Section 7.6

522 Chapter 7. Integration over submanifolds
Remark. It is also possible to prove Theorem 7.6.1 starting from
/ Df(x)vdx = / lim-(f(x + tv)- f(x))dx
J SI J SI '^° t
= lim-(/ f(x + tv)dx— I f(x)dx)
'^° t y^i' J SI '
= lim-( / f(x)dx— I f(x)dx].
Here v e R" is a fixed vector and
fi + ni = [x + tv \x € ^}.
Where Q+tv and Q overlap, the integrals cancel each other; what remains,
therefore, is an integral over a strip along the boundary. To first-order approximation in t,
the thickness of this strip at a point y e dQ equals t {v(y),v), where {v(y),v) < 0
corresponds to a region where the strip is to be counted negative. If, finally, it can
be shown that the integral over a strip along the boundary, of thickness S, to first-
order approximation (for ^ s|, 0) equals S times the integral over the boundary with
respect to the Euclidean density, the proof will be completed (see the Motivation in
Section 7.3, and Exercise 7.35.(iii)). The advantage of this proof is that it is more
transparent geometrically, and furthermore that it provides an interesting additional
interpretation of integration over an (n — l)-dimensional manifold with respect to
the Euclidean density. The price to be paid is, of course, that at several stages limit
arguments will have to be given, in which uniform convergence may be expected
to play an important role. A fully detailed proof along these lines will therefore be
of considerable length.
In our development of the theory the formula below will serve as justification
for this line of reasoning. Let the notation be that of Example 6.6.9. Let v e R",
and choose ^'(x) = x + tv, for (t, x) e R x R"; then i/f (x) = v and div i/f (x) = 0,
for X e R". By combining the transport equation (6.22) and Theorem 7.6.1 we find
d_
It
f f(x)dx= f Df(x)vdx= f f(y)(v(y),v)d„_iy. (7.52)
r=0 Jsi+rv J SI JdSl
7.7 Generalizations of the preceding theorem
In many of the applications envisaged, the conditions of Theorem 7.6.1, such as
those relating to continuity of / and Df on Q, or those relating to smoothness of
the boundary dQ, are not fully met.
The assumption concerning the continuity of Df on Q may be relaxed. It
is sufficient to assume that / is continuous on Q, and that x i-> Df(x)v, with
V e R", is continuous on Q and absolutely Riemann integrable over Q. Indeed, in

7.7. Generalizations of the preceding theorem 523
Formula (7.51)
may be replaced by
I
L
° 9(/ o *)
dt
The arguments from the proof can be applied to this integral, and in the resulting
identity we finally take the limit for e s|, 0. This leads to (7.43), without the
continuity of Df up to and including the boundary having been needed.
In addition, the assumption that 9^ is a C' hypersurface may be relaxed; it
is sufficient that the nondifferentiability of 9S^ be localized in a "relatively small"
subset S of 9S^. Begin by assuming that S is a closed (and hence compact) subset
of 9S^ such that
is in fact an (n — l)-dimensional C' manifold, with Q. at one side of W at each
point of W. The idea is to approximate / by functions whose support is disjoint
from S, because Theorem 7.6.1 does apply to such functions. In order to limit
the number of technical complications, we assume that / and Djf can both be
extended to continuous functions on Q., for 1 < _/ < n. Now assume the subset S
of 9^ satisfies the following condition:
for all e > 0 and for every open neighborhood f/of S in R"
there exist an open neighborhood f/' of S in R" and x £ C'(R") with
f/' C f/, 0 < X < 1, X = 1 on f/', supp(x) C U,
I XM dx < €, I II grad xMW dx < e.
(7.53)
Because supp ((1 — x) /) H S = 0, one has, by Theorem 7.6.1,
f Dj((l - X) f)(x)dx =/"((!- X) / ^j)(y)d„^iy. (7.54)
Let II ■ II be the uniform norm on the linear space of bounded functions on Q, with
\\g\\=sup{\g(x)\\xeQ},
then, by Leibniz' rule,
I f Dj(xf)(x)dx < f \(Djx)(x)\ \f(x)\dx + f x(x) \(Djf)(x)\dx
'Jn Jn Jn
< 11/11 f \(Djx)(x)\dx+\\Djf\\ f x(x)dx
<e (11/11+ l|£',/ll).

524 Chapter 7. Integration over submanifolds
As a result, the left-hand side in (7.54) converges to f^(Djf)(x) dx, if e ], 0; and
the right-hand side in (7.54) converges to /w(/ Vj)(y) di,^iy, if U shrinks to S.
We now wish to replace condition (7.53) on S by one that is more easily verified
in practice. This problem has a new aspect in that we now also need to control the
magnitude of the partial derivatives of the bump function x. and the support of x
as well. A number of pertinent remarks follow; in fact, these are part of a theory of
simultaneous uniform approximation of a function and of its derivatives (see also
Exercise 6.103).
Choose a C' function i/f e Cc(R") (see Definition 6.3.6) such that
f>0; fix)=0 (||x|| > 1); I f{x)dx=\.
Next define, for t > 0, the function fr : R" ^^ R by f,(x) = t~" f(t~^x). Then
f^(x) = 0 (||x||>f); f f,i.x)dx=( f{t~^x)d{t~^x) = \.
Jr" Jr"
Subsequently define, for every g e C(R") and f > 0, the function gr : R" —> R by
(compare with Example 6.11.5 on convolution)
gri-x) = g* f,{x) = g(x- y) i/r(y) dy=\ ft{x - y) g{y) dy. (7.55)
JR" Jr-
Note that the integration in (7.55) is in fact over a subset of supp(T/fr); that is why
gr is well-defined. It follows immediately that
\grM\< f \g(x-y)\tr(y)dy<\\g\\ I f,{y)dy=\\g\\ (xeR");
that is
ll^rll < \\g\\ {t > 0). (7.56)
Using the Differentiation Theorem 2.10.4 or 6.12.4 and Theorem 2.3.4 one proves
that g, is a C' function on R", with
{D^g,){x)= f (Djf,)(x-y)g(y)dy.
Jr"
Because {Djft){x) = t~" Dj{x i-> f{t~^x)) = t~"~^{Djf){t~^x), this leads to
\{Djg,){x)\ < f-"-' f \{Djf){t-\x - y))\ \g(y)\dy
Jr"
<r'\\g\\ f \(Djt)(t~\x-y))\d(t~'y)
Jr-
<r'\\g\\ f \\giadt(y)\\dy=:t~'\\g\\k.
Jr"

7.7. Generalizations of the preceding theorem 525
Therefore
WJ^M < r' \\g\\ k (1 <;•<«, r > 0). (7.57)
We now recall the sets S^ = {y £ R" | there exists x e S with Hx — 3?|| < ^ },
for ^ > 0, defined in Lemma 6.8.1. Again, every Sb is compact in R". Suppose U
is an open neighborhood of S in R", then by Lemma 6.8.1 there exists a ^ > 0 such
that S2S C U. Hence S C S^ C int(S25) C U, where int(S25) is an open covering
of Ss. Applying Theorem 6.7.3 we find a continuous function Xs : R" —> R with
0 < Xa < 1, xa = 1 on Ss, suppCxs) C Sib- (7.58)
Next we define, for 0 < f < ^ (compare with (7.55)),
XsA^) ■■= iXsUx) = f My)X5(x - y)dy (x e R").
Now let X e R" with Xs.ti-'^) ¥" 0. This can only occur if there exists y e R" with
ft(y) > 0, Xs(x -y)>0- that is, \\y\\ <t, x-y € Sis-
Consequently, there exists z ^ S with ||x — y — z|| < 2S, and so
\\x -z\\ = Ux-y-z) + y\\ <\\x-y-z\\ + \\y\\ <2S + t.
That is
supp(x5,r) C S2s+r; in particular, supp(DjX5,r) C Sis+r (1 < 7 < «)■
(7.59)
From (7.59), (7.57) and (7.58) it then follows, for I < j < n.
L
outer measure(S25+r)
.(.DjXsj)(.x)\dx < ||DjX5,rll-outermeasure(S25+r) <^ .
R« t
Setting ^ = I one obtains
f outer measure(S's^/9)
/ \(DjX5,5/2)(x)\dx<5k '''^\ (7.60)
Moreover, from (7.56), (7.58) and (7.59)
Xs,s/2(x)dx < outer measure(555/2). (7.61)
L
Definition 7.7.1. A compact subset S of R" is said to be (n — \)-dimensional
negligible if
outer measure(Sx) _
lim ;; -^ = 0. O
54,0 S

526 Chapter 7. Integration over submanifolds
One now obtains from (7.60) and (7.61):
Lemma 7.7.2. Let S C R" be compact in R" and (n — l)-dimensional negligible.
Then S satisfies (7.53).
One readily verifies that Si U S2 is an (n — l)-dimensional negligible set if Si and
S2 are; and further, that a compact subset S of an (n — 2)-dimensional manifold in R"
is an (n — l)-dimensional negligible set. Indeed, on the strength of Theorem 4.7.1
one has, locally at least,
S C {(hi(y), h2(y), y)€R"\y€Dc R""' open, /z,- e C(D, R) (1 < / < 2)}.
Because
\\(xi,X2,y)-(x[,x!^,y')\\ <S =^ |x,-x;|<^, ||>'-yil<^,
one certainly has
Ss C {(xi, X2, y) e R" I \xi - hiiy)\ < S (l < i <2), y € Ds}.
This implies
outer measure(S5) < / dy I dxi I dx2
Jds Jh,(y)-S Jh2(y)-S
= 4S^vol„^2(D5) = 0(S^), no.
As a result
outer measure(S5)
o
This also makes the finite union of compact subsets of (n — 2)-dimensional C'
manifolds in R" an (n — l)-dimensional negligible set, for example, the edges of a
cube in R^.
The following generalization of Theorem 7.6.1 has now been proved. Note that
on the strength of this generalization (the interior of) all known figures from the
box of bricks (rectangle, part of cylinder, part of cone, bridge etc.) qualify as Q.
Theorem 7.7.3. Let Q C R" be a bounded open subset with boundary dQ. Let
S be a closed subset ofdQ such that S is an (n — \)-dimensional negligible set,
d'Q = dQ\S aC^ manifold in R" of dimension n — 1, and Q lies at one side ofd'Q
at each point of d'Q. Let f : Q —> R be a C^ function such that f can be extended
to a continuous function on Q, and its total derivative Df : Q —> Lin(R", R) ~ R"
is continuous and absolutely Riemann integrable over Q. Let v and d„.-iy be as in
Theorem 7.6.1, but now defined with respect to d'Q. Then
f Df(x)dx= I f{y)v{y)'d„-iy

7.8. Gauss'Divergence Theorem 527
7.8 Gauss' Divergence Theorem
In vector analysis one applies variants of Theorems 7.6.1 and 7.7.3 to vector-valued
functions. Hence the following:
Definition 7.8.1. Let U be an open subset of R" and let f = (fi,..., f„) : U ^ R"
be a mapping. Such a mapping is often referred to as a vector field on U, particularly
when / is regarded as a mapping which assigns to x e f/ a tangent vector /(x) in
the tangent space TfU to U at x, where, for all x e f/, the latter space is identified
with R". In other words, the image vector f(x) e R" is seen as an arrow in R",
originating at x e f/ and with its head at x + /(x) e R". O
Example 7.8.2. Examples of vector fields are
•AC' diffeomorphism <!):[/—> V, with U and V open subsets in R".
• The gradient vector field (see Definition 2.6.2)
gmdg= J2 Djgej:U^R\
associated with a C' function g : f/ —> R. Note that the definition of grad g
seemingly depends on the choice of coordinates on R". However, this is not
actually the case, in view of the characterization of grad g(x) as the vector in
R" that points from x in the direction of steepest increase of the function g, and
whose length equals the rate of increase in that direction (see Theorem 2.6.3).
Exercise 3.12.(iii) contains formulae for the gradient vector field with respect
to arbitrary coordinates.
• The tangent vector field of a one-parameter group of diffeomorphisms (<I>')r6R
on R", as defined in Formula (5.21). liV
Let / : f/ —> R" be a C' vector field; one then has the total derivative Df(x) e
End(R"), for every x e f/. Further, we recall the trace tr Df(x) of Df(x), defined
as the coefficient of —A."""' in the characteristic polynomial, of Df(x)
det (XI - Df(x)) = X" - X"-^ tr Df(x) + ■■■ + (-1)" det Df(x). (7.62)
The definition of tr Df(x) is most obviously independent of the choice of
coordinates on R"; with respect to the standard basis in R" we find
trD/(x)= J2 ^jfj^"") (x€U).

528
Chapter 7. Integration over submanifolds
Definition 7.8.3. Let U be an open subset of R", and let / : f/ ^^ R" be a C^
vector field. Then we define the function div / : f/ —> R, the divergence of the
vector field f, by
div/ = trD/= J2 Djfj-
The operator V (this symbol is pronounced as nabla or del; the name nabla
derives from an ancient stringed instrument in the shape of a harp) is the n-tuple of
partial differentiations
In particular we have the formal notations, with g e C'(f/), and / e C^(U, R") a
vector field,
Vg = gmdg; ( V, /) = V . / = ((Di,..., D„), (/j,..., /„)) = div/;
A := ( V, V) = V • V = Y^ D].
Here A is the Laplace operator, or Laplacian, acting on g e C^{U) via
I — \
9xi
9
V 9x„ /
( D,\
\ D„ J
Ag = divigrsid g) = J^ DJg.
(7.63)
(See Exercises 3.14 and 7.60, and 3.16 and 7.61, for formulae giving the divergence
and the Laplacian, respectively, with respect to arbitrary coordinates.) O
Example 7.8.4 (Newton vector field). Define the vector field / : R" \ {0} -^ R"
by
1
fM = irir^' that is, fj(x) = -^ (!<;■< «)•
Using Example 2.4.8 we get Dj/j(x) = j-qpr— „;(,.;, for 1 < j <n. Consequently
div/(x) =
n n\\x\
||«+2
= 0 (x e R" \ {0}).
Note that if n > 2 (see Exercises 2.30.(ii) and 2.40.(iv))
/(x) = gradf -V and so a( A=0 on R"\{0},
■' ^ \2-n \\x\\'-^/ VII • ll"""V

7.8. Gauss'Divergence Theorem 529
while for n = 2
/(x)=gradlog||x||, and so A(log || • ||) = 0 on R'\ {0}.
Indeed, combination of the condition f{x) = grad g(x) and the assumption g(x) =
^o(lkll) gives, for 1 <J<n,
i7 = ^o(lkll)^=
hence
g'^(r) = r'~" and goir) =
The vector field
1
r^-", n>2;
2-n
log r, n = 2
X t-^ ■ X : R" \ {0} -^ R"
IS"-'MixII" ^' '
is said to be the Newton vector field; in this connection, see Exercise 7.21.(ii) for
|S"-'| :=hyperarea„_i(S"-'). tV
We employ the notations of Section 7.6. Henceforth consider a C' vector field
/ : S^ —> R", instead of a function / : S^ —> R, as we did in Section 7.6. For
y e dQ we introduce (/ v')(y) e End(R"), the matrix of which is equal to the
matrix product of the column vector f(y) e R" and the row vector v(yy e R";
that is
/ My)^i(y) ■■■ My)^„(y)\
if v')iy) = : ^ h ^^^^"' ^^
\ /«iy)L'liy) ■■■ fn(>')v„iy) }
This implies tr(/ v') = (/, v) : dQ —> R. Applying Formula (7.43) we find the
following identity of elements in End(R"), or matrices in Mat(n, R), where the
integration is performed by coefficients:
f Df(x)dx= f (fv')(y)d„^iy.
Jn Jan
Forming the traces on the left and on the right, we obtain the following:
Theorem 7.8.5 (Gauss' Divergence Theorem). Let Q c R" be as in Theorem 7.6.1
or 7.7.3. Let f : Q. ^ W be a vector field whose component fi^nctions ft, for
1 < z < n, satisfy the conditions from Theorem 7.6.L Then
I div f(x)dx= I {f,v)(y)dn-iy,
where the integration on the right-hand side is performed over dQ or W,
respectively.

530
Chapter 7. Integration over submanifolds
y + f(y)
(/(>'), 1^0'))
Illustration for Gauss' Divergence Theorem
(/(y). ^(y)) dn-iy is the volume of the cylinder whose base plane has area
d„_iy, and whose height equals the length of the normal component of the vector
field f{y), that is, equals the length of the component of /(>') orthogonal to the
base plane; this is an approximate description, of course
Definition 7.8.6. Let V C R" be a C' hypersurface for which a continuous choice
y i-> v{y) of a normal v{y) at the points y e V has been made. Let / : V —> R"
be a continuous vector field. Then ih&flwc of / across V with respect to the choice
of V is defined as (see Formula (8.32) for additional details)
L
{f.v)(y)d„-iy.
o
The Divergence Theorem can be formulated in words as follows. The integral
of the divergence of a vector field over an open set Q equals the flux of the vector
field through the closed hypersurface dQ with respect to the outer normal. The
Divergence Theorem explains why the divergence of a vector field is also known as
the flux of the vector field across closed surfaces per unit enclosed volume.
7.9 Applications of Gauss' Divergence Theorem
Example 7.9.1. Let S^ = {x e R" | ||x|| < 1}, then Q = B" and dQ = S""',
and v(y) = y is the outer normal to S"~^ at y e S"~^. Consider the vector field
/ : R" —> R" given by /(x) = x. Then div/(x) = n, and Gauss' Divergence
Theorem yields (compare with Exercises 7.21.(iv) and 7.45.(ii))
nvol„(B")= / div f(x)dx= I (i„_i)'= hyperarea„_j(S""~').
Jb" Js—i

7.9. Applications of Gauss'Divergence Theorem 531
More generally, let Q C R" be an open set that satisfies the conditions of
Theorem 7.6.1 and is contained in a ball of radius r. Then placing the origin at the
center of the given ball and noting that | (/, y )(>')| < r, for y e dQ, we obtain
n vol(S^) < r hyperarea„_j(9S^). ik
Example 7.9.2. We return to Example 7.5.5, that is
a^ b^ c^
and d{y) is the distance from the origin to the geometric tangent plane y + Ty{dQ.).
From Formula (7.41) it follows that
C 1 , 1 /" ,. , ^ s , 4n /ab be ca\
/ -7TTd2y = - (iwgmdg(x)dx = —[ \ h—).
Jasi d(y) 2Jsi 3 \ c a b /
Indeed,
/ 1 1 1 \ _ Anabc
A.W = 2t + ^ + ^), vol3(^) = ^-;
for volsC^^) use the substitution
^ : (r, a, 9) i-> r (a cos a cos 0, Z>sina;cos0, csin^),
satisfying det D^(r, a, 9) = aba r^ cos 0. In particular, iffl = Z> = c=l, then
Q. = B^, dQ = S^, and d(y) = 1, for all y e 9^. Thus follows (compare with
Example 7.4.10)
area(S^) = f d^y = —(1 + 1 + 1) = 477:. ik
JdQ. 3
Example 7.9.3. Let / : R^ ^^ R^ be the vector field with
f{x) = (Xixf, xfx2 - x|, lXiX2 + X2X3),
letfl > 0, and let V be the half sphere in R^ with V = {y e R^ | \\y\\ = a, y^ > 0}.
Then the following applies to the flux of / across V with respect to the choice
v(y) = iyJory e V:
(A^Ky)d2y = ^. (7.64)
'V
Indeed, we have
div/(x)=x| + xf+x|= \\xf.
Therefore, let Q be the open half ball in R^
I
Q = {x sR'' \ \\x\\ <a, X3>0]

532 Chapter 7. Integration over submanifolds
Then 9S^ = Vj U Vz U S, with
Vi = V, V2 = {y€B?\y\ + yl<a^,y^ = 0},
S = {>'eR3|>'? + >'| = fl^ >'3 = 0}.
Then S is a one-dimensional submanifold in R^, and therefore a two-dimensional
negligible set in R^. One has accordingly, with v{y) now the outer normal to 9S^
aly,
T f {f,^)(y)d2y= f \M\^dx.
Now we may write (see Example 7.4.11)
/ \\xfdx= da cosOdO / r'^dr = 2n\ smOy — =——.
Concerning the calculation of the integral over the subset V2 in the plane xj, = 0,
we note that for y e V2 one has v(y) = (0, 0, —1), and further d2y = dyi dy2. It
follows, therefore, that
/ {f,v)iy)d2y = -2yiy2dyidy2
yi / , yi dy2 dyi = 0,
•a ./ —*/fl-—vr
because the integrand in the inner integral is an odd function. This yields (7.64).
Note that we have avoided calculating the integral over V by means of a parametriza-
tion of V. i>[
Example 7.9.4 (Flux of Newton vector field). Let / : R" \ {0} -^ R" be the
Newton vector field from Example 7.8.4, with f(x) = ,' . .„ x, and div / = 0.
i"^ 111-^11
Further, let S^ be an open set in R" that satisfies the conditions of Theorem 7.6.1 or
7.7.3. Then (see Example 8.11.9 for a generalization)
f f 1, Oefi;
/ {f,v)(y)d,,-iy=\
Jsi^ [ 0, O^Q.
Indeed, if 0 £ fi, there exists a number r > 0 such that the sphere S"~^(r) in R"
about 0 and of radius r is entirely contained in Q. Let S^' C ^ be the open set
bounded by S"~^(r) and dQ. Because 0 ^ fi', we may write
0 = /" div fix) dx= f if, v)iy) dn^iy + f (/, i;)(>') d„.
Jn' Js"-\r) Jan
ly-

7.9. Applications of Gauss'Divergence Theorem 533
We now have, for y e S"~^(r), that v(y) = — iTt}', and so
Therefore
j5"-'(r) I'J K Js"-\r)
Example 7.9.5 (Heat equation). In physics the transfer of heat to a body S^ c R^
of constant mass density m, whose temperature at point x e Q and time f e R
equals T(x, t), is described, in first approximation, by the following laws.
(i) The amount of heat, A Q, required to raise the temperature of the part x + Ax
of Q, during the interval of time from t to t + At, from T(x, t) to T(x, t) +
AT(x, t), is proportional to the mass mAx of x + Ax, and to the temperature
difference Ar(x, ^). That is, there exists a constant ci > 0 such that
AQ = Cim Ax AT(x, t).
(ii) Let y + Ay be a part of the boundary ^Q. of Q., with outer normal v{y) at y,
and of area A2>'. Then the amount of heat, AQ, moving inward during the
interval of time from f to ^ + A^ across the part y + Ay of 9 S^ is proportional
to the following three quantities:
(a) the interval of time At,
(b) the variation of T in the direction of v{y), that is, to
D„(y)Tiy, t) = DyT(y, />(>') = (grad^, T(y, t), v(y)).
(This is because the flow of heat is caused by temperature differences,
and then takes place from hot to cold, its magnitude being proportional
to the component orthogonal to 9S^ at y of the opposite of the gradient
with respect to the spatial variable of T),
(c) the area A2y.
That is, there exists a constant C2 > 0 with
AQ = C2 Ar( grad, T(y, t), v(y)) Azy.
(iii) The total amount of heat, Qi, absorbed by Q during an interval A^, equals
the total amount of heat, Q2, which has moved inward in the course of the
same interval At across dQ.

534 Chapter 7. Integration over submanifolds
According to (i), therefore
Qi = cim I AT(x,t)dx.
And according to (ii),
Q2 = C2At / {grsidyT(y,t), v(y))d2y.
Jan
Assertion (iii) tells us that Qi = Q2. Thus we find the following equation for the
temperature T(x, t):
f AT(x,t) C2 f
/ T-—dx = / {giad Tiy,t), viy))d2y.
Taking the limit for At ], 0 and applying the Divergence Theorem, we obtain, with
citn'
r dT r
I —(x, t)dx = k I divj. gradj, T(x, t)dx.
Jn dt Ji2
Because this is valid for any body S^ C R^, one infers the heat equation for the
temperature T,
where A,, is the Laplace operator with respect to the spatial variable from (7.63).
This is an example of a partial differential equation for T, defining a relation
between different partial derivatives of T. ik
Example 7.9.6 (Green's identities). Applying Corollary 7.6.2 concerning
integration by parts in R", with / replaced hy Djf : Q —> R, for 1 < j < n, then
summing over j, we obtain Green's first identity,
f (g A/) (X) dx= f (g ^) (y) d,.^,y- f( grad /, grad g) (x) dx. (7.65)
Here g^, the derivative of f in the direction of the outer normal v, is defined by
^(y) = D„(y)f{y) = Df(y)v(y) = (grad /(>-), i;(>-)) (>- e dQ).
dv
Subtracting a similar identity from Formula (7.65), but with the roles of / and g
interchanged, we obtain Green's second identity,
fifAg-gAf)ix)dx= f (f^-g^)(y)d„^,y.
In the theory of the Laplace operator this identity is an important tool. ik

7.9. Applications of Gauss'Divergence Theorem 535
Example 7.9.7 (DiricWet problem). Let g and h be given continuous functions on
Q and dQ, respectively. The partial differential equation
A/ = g on Q,
together with the boundary condition
f\dQ = ^'
is said to be a Dirichletproblem on Q, for a C^ function /. We state without proofs
that, for functions g and h that can be differentiated sufficiently many times, and
for sufficiently well-behaved 9^, the Dirichlet problem on Q, has a solution /. See
also Exercises 6.99, 7.65, 7.69.(iii), 7.70.(v) and 8.23. What will be proved here
is the uniqueness of a solution / whose partial derivatives of order < 2 can be
continuously extended to Q,. Indeed, suppose that / is another such solution. Then
A(/-/) = A/-A/ = 0 on Q., f-f = 0 on dQ.
Therefore consider a function / which is harmonic on Q, that is, which satisfies
Afix) = 0 (X € ^);
and for which, in addition
f\aQ = 0. (7.66)
For such / one has, by Formula (7.65),
f \\gTadf(x)fdx=0.
Jn
Because the integrand is continuous and > 0, we conclude that grad f(x) = 0, for
X € Q. Therefore / is constant along any line segment contained in Q;
consequently, from (7.66) follows f(x) = 0, for every x € Q. i^
See, for example, Chapters 27 and 12, respectively, in: Treves, F.: Basic Linear Partial
Differential Equations. Academic Press, New York 1975; and Loomis, L. H., Sternberg, S.: Advanced
Calculus. Addison-Wesley Publishing Company, Reading 1968.

Chapter 8
Oriented Integration
For an open set at one side of its boundary one has a natural prescription for the
direction of the normal. However, this is not the case for a manifold of lower
dimension (consider a surface in R^, for example), and an orientation must then
be chosen. This choice plays a role in the oriented integration over the manifold,
introducing a dependence on the sense in which the manifold is swept out. The
generalization to R" of the second aspect of the Fundamental Theorem of Integral
Calculus on R, antidifferentiation of a fiinction of several variables, leads to the
concept of curl of a vector Beld. Oriented integration and curl together form the
ingredients for the integral theorems of vector analysis and the theory of complex
functions. Antidifferentiation of a function of several variables culminates in the
theory of differential forms. Through systematic use of linear algebra an orgy of
indices and partial derivatives is avoided.
8.1 Line integrals and properties of vector fields
The divergence is one of the infinitesimal invariants of a vector field. There are others
that are also needed in vector analysis. For an understanding of their meaning it
is of advantage to be familiar with the concept of line integral, which we therefore
introduce at the outset; the notion is also important for intrinsic reasons.
Definition 8.1.1. Let / C R be a closed interval, y •./—>■ R" a C' curve, and let
/ : im(y) —>■ R" be a continuous vector field. Define f (f(s), dis ), the oriented
line integral of / along y, by
j{f{s),d,s)= jifoy, Dy)(t)dt. (8.1)
'Y
537

538 Chapter 8. Oriented integration
The integral on the right-hand side contains the inner product of the vectors /(y (t))
and y'(t) in R". If y is a closed curve, the integral on the left is sometimes referred
to as the circulation of / around y. O
Example 8.1.2. Letx e R" and define y;, : / = [ 0, 1 ] ^ R" by y_^(t) = tx. Then
Dyx(t) = X, for all f e /, and accordingly we have, for every / e C(R", R"),
f (f(s),dis)= f (f(tx),x}dt. ik
Jyx jo
The term oriented line integral is used because the value of the line integral
changes sign when the curve is traced in the opposite direction, as is shown in the
following:
Lemma 8.1.3. The right-hand side in (8.1) does not change upon a reparametriza-
tion of the interval I that preserves the order of the endpoints. That is, let J =
[ y_, y+] ^'^d let ^jr : 7 —>■ I be a C^ mapping with I = [ ■tj/(j^), i/(j+) ]. Then
f {f(s),dis}= f{f(s),dis).
Jyo\J/ Jy
lyof Jy
If however, I = [■tj/(j+), i/f (_/_)], then
f {f(s),dis} = - f{f(s),dis}.
Jyo^lf Jy
Proof. On account of the chain rale, D(y o ■tj/)(t) = (Dy) o ■tj/(t) D'tj/it) =
i/'(t)(Dy) o ■tj/(t), and thus, by the Change of Variables Theorem 6.6.1 on R,
f {f(s),dis}= ({fo{yof),D{yof))(t)dt
Jyotjr J J
= {{foy)of,{Dy)of){t)f\t)dt=\ (foy,Dy)(t)dt.^
Jj- JfU-)
We now introduce some other invariants of a vector field, first giving motivations
and definitions, and then filling in the background.
Theorem 7.6.1 is a generalization of the Fundamental Theorem of Integral
Calculus on R, but there exists another variant which will also be needed. In the integral
calculus on R, one of the possible formulations of the Fundamental Theorem (see
Theorem 2.10.1), valid for continuous / on[a,b ], is
dx J a
/W = -^ / f(t)dt (x€[a,b]).

8.1. Line integrals and properties of vector fields 539
Consequently, on an interval every continuous function / has an antiderivative, that
is, every continuous / is the derivative of a differentiable function g on that interval.
The «-dimensional analog would be that a continuous vector field f : U ^f R"
possesses an antiderivative, integral, or scalar potential g : f/ —>■ R, that is, / is
the total derivative of a differentiable real-valued function g; another way of saying
is that / is a gradient vector field,
f = gradg, or, equivalently, fi = Dig (1 < i < n). (8.2)
Assume that this is the case for a C^ function g : f/ —>■ R, while n > 1. By
Theorem 2.7.2, the order of differentiation of the C^ function g is interchangeable;
hence
Djfi = DjDig = DiDjg = Difj (1 < i, j < n). (8.3)
Thus, in contrast to the case n = \, continuity of / as such is not sufficient for the
existence of an integral: / also has to satisfy the integrability conditions in (8.3),
which can be rewritten in matrix form as
A/(x) = 0 with Afix)!j = Djfiix)-D!fjix). (8.4)
Therefore the nonvanishing on U of the infinitesimal invariant Af of the vector
field / is an obstruction for finding an antiderivative for / on f/; moreover, it turns
out that the geometric properties of the set U C R" also play a role in this problem.
The most general result we shall formulate is Theorem 8.2.9.
Definition 8.1.4. Let f/ C R" be an open subset, and let / : f/ -> R" be a C'
vector field. According to Lemma 2.1.4, for every x e f/, the derivative Df(x) e
End(R") of / at X can be additively and uniquely decomposed into a self-adjoint
operator ^Sf(x) e End"'"(R"), with real eigenvalues, and an anti-adjoint operator,
^Af(x) e End~(R"), with purely imaginary eigenvalues; this is known as the
Stokes decomposition, or the (infinitesimal) Cartan decomposition of Df(x),
Df(x) = ^(Df(x) + DfixY) + ^(Df(x) - DfixY) =: ^Sf(x) + ^Af(x).
(8.5)
Here Df(xy is the adjoint linear operator of Df(x) with respect to the standard
inner product on R"; its matrix with respect to the standard basis is given by the
transpose matrix Z)/(x)' of Df(x). O
Thus the definitions of Sf(x) and Af(x) are independent of the choice of
coordinates on R", whereas they do depend on the choice of the inner product on
R". We note in addition
div/ = tr£)/=^tr5/- (8.6)
For the cases n =1 and 3 we now take a closer look at the anti-adjoint operator
Af{x) from the Stokes decomposition (8.5). We see immediately:

540 Chapter 8. Oriented integration
Lemma 8.1.5. For n = 2we have, in the notation used above,
0 £>2/i-£>i/2
with 7 = (J -J).
Here J is the matrix of the rotation in R^ by f in the positive direction.
Definition 8.1.6. Let U C R^ be an open subset and let / : f/ —>■ R^ be a C'
vector field. Define the function curl / : f/ —>■ R, the curl of the vector field f, by
curl / = div(7'/) = £>,/2 - £>2/i = {A/ e,, ^2 )■ O
Example 8.1.7. In the definition of curl / one has to make a choice for the sign
(this is done by introducing J). From the mapping Af(x) itself one can only deduce
det Af(x) = {D\f2 — D2fi)^(x). Under our convention, curl 7=2 for 7 : R^ —>■
R^,because 7'7 = /. Inotherwords, the curl of the vector field 7 .x i-> (—X2,xi),
of rotating in the positive direction, is positive. If / : R^ \ {0} —>■ R^ is given by
f(x) = ■jjjp-7x, then J' f(x) = j^x; hence it follows by Example 7.8.4 that
curl / = 0 on R2 \ {0}. ik
We now deal with the case n = 3.
Lemma 8.1.8. Let A = (Aij) e Mat(3,R) be an antisymmetric matrijc. Then there
exists a unique vector a e R^ with
Ah = a X h (h € R^), namely a = (A^2, A\^, A2\).
Furthermore, {Ah, k) = (a, h x k),forh, k e R^.
Proof. The uniqueness of a follows immediately. We further have
/ 0 -A21 An
A= \ A2i 0 -A^2
\ -An A^2 0
But this is seen to be the matrix of the linear mapping /z i-> a x /? : R^ —>■ R^ with a
as above. The second equality follows from the identity {A;, axh) = det(A; a h) =

8.1. Line integrals and properties of vector fields 541
det(fl h k) = (a, h x k), which holds for any triple of vectors a, h and k € B?
(see Formula (5.2)). □
Application of Lemma 8.1.8 in the case where A = Af(x) from the Stokes
decomposition (8.5) gives a vector a e R^ with j -th component given by
«; = Af(x)j+2,j+i = Df(x)j+2j+i - Df(x)j+ij+2
= Dj+lfj+2(x) - Dj+2fj + \(x),
where the indices 1 < y < 3 are taken modulo 3.
Definition 8.1.9. Let f/ c R^ be an open subset and let / : f/ —>■ R^ be a C'
vector field. Then define the vector field curl / : f/ —>■ R^, the curl of the vector
field f, by
curl / = (£>2/3 - D,f2, D,fi - £>,/3, D.A - £>2/i) = V x /, (8.7)
where V is the nabla operator from Definition 7.8.3 and where the cross product is
formally calculated. O
Corollary 8.1.10. Let U C ti^ be an open subset and let f : U ^ B? be a C^
vector field. Then, in the notation of (8.5) and (8.7), for x e U, h andk e R^,
Af(x)h = curl f(x) X h, {Af(x)h, k) = {curl f(x), h x k).
Example 8.1.11 (Curl of infinitesimal rotation). See Example 5.9.3 for the
notation. In particular, let 0^ = r^ : R^ —>■ R^ be the tangent vector field from that
example, that is
(pa(x) = (fl2-^3 — a^X2, a^x\ — a\x^, a\X2 — a2X\) (x e R^).
Since r2a is an anti-adjoint operator, the Stokes decomposition reads D<pa{x) =
fa = 2'"2a' f^^ all X € R^. Therefore
cml(j>a(x) =2a (x € R^). ilV
Remark. We have the following relation between the curl of a vector field in R^
and one in R^ If / : R^ ^ R^ is a C' vector field, define / : R^ ^ R^ by
f(x) = (f\(Xi,X2), f2(Xi,X2), 0).
Then
curl fix) = (0, 0, curi /(x,, xz)). (8.8)

542 Chapter 8. Oriented integration
Remarks of an analytical nature, motivating the definitions above. Let U C
R" be open and / : f/ -> R" a C' vector field. Let x e f/ and h, k e R", then we
obtain from the definition of differentiability, for t > 0 sufficiently small,
(fix + th) - fix), tk) = t^{Dfix)h, k) + a(f2), t I 0,
{fix + tk)-fix), th) =t^{Dfix)'h, k)+ait'^), t ^ 0.
Subtracting these identities we find
{fix), th} + {fix + th), tk)-{fix + tk), th)-{fix), tk)
= t^(Afix)h,k)+ait^), tiO.
In terms of Definition 8. LI we recognize the expression on the left as an
approximation of
f {fis),d,s},
the oriented line integral of the vector field / along y (x, h, k, t), the boundary of
the parallelogram based at the point x and spanned by the vectors t h and tk. In
other words, y(x, h, k, t) equals the union of: the line segment in R" from x to
x + t h, the line segment fromx +1 h to x +t h +1 k, the line segment fromx+tk to
X +t k +1 h traced in the opposite direction, and finally, likewise, the line segment
from X to X +1 k. Thus we find an interpretation of Afix), namely
{Afix)h,k}=lim^ f {fis),dis). (8.9)
'4-0 t Jy(x,hXt)
The arguments above are infinitesimal; in the following proposition, which for
« = 3 is a special case of Stokes' Integral Theorem 8.4.4, they are made global.
Proposition 8.1.12. Let I = [ 0,1 ], let U C R" be open, and assume T e
C^iI^,U) has continuous mixed second-order partial derivatives D2D\T and
D1D2T : /2 -> R". Then the following holds for a C^ vector field f : U -> R".-
I {fis),dis}= j {iAf)oT-DiT,D2T)ix)dx.
(8.10)
Here dil'^) is the boundary of the square I^, given by, successively, withxi, x^ € /,
xii->(xi,0); X2i->(l,X2); x\v^ i\ — x\,\); X2 i-> (0,1 - X2).
The expression on the left-hand side in (8.10) equals, by Definition 8.1.1,
[{f°'^, £>,r)(x,,o)rfx,+/"(/or, D2r}ii,x2)dx2
-A/or, £),r)(x,,i)rfx, - A/or, D2r)io,x2)dx2.
Furthermore, the notation ■ on the right-hand side in (8.10) signifies application of
a linear mapping to a vector

8.1. Line integrals and properties of vector fields 543
Proof. On I^ we have the following identities of functions:
£),(/or, D2T) = {{Df)oT-DiT, D2T) +{foT, A^zF),
D2{foT, DiT) = {{Df)<oT-DiT, D2T) + {foT, D2DiT).
One has, by Theorem 2.7.2, that D\D2T = D2D\T. Subtraction therefore yields
£>,{/or, D2T) - D2{f oT, D^T) = {{Af) oT ■ D^T, D2T).
Integrating this identity over I^ and using Corollary 6.4.3, we find
/" fDiif or, D2r)(x)dxidx2- f f D2{for, £>,r)(x)^x2rfx,
= f {(Af) or-DiT, D2T){x)dx.
When we apply the Fundamental Theorem of Integral Calculus 2.10.1 on R to the
left-hand side we obtain
/
{{f oT, D2T){1, X2)- {f oT, D2T){<d, X2))dX2
- j{{foT,D,T){xu l)-(/or, £>,r)(x,,0))^x,.
Remark. By means of differentiation under the integral sign and of integration
by parts we find the following result, which is related to Formula (8.10):
Di fifo r, DzF }(x)dx2 = f{ (Af) o r . £),r, DzF )(x) dx2
JI JI (8.11)
+{/or, £),r)(x,,i)-(/or, £),r)(x,,o).
This formula gives the derivative of a line integral along a curve X2 i-> r(xi, X2)
that has an additional dependence on a parameter xi, with respect to that parameter.
Remark in preparation for Section 8.6. In the preceding proposition curves
y : / —>■ f/ were seen to play a role, and, in addition, scalar functions y i->
{(/ o y)(y), Dy(y)), which occur as a result of pairing the vector f{y(y)) € R"
and the tangent vector Z)y (}') e Ty^y-jU,foTy e / (see Definition 5.1.1). Obviously,
then, the vector f(x) e R" induces the mapping
m(x) = o)f(x) : h H> {f(x), h} in Lm(TxU, R).
The mapping m (x) then is an element of
1
/\ T^U := T;U := Un(T,U, R),

544 Chapter 8. Oriented integration
the linear space of all linear mappings from the tangent space Tx U to R, which
is isomorphic with R". A mapping m which assigns to every x e f/ an element
ct)(x) e /\ T*U is said tohe a. dijferentiall-form on U, and one writes CO e Q^(U)
(see also Exercise 5.76).
Likewise, Af(x) e End(R") induces a mapping
a)(x) = coAfix) € Lin^CTvf/, R) with a)Af(x)(h, k) = (Af(x) h, k).
This mapping cl>{x) then is an element of
2
/\ r;f/ = {7? € Un'-iT.U, R) I r)(h, k) = -r)(k, h)},
by definition the linear space of all antisymmetric bilinear mappings from the
Cartesian product of T^U with itself to R. Indeed, Af{xy = —Af(x) implies
a)(x)(k,h) = {Af(x)k, h) = {k, AfixYh) = -a)(x)(h,k).
A mapping co which assigns to every x e f/ an element to (x) e /\ T*U is said to
be a differential 1-form on U, and one writes u) e Q,^{U).
In this context the differential 2-form to a/ is said to be the exterior derivative of
the differential 1-form co/. The linear mapping oif i-> to a/ : ^'(f/) —>■ Q,^{U) is
said to be the exterior differentiation, and is often denoted by d : Q^(U) —>■ Q^(U)
instead of A (from antisymmetric) as we have done, that is, dcof = ma/- The
condition Af = 0 for / then becomes do)/ = 0 for Mf, and Mf is said to be a closed
differential l-form. Conversely, a closed differential 1-form m is not necessarily of
the form ftJgradg, for a function g; but if it is, m is said to be an exact differential
l-form.
Remarks of a geometrical nature, motivating the definitions above. The
notation is that of Section 5.9. Assume that (<l>')/sR is a one-parameter group of
diffeomorphisms of R" with tangent vector field (p : R" —>■ R". For fixed x e R",
we wish to study the images ^'(x + v), for small values of f e R and v e R".
Let w e R" be arbitrary and define / : R^ ->■ R" by f(t, u) = <l)'(x + uw).
Then we have DzDi/CO, 0) = D(j>(x)w, if we assume that D(j>(x) e End(R"), the
total derivative of (p at x, exists (note that (p does not depend on t). The identity
lim(,^u)_^(o^o) r(t, u) = D2Dif(0, 0) from Formula (2.19) now implies
1
lim —((i>'(x+ uw) — (i>'(x) — uw) = D(p(x)w.
(r,H)->(0,0) tu
Hence, setting uw = v we obtain, for small values of f e R and t; e R",
^(v) := <D'(x + v)- <D'(x) = (/ + tD<pix))v + aitM) = (I + tD<pix))v.
_ (8.12)
This i>i is said to be the local effect near x of the flow <!>'.

8.1. Line integrals and properties of vector fields 545
The Stokes decomposition of D<p{x) from Formula (8.5) asserts
D(j){x) = -S(p(x)+-A(p(x), S(p(x) € End+(R"), A(p(x) € End^CR").
(8.13)
In view of the Spectral Theorem 2.9.3 the operator S(j)(x) can be diagonalized,
having eigenvalues A.i,..., A.,, e R, and corresponding eigenspaces spanned by
eigenvectors Vi,... ,v„ e R", say. Therefore S<p{x) is an anisotropic (r] TpoTtV] =
change) linear dilatation in R"; in the direction of the eigenvector Vj its action is
that of multiplication by Xj, for I < j < n. (In other words, S<p{x) maps a ball
about 0 to an ellipsoid.) According to Example 2.4.10 we havee2^*W € Aut(R"),
for f e R. It is an anisotropic dilatation in R" with eigenvalues e^^"\ ..., e^^"\ and
under this transformation volumes change by a factor e2-^1 . • • gi^". It now follows,
by Formula (8.6) (also compare with Formula (5.33)), that
(jetgi^^W = g5P'i+-+^^) = gitr5<^(A-) ^ grdiv<^(A-) (8.14)
Since A<p{x) is anti-adjoint. Exercise 4.23.(iv) implies that e2'*'^(^) e Aut(R"), for
t e R, is a rotation in R". In particular, therefore, this transformation is volume-
preserving. Formulae (8.12) and (8.13) now yield
$', = {1+^- S(t>(x))(I + '- A(j>(x)) = e5^*We5^*W. (8.15)
This means that the local effect ^'^ near x of the flow <!>', in the approximation of
small values for t, can locally be written as a composition of the rotation e2'*'^(^)
in R" and the subsequent anisotropic dilatation eS'^*^^) in R". The local effect O^.
is approximated by a linear mapping; denoting the latter also by O^, we have, by
Formula (8.14), _
det <D^ = e"*'^*W
Thus, this formula gives the geometrical interpretation of div0(x), for a tangent
vector field (p associated with a one-parameter group (<l>')/sR acting on R": it equals
the rate of change of volume at time t = 0, resulting from the local effect i>'^ near
X of the flow <!>'. This conclusion also follows from Formula (5.31), that is
d_
dt
detZ)<l)'(x) = div0(x).
/=0
In particular, for « = 3 it follows from Corollary 8.1.10 and the theory of
rotations in Exercise 5.58 that t i-> eJ'**^*) is the one-parameter group of rotations
t i-> Rf icuriAfv) of the space R^, about the axis spanned by curl0(x) e R^ and
with angular velocity 11| curl (j> (x) \\. In combination with Formula (8.15) this leads
to the geometrical interpretation of curl (p (x), for a tangent vector field (p associated
with a one-parameter group (<l>')/sR acting on R^, that is, curl 0(x) determines the
"rotational component" of the local effect i>\ near x of the flow <!>'.

546 Chapter 8. Oriented integration
Example 8.1.13. In the terminology of Section 5.9, the vector fields J and /
from Example 8.1.7 both are tangent vector fields of one-parameter groups of dif-
feomorphisms of R^, having concentric circles about the origin as orbits. Now
II7 (x) II = ||x||, while ||/(x)|| = tj^. Thus, as x moves away from the origin, the
orbits under the flow associated with / are traced at decreasing rates. Evidently,
curl / = 0 implies that this phenomenon exactly compensates the rotation of x
under the influence of the flow associated with J. ik
8.2 Antidifferentiation
We return to the problem of finding an antiderivative (see Formula (8.2)).
Definition 8.2.1. Let U C R" be an open subset and let / : f/ -> R" be a C' vector
field. We say that / satisfies the integrability conditions if (see Formula (8.4))
Af = 0, that is, Djfi = Difj (1 < i, j < n).
The vector field / is said to be divergence-free, source-free or incompressible on
U if div / = 0. Furthermore, / is said to be harmonic on U if both Af = 0 and
div / = 0.
In particular, let« = 3. Then / is said to be curl-free, vortex-free or irrotational
on U if curl / = 0 (and therefore also Af = 0). O
Example 8.2.2 (curl grad and div curl). Let U C R" be an open subset and let
g € C^iU), then (see Formula (8.3))
A(gmdg) = 0. (8.16)
Indeed, D(gj:adg)(x) is self-adjoint due to the symmetry in the indices of the
second-order partial derivatives of g (see Theorem 2.7.2). A gradient vector field
gradg therefore satisfies the integrability conditions. The gradient vector field
grad g of a harmonic function g, that is, a function with div grad g = 0, is harmonic.
The component functions of a harmonic vector field are harmonic functions. The
Newton vector field from Example 7.8.4 is harmonic.
One has in particular, for « = 3 and/? : U —>■ R^ a C^ vector field with f/ C R^,
curl grad g = V X (Vg) = 0 and div curl/z = V • (V x/?,) = 0. (8.17)
Indeed, the matrix of Z)(curlh)(x) is antisymmetric, for every x € U. ik

8.2. Antidifferentiation 547
Definition 8.2.3. Let U C R" be an open subset and let f : U ^ R" be a
continuous vector field. A C' function g : f/ —>■ R is said to be an antiderivative,
integral, or scalar potential for f on U if f = grad g.
And, if « = 3, a C' vector field /z : f/ —>■ R^ is said to be a vector potential for
f on U if f = curl/z. O
From (8.16) and (8.17) it follows that A/ = 0 (or div / = 0 if « = 3) on f/ is
a necessary condition for the existence of a scalar potential (or a vector potential,
respectively) for f onU. And, under an additional condition on f/, these conditions
on / are also sufficient, as shown in Lemma 8.2.6 below. The necessity of additional
conditions on U is apparent from the following. Assume that the continuous vector
field / : f/ —>■ R" possesses a scalar potential g : f/ —>■ R. Further, letx and y € U
be fixed. Then, for every C' curve t h> y(t) : I ^>- U from x to }', the integral
/ {f(^)^d\^) is independent of the choice of the curve y, as long as the latter runs
from the fixed point x to the fixed point y. Indeed, the value of the integral is given
by the potential difference
j{ (gradg) o y, Dy )(t)dt = ^ ^^^^(t)dt = g(y) - g(x). (8.18)
For a closed C' curve y one has in particular f {f(s),dis) = 0.
Example 8.2.4. Let U = R^ \ {0}, and consider the vector field / : f/ ^ R^
from Example 8.1.7, given by f(x) = j^Jx; then we know that curl/ = 0
on U. Define y : ] —jt, jt [ —>■ f/ by y(t) = (cost, sint). Then Dy(t) =
(— sint, cost) = Jy(t) and therefore
{foy,Dy)(t)= " 7 " =1; hence \ {f{s),d,s) =ln.
This result can also be derived from Example 7.9.4, because J' f : x i-> j^a^ is
the Newton vector field on U (neglecting the constant 2n).
As a consequence, / can not have an antiderivative on U. On the other hand, /
does have an antiderivative on f/' = R^\(] —oo, 0] x {0}), because / = grad argon
U', where arg : f/' —>■ R is the argument function, defined by (see Examples 3.1.1
and 2.4.8)
arg(x) = 2arctan ( r^r-r).
Note that U' is the maximal domain of definition for the function arg.
According to Example 7.8.4 the vector field J' f has an antiderivative on JJ,
and so
[{J'f(s),dis}=0. t\
Jy

548 Chapter 8. Oriented integration
Definition 8.2.5. A set f/ C R" is said to be star-shaped if there exists a point
x° € U such that for all x e U the line segment from x° to x lies in U, that is,
im(y;,) C U with y^ :[0,\]^ U defined by y^(t) = x° + t(x - x°). O
Lemma 8.2.6 (Poincare). Let U C R" be star-shaped and open, and let f : U ^^
R" be a C^ vector field. Then the following two assertions are equivalent.
(i) Af = OonU.
(ii) There exists g € C^{U) which is a scalar potential for f; for example the
following, with y^ as in Definition 8.2.5:
g(x)= f {f(s),d,s) (x€U). (8.19)
Jy.1
For n = 3, other equivalent assertions are as follows.
(iii) div f = 0 on U.
(iv) There exists a C' vector field /?. : f/ —>■ R^ which is a vector potential for f;
for example
h{x)= I ((/ o Kv) X y,){v) dv {X € U).
Jo
Here the integration is carried out by components.
Proof. In order to simplify the formulae we assume that x° = 0.
(i) ^ (ii). According to Example 8.1.2 one has g{x) = /g k(v, x) dv, where
k:[0,\]x U ^R with k(v,x) = (f(vx), x) = fivxYx.
From Af = 0 and Formula (8.5) it follows that Df(vxy = Df(vx), and therefore,
if gradj, is the gradient with respect to the variable x € U,
gradj, k(v, x) = v Df(vxyx + f(vx) = v Df(vx)x + f(vx)
= v—f(vx) + f(vx) = —(v f)(vx).
dv dv
By means of differentiation under the integral sign we find
gradg(x)=/ —ivf)ivx)dv = fix).
Jo dv
(iii) ^ (iv) is proved in an analogous manner. Begin by h(x) = f^ m(v,x)dv,
where
m : [0, 1 ] X f/ —>■ R^ with m(v,x) = f(vx) x vx = —v(rx o f)(vx).

8.2. Antidifferentiation 549
Here r^ € End(R^) is given by r^iy) = x y. y. This yields
D^miv^x) = vrf(^^) - v^r^ o Df(vx).
Because rf^yx) and Kx are anti-adjoint operators, it follows that
Axm(v,x) := Djcm(v,x) — Dxm(v,xy
= 2t; r/(„;,) - v^(rx o Df(vx) + Dfivx)' o r^).
Application of the remark below, with L = Df(vx), gives
Axm(v, X) = 2v r/(„j:) + V rDf(xv)x = r2vf(,vx)WDf(,vx)x-
Consequently, if curl_v is the curl with respect to the variable x e f/,
curlj; m(v, x) = 2vf(vx) + v^Df(vx)x = —(v^ f)(vx).
dv
By differentiation under the integral sign we find
curl/2(x)=/ —{v^f){vx)dv=f{x). □
Jq dv
Remark. Let L e End(R^), with tr L = 0. Then, for all x e R^
rx o L + L' a rx = -rix-
Indeed, the fact that tr L = 0 implies, for all h, k e R^,
det(Lx h k) + det(x Lh k) + det(x h Lk) = 0.
This can be shown by calculating the coefficient of A.^ in (see Formula (7.62))
det(Lx -Xx Lh- Xh Lk - Xk) = det(L - XI) det(x h k).
In view of det(p q r) = (p x q,r), for p, q and r e R^ (see Formula (5.2)), this
becomes
{Lx xh.k) + {x X Lh,k) + {x x h, Lk) =0.
Therefore
{(rix+rxoL + L' orx)h,k)=0.

550 Chapter 8. Oriented integration
Remark. From Lemma 8.2.6 it follows immediately that a harmonic vector field
on R^ possesses a scalar potential that is itself a harmonic function.
We now formulate the property of U being simply connected, which is weaker
than that of being star-shaped (see Definition 8.2.5), but which will still guarantee
that a vector field satisfying the integrability conditions on U possesses a scalar
potential on U (see Theorem 8.2.9). This concept originates from homotopy theory,
a subject in algebraic topology.
Definition 8.2.7. An open set U C R" is said to be simply connected if, for every
point X € U and every C' curve y : I ^>- U beginning and ending at x e U, there
exists a C' homotopy between y and the constant curve y^ : I —>■ {x}, that is, if
there exists a C' mapping (with / = [0, 1 ], for simplicity)
r-.i'^^u with r(o, t) = y(t), r(i,t) = r(s,o) = r(s,i) = x,
forwhichthemixedsecond-orderpartialderivativesZ)2Z)ir andZ)iZ)2r : /^ —>■ R"
exist and are continuous. O
For example, R", as well as every star-shaped open set U C R", are each simply
connected, because r(s, t) = s x + (1 — s)y{t) € U, in view of the definition of
being star-shaped. Note that DxD2T{s, t) = D2DiT{s, t) = -y'(t).
Lemma 8.2.8. Let U C R" be simply connected and open, and let f : U ^^ R"
be aC^ vector field with Af = 0. Then, for every closed C' curve y : I ^>- U,
f(f(s),dxs)=0.
Jy
Proof. Assume that y begins and ends in x e U, and let F be the corresponding
C' homotopy. On the strength of Definition 8.1.1, the conclusion follows from
Proposition 8.1.12. Indeed, Z)2r(0, t) = Dy{t), while D2T{\, t) = DxT{s, 0) =
DiT{s, 1) = 0 because r(l, t) = r(s, 0) = r(^, l)=x. □
Tlieorem 8.2.9. Let U C R" be simply connected and open, and let f : U ^^ R"
beaC^ vector field with Af = 0. Then there exists a C'^ function which is a scalar
potential on U for f.
Proof. Letx° e f/be fixed. Byanalogy with Formula (8.19), we define the function
g:U^Rhy
g(x)= f (f(s),d,s};
JVx

8.3. Green's and Cauchy 's Integral Theorems 551
here y^ : / —>■ f/ is an arbitrary C' curve in U from x° to x e U. Then g is
well-defined on U. Indeed, two different curves y^ and y\., meeting in C' fashion
at X, together form a closed C' curve y which begins and ends at x°. Application
of Lemma 8.2.8 to y then yields the result that the oriented line integral of / along
yx equals the oriented line integral of / along yx. One then notes that, for all y,
X € U,
8(y)-8M= f {f(s),dis}.
W= f
where y^y is a curve in U from x to y. But now one immediately concludes that
grad g(x) = f(x), for all x e f/. □
8.3 Green's and Cauchy's Integral Theorems
Gauss' Divergence Theorem 7.8.5 can be used to prove other integral theorems:
that of Green for open sets in the plane R^, that of Cauchy for open sets in the
complex plane C, and that of Stokes for surfaces in R^.
We now introduce the concept of a positive parametrization of the boundary of
an admissible open set in R^. First we consider an example. The unit circle 5'
equals dQ, with Q the bounded open subset {x e R^ | ||x|| < 1}. Consider
the C°° parametrization ] —n, jt [ ->■ S^ \ {(-1, 0)} of S^ given by t i-> y(t) =
(cost, sint). One then has v{y(t)) = y(t) = (cost, sin t), for the outer normal to
S^ at y(t), and Dy(t) = (—sint, cos f) (see Theorem 5.1.2), for the tangent vector
to 5' at y(t). Accordingly, the direction of the tangent vector is obtained from that
of the outer normal by application of J, the rotation in R^ by j in the positive
direction, whose matrix with respect to the standard basis in R^ is
J = C, -I). (8.20)
In addition we find
nr\c t — cin /
= 1 >0.
det(i; oy \ Dy)(t) =
cos t — sin t
sin t cos t
In linear algebra this is precisely the condition for v o y(t) and Dy(t) to form a pair
of positively oriented vectors in R^. The geometrical equivalent of the positivity of
the determinant is that the point y(t) e 9^ moves counterclockwise in the plane
R^ when t in R ranges from —n to n. As this goes on, Q, constantly lies "to the
left", that is, in the direction of the inner normal, and the complement of Q, lies "to
the right", that is, in that of the outer normal. This suggests the following:

552 Chapter 8. Oriented integration
Definition 8.3.1. Let ^ be a bounded open subset of R^ having a C' boundary
dQ and lying at one side of dQ. Assume that / —>■ dQ with t i-> y(t) is a C'
parametrization of 9^ by the disjoint union / of intervals in R, with
det(i; o y \ Dy)(t) > 0 (f € /).
We then say that t i-> ^(f) is a positive parametrization of dQ,, or, alternatively,
that the parametrization t i-> y{t) endows 9^ with Si positive orientation. O
Remarlt. If f i-> ^(f) is a positive parametrization of 9^, the direction of the
tangent vector Dy(t) to dQ at y(t) is uniquely determined by the outer normal
V o y(t) to dQ at y(t), by means of
\\Dy(t)\r'Dy(t) = J v(y(t)) (t € /). (8.21)
Here J is the matrix from (8.20). In particular, there is no longer a freedom of choice
as regards sign. In other words, every positive parametrization of 9^ induces the
same unit tangent vector field r on 9^, namely
T = J ov.d^^R^. (8.22)
Example 8.3.2 (Annular domain). Let 0 < r < R and consider the annular
bounded open set ^ = {x e R^ | r < ||x|| < R}. Then 9^ is the (disconnected)
set {x e R^ I ||x|| = r} U {x e R^ I ||x|| = R}. Furthermore, t i-> y(t) is a
positive C°° parametrization of 9^ if
IR(cost, sint), (0 <t < 2n);
r(cost, — sinf), (4n <t< 6n).
Omitting from Q the points on a fixed line through the origin, one obtains two sets
^1 and ^2. Check that both dQi and 9^2 are connected sets. Now assume that
positive parametrizations are chosen for both of these. Then the line segments in
the intersection 9^i (1 9^2, considered as subsets of 9^i, are traced in a direction
opposite to that in which they are traced as subsets of 9^2. ^
We repeat the definition of line integral (compare with Definition 8.1.1) in the
present context.
Definition 8.3.3. Let ^ C R^ be as in Definition 8.3.1, and let t i-> y(t) be a
positive parametrization of dQ. Let / : 9^ —>■ R^ be a continuous vector field.
Then define f^Qifiy), diy), the oriented line integral of the vector field f along
9^, by
f {f(y),d,y}= f(foy,Dy)(t)dt. (8.23)
Jbq Ji

8.3. Green's and Cauchy's Integral Theorems 553
Here the right-hand side contains the inner product of vectors in R^. The expression
on the left is also known as the circulation of f around 9 Q with respect to the choice
o/T (see (8.22)). O
Lemma 8.3.4. Let Q Cti^ be as in Definition 8.3.1, and lett i-> y(t) bea positive
parametrization ofdQ.
(i) A parametrization t i-> ^(t) : / —>■ 9^ is a positive parametrization of dQ
if and only if
detZ)(>'-'oy)(?) >0 (T€7).
(ii) The integral on the right in (8.23) does not change when a different choice is
made for the positive parametrization t i-> y{t) ofdQ; therefore the integral
on the left is defined independently of the choice ofa positive parametrization.
Proof. Lety(t) = y(t), witht e /and?e 7; thenr = (y"^ oy)(t) =: >!'(?).
Application of the chain rale toy = y o >!' on / therefore gives Dy(t) = det D^P (?) Dy(t)
for t € I and f = >!'(?) e /. Assertion (i) now follows from
det(i; o y \ Dy)(t) = det D^(t) det(i; o y \ Dy)(t).
To prove (ii) we apply the Change of Variables Theorem 6.6.1 with 'P :/—>■/;
then we obtain
jyoy,Dymdt = fjfoy,Dy}(t)^-^^dt. □
Remark. See Lemma 8.L3 for another formulation of Lemma 8.3.4.
Furthermore, Lemma 8.3.4.(ii) can also be proved in a different way. Formulae (8.21) and
(8.22) yield
f {f(y),diy) = f{foy(t), \\Dy(t)\r'Dy(t))\\Dy(t)\\dt
Jan Ji
= I {f, Jv){y)d,y= f (f T}(y)d,y.
(8.24)
Thus the oriented line integral of the vector field / along 9^ is seen to equal the
integral over 9 Q with respect to the arc length on 9 ^ of the function y h> {f r ) (y);
and according to Definition 7.1.2 - Theorem the latter integral is independent of
the choice of the parametrization.

554 Chapter 8. Oriented integration
Theorem 8.3.5 (Green's Integral Theorem). Let Q c ti^ be a bounded open
subset having a C^ boundary dQ and lying at one side of dQ. Assume that t i-> y(t)
is a positive parametrization ofdQ. Let f : ^ —>■ R^ be a C^ vector field such
that f and its derivative Df : Q —>■ End(R^) can both be extended to continuous
mappings on Q. Then
f {f(y), diy)= f curl f(x) dx. (8.25)
Proof. Note that it follows by Formula (8.24) that
I {f(y),do'}= I {J'f,v){y)d,y.
Jbq Jbq
Here J' is the transpose of the matrix J from (8.20). Formula (8.25) now follows
by means of the Divergence Theorem 7.8.5, because div(y'/) = curl /, according
to Definition 8.1.6. □
Example 8.3.6 (Descartes' folium). From Example 8.1.7 we know that curl J =2.
We therefore immediately conclude
area(^) = i dx = - i {Jy,d^y) = - i (y, y'^ - y2 y[)(t) dt
Jq ^ Jbq ^ JI
(8.26)
1
= /
det(y\Dy)(t)dt.
I ^
Note that ^det(y \ Dy)(t) is the area of the triangle with vertices 0, y(t) and
y + Dy(t).
Descartes' folium (compare with Example 5.3.7) is the curve in R^ defined by
3at
yit) = - -(1,0 i-oo<t <oo, ty^-1), with >',(±oo) = 0.
F + 1
We know that its points satisfy the equation y^ + yl = 3ayiy2. One has
Let ^ C R^ be the bounded subset which is bounded by the part of the folium
parametrized by ] 0, oo [. We have
3a^ r°° 3t^ 3a^ r°° 1
area(^) = —j^ ^^^-^^dt = — J^ ~du =
3a^

8.3. Green's and Cauchy's Integral Theorems 555
This result can also be obtained by means of the following C' diffeomorphism
'P : V ->■ ^. Here V is the triangle
V = {y € R^ I 3'i + 3'2 < 3a }, Q = {x € R^ | Xj + x| < 3axiX2},
Then det ZXP (y) = |, and hence area(^) = f^ ^dy = ^3a^ = ^. iic
As preparation for Cauchy's Integral Theorem we make the following remarks.
We identify C with R^ as in (1.2). In this context, the matrix J from (8.20) is also
known as the complex structure on R^, because this linear mapping yields the action
of multiplication by i = V—T on C identified with R^. As in Formula (1.2) we
identify a function / : C —>■ C with the vector field / = (/,, /2) : R^ —>■ R^ by
means of
f(xi+ix2) = Re f(xi +ix2)+iIm f(xi +1x2) <—> (/i(xi, X2), f2(xi, X2)).
Definition 8.3.7. Let ^ C C ~ R^ be as in Definition 8.3.1, and let t i-> z(t) be
a positive parametrization of 9^. Let / : 9^ —>■ C be a continuous function. In
view of the equality
(/ o zmozit) = (/i + if2)iz(t))(Dzi + iDz2m,
we define f^^ f(z) dz, the complex line integral of the function / along dQ, by
f f(z)dz= I (J{z),d,z) + i I {{Jj){z),d,z), (8.27)
JdQ. Jan Jan
where / = (/i, —/z) is the complex conjugate function of /; and therefore J f =
(/2,/l). " " " O
As in Lemma 8.3.4.(ii), one proves that the expression on the left does not
change upon choosing another positive parametrization of 9^.
Example 8.3.8. If z = xi + ix2, then g(z) := i = -^ ^ (-^, ■^). Thus,
with / as in Example 8.2.4,
Ik II
and from the integrals in that example we obtain at once, where y denotes the unit
circle with positive orientation.
/
- dz = Ini. ik
z

556 Chapter 8. Oriented integration
As in complex analysis we have the following:
Definition 8.3.9. Let ^ C C be open and let / : ^ —>■ C. Then / is said to be
holomorphic or complex-differentiable on ^ if / is continuously differentiable over
the field R and the following limit exists, for every ^ e ^:
The next lemma essentially follows from the definitions.
Lemma 8.3.10 (Caucliy-Riemann equation). For f = f^-\- if2 : ^ —> C and
(/i' fi) '■ ^ ~^ R^> respectively, one has the equivalent assertions.
(i) / is holomorphic.
(ii) The C^ vector field (/i, /2) satisfies the Cauchy-Riemann equation
(£>/) oJ = JoDf€ End(R2),
that is, Di fi = 02/2 and D\f2 = —D2f\.
(iii) The C' vector field f = (/i, —fi) i^ harmonic (see Definition 8.2.1), that is
curl 7=0, div 7 = curl (7/) = 0-
In other words, Af = A(J f) =0.
(iv) The real-differentiable C^ function f satisfies
df d 1
^ = 0, where ^ =-(Di+iD2)', that is, iDif = 02/-
dz dz 2
Proof, (i) 4» (ii). The identity, valid for /z e R,
,. f{z + h)-f{z) ,. f{z + ih)-f{z)
lim ; = lim ;
/!->o h /i->o ih
corresponds to Di/i + zDi/z = Dz/z - 2^2/1 (since \{fi + z'/z) = /z - if\); in
turn, this is equivalent to {Df) o J = J o Df. □
The notation in part (iv) of the lemma above is explained as follows. For
Z = Xi + ix2 € C we have Xj = ^(^ + 2) and X2 = —\iz — z) € R. Regarding z
and I as though they were independent variables, we find
9xi 9xi 1 9x2 9x2 /
9^ Sz 2' 9^ &z 2

8.4. Stokes' Integral Theorem 557
Then the chain rale imphes
az az az 2.
df dx\ dxi 1
■^ = DJ-^ + £>2/^ = -iD,+iD2)f.
az az az 2.
Theorem 8.3.11 (Cauchy's Integral Theorem). Let ^ c C ~ R^ Z>e a bounded
open subset having a C' boundary dQ and lying at one side ofdQ. Let / : ^ —>■ C
be a holomorphic function such that the vector field f and its derivative Df : Q —>■
End(R^) can be extended to continuous functions on Q. Then
f f(z)dz = 0.
Proof. The conclusion follows immediately from Lemma 8.3.10.(iii) and Green's
Integral Theorem. □
By means of Definition 8.1.1, a version of (8.27), Lemma 8.2.8, and finally
Lemma 8.3.10.(iii) we find the following variant (see Exercise 8.11 for another
proof):
Theorem 8.3.12 (Cauchy's Integral Theorem). Let U C C — ^ be a simply
connected and open subset. Let f : U —>■ C be a holomorphic function. Then, for
every closed C' curve y : I ^>- U,
f f(z)dz = 0.
Jy
In fact, the theorem is valid for every closed piecewise C' curve y : I ^>- U.
8.4 Stokes' Integral Theorem
We now prepare the formulation of Stokes' Integral Theorem. Let Q C R^ be
as in Green's Integral Theorem, and let f i-> y(t) : I —>■ 9^ be a positive C'
parametrization of dQ. Further, let
0 : ^ —>■ R^ with y t-^ (p(y) = X,
be a C^ embedding whose image is the C^ surface in R^

558 Chapter 8. Oriented integration
According to Formula (5.5),
-1 /r. A „ r. A\/,A ^ T>3
V o ct>(y) = \\(D,ct> X D2ct>)(y)\r {D,(j> x D^^Xy) € R
is a normal to H at ^(y). We assume that <p and the total derivative D<p : ^ —>■
Lin(R^, R^) can be extended to continuous mappings on ^. In particular, we may
then define
9S =(j)(dQ).
The extension of 0 to 9^ is not necessarily injective on dQ, that is, a subset S C da
may exist with S = <j>(Vi) = <pi^2) for V\ and V2 C 9^, while V\ ^ V2. A part
S such as this does not contribute to 93, if under the positive parametrization of
dQ, the sets <p{V\) and <p{V2) are traced in opposite directions (these parts of 9S
then "cancel each other out"). One has that 9 H, less the above-mentioned sets, is
a piecewise C' submanifold in R^ of dimension 1 that does indeed play the role of
the boundary of S. The mapping
y.th^ i<poy)(t) -.I^da,
is a parametrization of 93. Considering Theorem 5.1.2, we see that Dy(t) is a
tangent vector to 9 H at y (t).
Remark. Let the notation be as above. Let / : 93 —>■ R^ be a continuous
vector field. In Definitions 8.1.1 and 8.3.3 we have introduced fg^(f(s), dis ), the
oriented line integral of the vector field / along 9 3, by
f (f(s),dis)= f{foy,Dy}(t)dt. (8.28)
^3S Jl
The integral on the left is sometimes referred to as the circulation of / around 9 3
with respect to the choice of r, the unit tangent vector field along 9 3. The right-hand
side in (8.28) does not change when a different positive C' parametrization of dQ
is chosen, nor does it change when another choice is made of a C^ parametrization
(j) : Q ^>- 3, provided detZ)(0~' o (j))(y) > 0, for all y € Q. Consequently, the
integral on the left is defined independently of the choice of the parametrization y
of 9 3, provided the positivity conditions are met.
Proposition 8.4.1. In the notation used above we may write
f {f(s), dis}= f (((Dct>y ■ (f o ct>))(y), do'), (8.29)
Jas Jan
where D<p{yy e Lin(R^, R^),/or y e 9^, while ■ means application of a linear
mapping to a vector Thus we have here the vector field
(Dct>y-(foct>) : >- ^ (foct>)(y) ^ DHyY{f o(j>){y) : 9^ ^ R^ ^ Rl (8.30)

8.4. Stokes' Integral Theorem 559
Proof. By virtue of Formula (8.28) we find
fjf(s),d,s) = j{(fo<p)oy, {D(j>)oy.Dy){t)dt
= j^a(Dct>y-(foct>))oy,Dy}(t)dt,
where we have taken the adjoint. According to Formula (8.23), the expression on
the right equals that in Formula (8.29). □
Formula (8.29) tells us that the line integral along 9 S can be "pulled back" to
a line integral along dQ. We then wish to apply Green's Integral Theorem to the
latter, to convert it into a surface integral over Q. Accordingly, we calculate the curl
of the vector field in Formula (8.30) by means of the following:
Lemma 8.4.2. Assume that / : H —>■ R^ w a C' vector field. In the notation above
we have the following identity of functions on Q:
curl ((£)0)' . (/ o 0)) = ((curl /) o <p, Di(j> x D2<p).
Proof. For the vector field in (8.30) we have
( {D,4>, fo4>)\
g := {D<j>)' • (/ o 0) = : ^ ^ R^
\ {D2<t>, fo<j>) I
Therefore
Dg = S + (D<py ■ (Df) o<p.D<p with S = {{DjO^ip, / o 0)).
Theorem 2.7.2 implies that 5'(>') € End(R^) is self-adjoint, for y e Q. Furthermore,
the adjoint of (Dcj))' ■ (Df) o cp ■ D(j> equals (Dcj))' ■ (Df)' o cp ■ D(j>; and so
Ag = Dg- (Dg)' = (D<py ■ (Df - (Df)') o<j>.D(j> = (D(j>)' ■ (Af) o<j>.D(j>.
Using Definition 8.1.6 and Corollary 8.1.10 we then find
cm\g={Agei, e^) = {(Af)o(j>.D(j>ei, D(j>e2) = ((curl/)o0, £),0x£>20>-Q
With the help of Lemma 8.4.2 we now apply Green's Integral Theorem to the
right-hand side of (8.29); this gives
I {f(s),dis)= f {(cmlf)o<p, Di,pxD2<P)(y)dy. (8.31)
Finally, because the left-hand side of this formula is expressed in terms of H, we
want to recognize the surface integral over Q on the right as a surface integral over
S which has been "pulled back"; hence the following:

560 Chapter 8. Oriented integration
Definition 8.4.3. Let g : S —>■ R^ be a continuous vector field. Then define
f^(g(x), d2X ), the oriented surface integral of the vector field g over H, by
\ {gix),d2x)= I {go,p, DicPxD2(p)(y)dy= f det (g o ,p \ DcP)(y) dy .O
Ja Jn Jn
The right-hand side does not change when we make a different choice (p :
J^ —>■ H for the parametrization of H, provided that detZ)(0~' ° <j>)(y) > 0, for
y € Q. Indeed, the second integral in Definition 8.4.3 is recognized by means of
Formulae (7.37) and (7.20); this yields
/ {g(-x),d2x)= / (go(j), vo(j))(y)M^(y)dy= / (g, v}(x)d.2X, (8.32)
^s Jn Ja
where the last integral is the integral with respect to the Euclidean area on H of
the function x h> (g, v )(x) on H. Formula (8.32) also shows that f„{g(x), d2X)
equals the flux of g across S with respect to the choice of v (see Definition 7.8.6).
Formula (8.31) and Definition 8.4.3 imply the following:
Tlieorem 8.4.4 (Stolies' integral tlieorem). Let Q cR^ be as in Green's Integral
Theorem, let t i-> y(t) be a positive C^ parametrization ofdQ, and let <p : Q, ^^
<p {Q.) = 3 be a C^ embedding in R^ such that <p and the total derivative D<p can
be extended to continuous mappings on Q,. Let f : a ^>- Vt? be a C^ vector field
such that f and the total derivative Df can be extended to continuous mappings
on H. Then
I {fis), dis) = I {curl f(x), d2X ).
J da V a
Stokes' Integral Theorem can be worded as follows. The circulation of a vector
field around the closed oriented curve 9 S C R^ equals the flux of the curl of that
vector field across the oriented surface H C R^. The theorem explains why the curl
of a vector field is also known as the circulation of the vector field around closed
curves per unit enclosed area.
Further note that Formulae (8.9) and (8.10) for « = 3 immediately follow from
Stokes' Integral Theorem.
Remarlt on compatible orientations. A phrase commonly encountered is the
requirement that the orientation of the surface S and the orientation of the boundary
9 S be compatible with each other. Here we give an exact formulation. According
to Lemma 7.5.4 we may assume
^nU = {y €R^\yi<0}riU.
Then v(y) = e\, for }' e 9^ (1 U, and y : f i-> ^^2 is a positive parametrization of
9^ n f/. Consequently, Di(j){y(t)) is a tangent vector to S at x = (0 o y)(t) e 9 S

8.5. Applications of Stokes' Integral Theorem 561
"pointing away from" H, that is, a differentiable curve S in R" with 5(0) = x
and DS(0) = D\<p{y{ty), leaves H via the boundary. Furthermore, D2<j>{y(t)) is
a tangent vector to H, tangent to 9 S at x, while S "lies at the left of" D2<j>{y(t)).
Now the vectors
Di<Piy(t)), D2<Piy(t)), {D,(j> x D2<j>){y{t))
form a positively oriented triple in R^. Thus it is evident that, in the formulation
of Stokes' Integral Theorem given above, the orientations of H and of 93 are
compatible in the following sense:
• a tangent vector to H, pointing outward from S,
• the "positive" tangent vector to 9 H (determined by the positive orientation of
9H),
• and the normal to H,
form a positively oriented triple of vectors at every point of 9 S.
8.5 Applications of Stokes' Integral Theorem
Example 8.5.1. Define the vector field
/ : R^ ^^ R^ fix) = (X3,xi, xz), then curl f{x) = (1,1, 1).
If ^ = ] —n, JT [ X ] i/r, f [ C R^, one has, less four points, that dO, equals
(] -TT, TT [ X {1/r}) U {{n} X ] l/r, ^ [) U {] -TT, TT [ X {^}) U {{-u} X ] l/r, ^ [),
where under positive orientation this set is traced counterclockwise. Further, let
S = V = <p{Q,) C R^ be the spherical surface from Example 7.4.6, described by
<p : (oi,9) i-> /?(cosQ!cos0, sinacosS, sin9) {(oi,9) e Q).
It follows that
d(j) d(j) J J J
— X —(oi,9) =R (cos a cos 9, sin a cos 9, sin 0 cos 0),
da 39
I d(j) d(j) \ 9, 9 X
((curl/)o0, —X —)ioi,9) =/? ((cosa + sinQ!)cos 9 + sin 9 cos 9).
\ da d9 I

562 Chapter 8. Oriented integration
Therefore
/ (curl/(x), d2x)
= R^ I ((cos a + sin a) cos^ 6 + sin 6 cos 6)dad6
TT — —
= R^ (cosa + sma)da j cos^6dO + InR^ / smQcosQdO
J —jT J ^j/- J "if/-
= 7tR^ [ sin^ 6> ] J = TzR^il - sin^ ir) = nR'^ cos^ f.
On the other hand, 4> (] —jt, jt [ x {i/f}) C 9H has the parametrization
y : a i-> /?(cosQ!cosT/f, sinacosi/r, sini/r) (—jt < a < n).
And thus
/oy(Q!) =/?(sinT/f, cosacosi/f, sinacosi/f),
Dy(a) =/?(—sinacosi/f, cosacosi/f, 0),
(/ o y, Dy ) (a) = R^(— sin a cos i/f sin i/f + cos^ a cos^ i/f).
Furthermore, we have (p Q —tt, jt [ x {^j) = {(0, 0, 1)}, and because this is a zero-
dimensional manifold, this part of 9 H does not contribute to the line integral along
9S. Finally,
71
(p{{7T} X ] i/f, — [) = {x e R^ I xi < 0, X2 = 0, X3 > sim/f, ||x|| = 1
= 0{{-7r}x]^,^[),
but the points x = <p{y) of these two image sets under <p are traced in opposite
directions when y ranges counterclockwise over the boundary 9^. Therefore, these
parts of 9 H do not contribute to the line integral along 9 S. As a result
/ (/(■^)> '^i'^) = R^ I (— sin a cos rjr smrj/ + cos^ a cos^ ijr) da
J da J —7T
/^ 1
-(1 + COS 2a) da = nR^cos^ f,
-JT -^
as expected on the basis of Stokes' Integral Theorem. ik

8.5. Applications of Stokes' Integral Theorem 563
Remark. In certain applications of Stokes' Integral Theorem, or of Gauss'
Divergence Theorem, one is only given a closed C' curve C, or a closed C' surface
D, in an open subset U C R^; it then remains to determine a suitable C' surface H,
or an open set Q,m U such that C = 9 H, or D = dQ, and such that the theorem
can be applied to the pair (H, C), or the pair (Q, D), respectively. Whether or not
this can be achieved is a subject of study in homology theory, which is a branch
of algebraic topology. In addition, one has to be careful in defining the integrals,
especially with regard to the orientations of curves and surfaces.
Formula (8.28) recalls the definition of the oriented line integral of a continuous
vector field / along a C' curve in R^ forming the boundary of an embedded C^
surface in R^. In that case the data define a preferential direction for a unit tangent
vector field along the curve. But the same definition is also meaningful for an
arbitrary embedded C' curve y : I —>■ R^, if we fix a continuous choice for a
tangent vector to that curve; in other words, y is such that Dy(t) points in the
prescribed direction, for all f e /.
This also applies, mutatis mutandis, to Definition 8.4.3 of the oriented surface
integral of a continuous vector field g over an embedded C' surface in R^. There
the data also define a preferential direction for a normal to the surface. And again
this definition is meaningful for an arbitrary embedded C' surface 0 : D —>■ R^, if
we fix a continuous choice for a normal to that surface; in other words, (p is such
that (Di(j) X D2<j>)(y) points in the prescribed direction, for all y € D.
In both cases we say that we have oriented the curve, or the surface, by choosing
the direction of the tangent vector field, or the normal vector field, respectively.
We therefore need, in addition, the following:
Definition 8.5.2. Let D C R"~' be open, let 0 : D -> R" be an oriented C'
embedding, and assume / : im(0) —>■ R" to be a continuous vector field. Define
f,(f(x), d„^ix ), the orientedhypersurface integral of / over (p, by
f {fix), rf„_ix) = /" det (/ o 0 I D,pXy) dy. O
J(j> J D
The definition is independent of the choice of the embedding, provided the
total derivative of the reparametrization has a positive determinant. Indeed, if
(j)(y) = <p(y), with y € D and y € D, then y = ((j)~^ o (j))(y) =: 'i^(y), see
Lemma 4.3.3. Application of the chain rale to the identity <p = <p o^ on D now
yields D^(y) = D(j>(y) D^(y). Given y € D, the function End(R"-') -^ R
given by
D«P(30 h^ det(/ o 0 I D$)(y) = det(/ o<j>(y) \ D<j>(y) D«P(>0)
is antisymmetric and (n — l)-linear in the n— I column vectors in R"~' of D^P (y).
It is therefore given by D'i^Cy) i-> cCy) det D'i^Cy), where cCy) = det (/ o(j)(y) \
D(j)(y)). This yields
det (/ o 0 I D$)(y) = det D^(y) det (/ o 0 | D(p)(y).

564 Chapter 8. Oriented integration
Applying the Change of Variables Theorem 6.6.1, with 'P : D —>■ D, we now find
\det D^(y}\
det D^iy)
f det(/ o 0 I Dct>)(y)dy= I det(/ o 0 | D^)(y) ^fl^Z^l^ ^y.
Jd Jd
if
WlilBl
Illustration for Example 8.5.3: Mobius strip
Example 8.5.3 (Mobius strip). A C' surface in R^ cannot always be oriented; as a
case in point, consider the Mobius strip. This can be obtained as H with H = im(0),
where 0 : ^ —>■ R^ is the following C°° embedding:
1 1
It ^\ t a. a\
(p(t,a) = I cosa^l +fcos — j, sma^l +fcos — j, fsm — 1.
For a positive parametrization of Q, the boundary 9 H is successively parametrized
by
<pit,-n)
1
0(2'")
4>(t.,n)
1
0(-->«)
= (-1,0,-0
=: y+{a)
= (-1,0,0
=: y^{a)
(— <t < -),
2 2
{—71 < a. < TZ),
(5 >' > -5)>
(tt > a > —7i).
The surface S itself is orientable. The nonorientability of H is a consequence of
the fact that different parts of dO, (that is, ] —^, ^ [ x {—n} and ] —^, | [ x {n})
have the same image under 0 in 9 S (that is, the "vertical part" of 9 S), whereas the
orientations on these images are equal (both are traced from the top down). As a
result, these parts "do not cancel each other out" (in contrast to what happens when
9^ is glued in such a way as to form a cylinder).

8.5. Applications of Stokes' Integral Theorem 565
Stokes' Integral Theorem can be applied to H, provided one does not equate the
overlapping parts of 9 H with each other, but considers each with its own orientation.
For example, consider the vector field
f:{x€R^\x^ + xly^O}^R^ with f(x) = — ^(-X2,xi,0).
-A. 1 I Jiry
By means of Formula (8.8) and Example 8.1.7 we see that curl / = 0. By virtue
of Stokes' Integral Theorem,
f {fis),dis)+ f {fis),dis)=0,
Jy.,. Jy-
because the integral over (twice) the vertical part of 9 S vanishes.
On the other hand, consider the C°° embedding 0 : J^ —>■ R^ with the property
S = S, if S = im {(j)(Q)), given by
J^ = ] 0, - [ X ] —Itt, In [, <p{r, a) = <p{r, a); then
1
(j>ir, -In) = (1 - r, 0, 0) (0 < r < -),
~ 1
(j>{-, a) =: y{a) {-In <a < In),
$(r, In) = (1 - r, 0, 0) (2 ^ '' ^ ^'*'
0(0, a) =: Y^ioi) {In > a. > —In).
Here ya{oi) = (cos a, sin a, 0). In this case the nonorientability of H is evident
from the fact that the "central circle", the image under yo, is traced twice in the
same direction. Because the integrals over the "horizontal part" of 9 H cancel each
other out, we find
f{fis),dis)=4n, since j {f{s),dis) =-An.
Jy Jyo
Now im(y) is the disjoint union of im(y_|_) and im(y_); and, furthermore, y and
y_i_ have the same orientation on their intersection, whereas y and y_ have opposite
orientations on their intersection. Consequently
f(f(s),d,s)= f {f(s),dis}- f {f(s),d,s},
and so / {f{s), dis) = i:2n.
Jy±
This example shows, moreover, that a closed C' curve in R^ that is not "knotted",
to wit y, can be the boundary of a nonorientable C' surface, to wit S. ik

566 Chapter 8. Oriented integration
Remark. We now study the global properties, as opposed to the local ones from
(8.16) and (8.17), possessed by a vector field in the presence of a potential.
Assume that the continuous vector field / : f/ —>■ R" has a scalar potential
g : f/ —>■ R. Further, let x and y € U he fixed. Then we know that for every C'
curve t H> y(t) : I ^ U from x to y, the integral f (f(s), dis ) is independent
of the choice of the curve y. According to Formula (8.18), its value is given by the
potential difference
f{f(s),dis}=g(y)-g(x).
Jy
Now assume that« = 3 and that / possesses a vector potential /z : f/ —>■ R^;
further, let the closed oriented C' curve y in f/ be fixed. Let H C f/ be an oriented
C^ surface with da = y, while at every point of y a tangent vector pointing
outward from H, the tangent vector to y in the positive direction (determined by the
orientation of y), and the normal to H (determined by the orientation of S), form a
positively oriented triple of vectors. Then, by Stokes' Integral Theorem
f {f(x),d2X}= [{h(s),dis:
J a J y
That is, the oriented surface integral of / over H is independent of the choice of
the surface H, provided the latter has the fixed curve y as its boundary, and the
orientations on S and y are compatible.
Definition 8.5.4. Let U C R" be open and let / : f/ —>■ R" be a continuous vector
field.
Then / is said to be conservative on U if, for any two points x and y € U, the
oriented line integral of / along a C' curve y in U from x to y is independent of
the choice of y.
Furthermore, / is said to be solenoidal (6 cwXt)v = tube) on U if the oriented
hypersurface integral of / over every compact oriented C' submanifold in U of
dimension n — \ equals 0. O
The terminology solenoidal is related to the fact that for every / having that
property the oriented hypersurface integral over hypersurfaces is preserved if those
hypersurfaces can be "connected" by a hypersurface T <Z U "consisting of
streamlines for /", that is, for which the normal to T at every point of T is perpendicular
to /. In other words, the integral of / over a hypersurface is preserved if that
hypersurface has transverse intersection with the same "stream tube". More precisely,
/ is solenoidal on U if the oriented hypersurface integral of / over Si is equal
to the opposite of that over H2, for all C' hypersurfaces Hi and H2 in U with the
following properties: there exists a C' homotopy F : / x I"~^ —>■ U such that
imr(0, •) = Hi, imr(l, •) = H2, moreover 9(imr) \ (Hi U H2) is aC'
submanifold in f/of dimensions—1, and/(x) e rt(9(imr))ifx isin9(imr)\(HiUH2),
while the orientations on Hi and H2 coincide with the outer normal on 9(im F).

8.6. Differential forms and Stokes'Theorem 567
Theorem 8.2.9 and Lemma 8.2.6 now yield:
Proposition 8.5.5. Let U C R" be open and let f : U ^ R" be a C^ vector field.
(i) Then f is conservative on U if U is simply connected and Af = 0, in
particular, for n = 3, if f is curl-free.
(ii) Letn = 3. Then f is solenoidal on U if U is star-shaped and f is source-free.
For arbitrary n e N it is nontrivial to prove that a connected compact oriented
C'' submanifold H in R" with A; > 1 of dimension « — 1 is the boundary of a bounded
open set Q that lies at one side of its boundary (see Example 8.11.10 below on the
Jordan-Brouwer Separation Theorem). Moreover, it depends on the properties of
U whether Q can be found lying inside of U. If all of this is the case for every S
as above. Gauss' Divergence Theorem 7.8.5 implies that a divergence-free vector
field on U is solenoidal on U. Here the relation between these two notions will not
be discussed systematically.
Example 8.5.6. It is possible for a vector field to be curl-free, or source-free, on an
open set f/ C R^, while the oriented integral along a closed oriented curve, or over
a closed oriented surface in U, respectively, differs from 0. We limit the discussion
to a few characteristic examples only.
The set Ui = R^ \ {0} is not simply connected. Let fi be the vector field /
from Example 8.2.4; that example then tells us that /^i (fi (s), dis ) = 27r ^^^ 0, if
S' C f/i is the unit circle with positive orientation.
Likewise, f/2 = R^ \ {0} is not simply connected either. Let /2 be the Newton
vector field on U2. According to Example 7.8.4, the vector field /2 is source-free on
U2, but, by Formula (8.32) and Example 7.9.4, one has f^iifiix), d2X ) = 1 ^ 0,
if S^ C U2 is the unit sphere oriented according to the outer normal.
On the other hand, it can be shown that every closed oriented C^ curve y
in f/2 is the boundary of an oriented C' surface S in U2 with 9 3 = y and with
compatible orientations on S and y; therefore, Stokes' Integral Theorem then yields
/^{/2(^),^,^)=0. tV
8.6 Apotheosis: differential forms and Stokes' Theorem
(iQ dKo6ewci^ = deification). The theory of differential forms is an extension of
the foregoing, and provides a natural framework for the theory of integration over
oriented manifolds. Moreover, this generalization leads to unification, because
Gauss' Divergence Theorem 7.8.5 and Stokes' Integral Theorem 8.4.4 are special
cases of Stokes' Theorem 8.6.10 below.
Infinitesimal changes dx of a variable x are described by vectors in the tangent
space at x to the space of those variables. Now let / be a function of x. The

568 Chapter 8. Oriented integration
differential df(x) of / at x is the linear part of the infinitesimal change in / when
an infinitesimal change dx occurs in x. This implies that df (x) should be interpreted
as a linear function acting on tangent vectors at x. Thus df itself can be regarded
as a function that gets evaluated along parametrized curves, by averaging over x on
the curve the values of df(x), acting on the tangent vector that is determined by
the parametrization, at x to the curve. Indeed, we know that the rate of increase
of a function along a curve depends linearly on the tangent vector to the curve.
By integrating we obtain differences between values of the function as oriented
line integrals. In the development of the calculus for differentials, one introduces
differential forms, that is, linear combinations of differentials, having functions
as coefficients; and an expression of that kind can only be the differential of one
function if integrability conditions analogous to those from Definition 8.2.1 are met.
Higher-order differentials, products of differentials of functions, occur as
obstructions against the integrability of the differential forms above. In fact, in the
Remark following Proposition 8.1.12wesaw that Af (x), the obstruction at x against
the integrability of a vector field /, led to an antisymmetric bilinear mapping acting
on two infinitesimal changes. In analogy with the situation for df, this property
enables the function Af to be evaluated along parametrized surfaces, by averaging
over X on the surface the values of Af(x), acting on the pairs of tangent vectors
that are determined by the parametrization, at x to the surface. Thus we see that
higher-order differentials, that is, products of differentials of functions, occur in
particular in the measurement of oriented volumes of infinitesimal parallelepipeds
spanned by tangent vectors. One desires these volumes to linearly depend on those
tangent vectors, and to vanish when the tangent vectors become linearly dependent.
Together these considerations render it plausible that the higher-order differentials
are generally defined as antisymmetric linear functions of the infinitesimal changes,
that is, acting on collections of tangent vectors.
In order to arrive at a rigorous formulation of the foregoing we begin by defining
differential forms. Then we give the proof of Stokes' Theorem, where we also
introduce the exterior derivative of a differential form; this concept thus finds a first
justification in the fact that it plays an important role in Stokes' Theorem. In addition
we develop some multilinear algebra in order to derive the usual expressions for
the exterior derivative. Our treatment heavily uses properties of determinants of
a matrix, but other approaches are possible, at the cost of more abstract algebraic
constructions.
Given a vector space V we write V* for the dual linear space of V consisting
of the linear functional on V, that is, V* = Lin( V, R). Note that R"* := (R")* is
linearly isomorphic with R" in view of Lemma 2.6.1.
Definition 8.6.1. Let V be a vector space. Define /\ V* = R. Fori; > 0 we define
/\ V*, the k-th exterior power of V*, as the linear space of all antisymmetric k-
linear mappings of the A;-fold Cartesian product V'' of V with itself, into R, that is.

8.6. Differential forms and Stokes'Theorem 569
oi'in f\ V* if and only if, in the notation of Definition 2.7.3,
ftj e Lin^'CV, R), ftj(t;„(i),..., t;„(fc)) = sgn(o-)ftj(t;i,..., t;fc) (a e Sfc).
Here Sk is the permutation group on k elements, and sgnCcr) is the sign of cr, that
is, sgnCcr) = (— 1)'" if cr = ctj ■ • • a,„ is the product of m transpositions. O
Note that /\ V* = V*. Considering cl>{v\, .. .,Vi + Vj,... ,Vj + vi,..., Vk)
we see that the condition of to e \lnl^{V, R) being antisymmetric is equivalent to
the condition, where t; e V occurs in z-th as well as in y-th position,
coivi, ...,v, ...,v, ...,Vk) = Q (i <i < j <n).
Definition 8.6.2. Assume ai,..., a^ to be arbitrary elements in V*. One readily
sees that ai A • • • A a^ e /\ V*, if we define, for all t;i,..., t;^ e V,
(ai A • • • Aak)(vi, ...,Vk)= det(Q!,(t^;))i</,;<fc. O
Definition 8.6.3. Let i < ii < • ■ ■ < ik < n and t;,-[,..., t;,-^ e V, then we use the
notation
/ = (ii,..., ik), Ik = {(ii, ■■■,ik)\i <ii < •■■ <ik <n}.
From now on the notation "^ indicates that the variable underneath is omitted. In
particular, if A; = « — 1 we set
T =(l,...,T,...,«)€i"-\
vr =(vi,...,v;,...,Vn)€V"'' (l<i<n). O
Lemma 8.6.4. Let dim V = n and let (ei,..., e„) be a basis for V. Further, let
(fip ..., e*) be the dual basis for V* given by ef(ej) = Sij,for 1 < /, j < n. Then
the elements
4 := e* A ■ • ■ A el (I € i,) (8.33)
form a basis for /\* V*. In particular, dim /\'' V* = Q); and so dim /\* V* = 0,
fork > n. Furthermore, dim/\" V* = 1; and for m e /\" V* we have
ftj = ftj(e(i,...,„)) det. (8.34)

570 Chapter 8. Oriented integration
Proof. The linear independence of the vectors in (8.33) readily follows from the
fact that, for / and 7 e i^
f 1, ifI = J;
e^ej) = (8.35)
I 0, in all other cases.
Now let CO € /\ V* he arbitrary, and write coj = ct)(ej). We assert that
a)=J2a)i e*j. (8.36)
Indeed, both members of (8.36) have the same value on all A;-tuples ej, for J e Ik,
and thus on V'', since m and the e^ are all contained in /\ V*. □
Example 8.6.5 (Relation to cross product of vectors). Let V = R" and let
(ei,..., e„) he the standard basis. In this case, the cross product of « — 1
vectors in R" (see the Remark on linear algebra in Example 5.3.11) is closely linked
to the results in Lemma 8.6.4. Since
dimA R"* = dimA"~ R"*
= n, we can
introduce an isomorphism between R" and each of these two exterior powers. In
fact, with V e R" we may associate bit; e A R"* = R"* ^^ i"^ Lemma 2.6.1, that
is, we define (l)iv)h = {v, h), for all /z e R". For the other isomorphism we set
H-l
b„_it; = z„ det e /\ R"*, (z„ det)(t;i,..., t;„_i) = det(t; t;i ■ • • t;„_i),
for any choice of vectors t;i, ..., t;„_i e R". Here z„ det is called the cowfracfzow of
det e A" R"* with the vector v. It is a direct consequence of Formula (8.36) that
b„_it; is completely determined by its action on the e^, for 1 < i < n. Expanding
the n X n determinant by its first column we see (b„_it;)(^) = (iudet)(ef) =
det(t; ei ■ ■■ e^ ■ ■ ■ e„) = (—l)'~'t;,-. Thus we have obtained
H-l
(The relation between b„_i t; and contraction with v explains the minus signs in this
formula. For more details on bi and b„_i see Example 8.8.2.)
Now we claim, for arbitrary Vi, ..., Vf,_i
bit;i A • ■ • A bit;„_i = b„_i(t;i x • • ■ x t;„_i).
Indeed, testing both sides of this equality on ef, for 1 < i < n, and applying
Definition 8.6.2, we obtain
det({vi,,ej)) i<i,^^„ = (-l)'~'det(e,-t;i •■■ t;„_i) = (-l)'~'(t;i x-■-xt;,,-!),-,
which agrees with Formula (5.2) for the components of the cross product. This
proves the claim. ik

8.6. Differential forms and Stokes'Theorem 571
In the following we pointwise apply Definition 8.6.1.
Definition 8.6.6. Let U C R" be an open subset and Q <k <n. We consider
k
1[/\t:u,
the disjoint union over all x e f/ of the k-Xh exterior powers /\ T*U of the dual
spaces T*U associated with the tangent spaces TxU = R" to U at x. The fact
that TxU = R" follows from Theorem 5.1.2, because dim U = n. Here we ignore
the fact that all these exterior powers actually equal the same space /\ R"*, taking
for every point x e f/ a different copy of this. A differential k-form to on f/ is a
mapping
0)
U ^]J/\ t;u with (oix) € /\ T*U (x € U).
xsU
A differential k-foim on U therefore acts on a A;-tuple (t;i,..., t;^) of vector fields
V; : U ^ R", if the V; are considered as mappings U B x h> ^i(x) € R" = T^U.
Hereafter, a differential A;-form on U will frequently also be made to act on a A;-tuple
of vectors (vi,..., Vk); these t;,- e R" then are to be regarded as translation invariant
vector fields on U, that is, as constant vector fields U B x h> Vi € TxU.
We write
Q''(U)
for the linear space with respect to the pointwise operations of differential A;-forms
on U. Further, m is said to be a C' differential A;-form on U, with / > 0, if
X i-> m(x)(vi, ..., t;^) : f/ —>■ R is a C' function on U, for all vi,... ,vic € R".
The derivative of this function is an element of Lin (R", R) that we denote by
vo i-> Da)(x)(vo)(vi, ...,Vk). (8.38)
O
Definition 8.6.7. Assume D C R'' and U C R" are open subsets and 0 < k < n.
Let (p € C^D, U) and a, € ^*^(f/). Then (J)*cl, € ^*^(D) is defined by, for y € D,
V, € TyD ~ R'',
i(j>*cS)iy)iv,, ...,Vk)= coi<P(y))iD<P(y)vi,..., D(j>iy)Vk).
The mapping <p* e Lin {QJ^{U), Q'^(D)) is said to be the operator of pullback under
the mapping <p. O

572 Chapter 8. Oriented integration
Let V C R''be open and let / e C^(U, V). Applying the chain rale one readily
proves
(/ o 0)* = r ° /*■ (8.39)
A generalization of Definitions 8.1.1, 8.4.3 and 8.5.2 is the following:
Definition 8.6.8. Assume 0 <k < n, and D C R*^ and U C R" are open subsets.
Let (j) e C^(D, U) and let m e Q!^{U) be a continuous differential A;-form. Then
define /, u), the oriented integral of u) over <p, by
! c= i (ro))(y)(e^i,...,k))dy = I cD{(j>{y)){D,(j>{y),..., Dk<P(y))dy,
J(i> Jd Jd
provided the integral in the last term converges. Here (ei,..., ek) is the standard
basis for R*^. O
Remarlt. The value of the oriented integral of m over (p does not change upon a
reparametrization with positive orientation of the set D. That is, if 'P : D —>■ D is
a C' diffeomorphism with det D^P > 0, then
J(j> J(j>0^
oi. (8.40)
Indeed, application of Formula (8.34) to ni'i'iy)) e f\!^ T^{y)D yields
^*rj(y)(e^,_k)) = rj(^(y))(Di^(y),..., Dk^(y))
= 7?(«P()>))(£(,....,,))I det D«PW|.
Thus the assertion follows from the Change of Variables Theorem 6.6.1.
Further, let V C R'' be open, let / e C^(U, V), and m e Q''(V) a continuous
differential A;-form. An immediate consequence of Definition 8.6.8 and (8.39) then
is
f 0)= f f*co.
J fo<f> J (j>
Finally, assume in particular that 0 is a C' embedding. Then X = <p (D) is a C'
submanifold in R" of dimension d, on account of Corollary 4.3.2. If 0 : D —>■ R"
is another C' embedding with X = <p{D), it follows from Lemma 4.3.3.(iii) that
vp = 0-' o 0 is a C' diffeomorphism, with <p = <p ° ^- Under the assumption
det £>(0~' o <p) > 0 one verifies, analogous to the proof of Formula (8.40),
J<P J(t>
0).

8.6. Differential forms and Stokes'Theorem 573
In this case, therefore, f^ m, the oriented integral of o) over the manifold X, has
been defined by
Jx J(t>
CO.
We are now in a position to prepare the proof of Theorem 8.6.10 below, known
as Stokes' Theorem; this proof is a direct generalization of the proof of
Proposition 8.1.12.
Let A; > 1; assume that D C R*^ and U C R" are open subsets, and that
0 € C^(D, U); finally, let to be a C' differential (k — l)-form on U. In the notation
from Definition 8.6.3 consider fi : D ^ R,for I < i < k, given by
fiiy) = (ct>*a))(y)(er) = oy{(j>{y)){D,(j>{y),..., D^),..., Dk4>{y)).
Then we have, with Deo defined in (8.38) and by using Proposition 2.7.6,
Difiiy) = Dco{(j>{y)){Di(j>{y)){Di(j>{y),..., D^),..., Dk,p(y))
+ Yl oj{<j>(y)){Di<j>(y),...,D;^),...,D,Dj<P(y),...,Dk<P(y)).
In the last sum the summation is over j e {I,... ,i — 1,1 + I,... ,k}, while
DiDj(p(y) can also occur to the left of Di(p(y), specifically, if j < i. Applying
alternating summation over I <i <km order to ensure antisymmetry, one obtains
J2(-iy''D!fi(y)
J2 (-ir'Doj{<j>(y)){Di<j>(y)){Di<j>(y),..., D^),..., Dk<P(y))
l<i<k
l<i<k
+ ^(-1)- j:
l<i<k l<j<k,J^i
oj{<j>(y)){Di<j>(y),..., D^),..., D,Dj(j>{y),..., DuHy))
: Y, (-ir'Doj{<j>(y)){Di<j>(y)){DMy), ■■■, D^),..., Dk<P(y))
\<i<k
\<i<j<k
co{(j>{y)){DiDj(j>{y) - DjDi(j>{y), D^y), ■■■, D^),..., Dk,p(y))
J2 {-iy''Doj{(j>{y)){Di(j>{y)){DMy), ■■■, D^Hy),..., DkHy)),
\<i<k
(8.41)

574 Chapter 8. Oriented integration
on the strength of Theorem 2.7.2. Note that
induces a mapping belonging to hin''(T^U, R).
Definition 8.6.9. On account of Lemma 8.6.13 below, the mapping in Lin*^ (TjcU, R)
given by
(vu...,Vk) i-> ^ (-iy~^ Da)(x)(Vi)(vi,... ,vi,..., Vk)
l<i<k
belongs to /\ T*U; we denote it by
(t;,,..., Vk) i-> dM(x)(vi,...,Vk).
Here dco e Q'^(U) is said to be the exterior derivative of co € ^*^~'(f/). The
operator
d€Uni^''-\U), Q''(U))
is known as exterior differentiation. We say that m e Q''~^(U) is closed on U
if dco = 0. Furthermore, rj e Q^(U) is said to be exact on f/ if there exists an
CO € ^*^-' (f/) with ?? = rfftj. O
By the use of Definitions 8.6.9 and 8.6.7, Formula (8.41) may be converted into
the following equality of functions, valid for y € D:
d(ct>*c^)(y)(e^i_k)) = Yl (-iy''D,(ct>*c^)(y)(ef)
isisk (8.42)
= dco{(j>{y)){Di(j>{y),..., DkHy)) = (r(dco))(y)(e^i,...,k)).
Note that (8.42) implies the following identity of differential A;-forms on D:
d(<j)*co) = <j)*(dcoy, (8.43)
in other words, exterior differentiation and pullback to D C R*^ commute, acting
on^*^-'(f/).
Now let 7*^ = [ 0, 11*^, a closed A;-dimensional hypercube in R*^, and assume that
(p : l'' ^ R" is the restriction of a C^ mapping D —>■ R" for an open neighborhood
D of l''. Integrating the equality between the second and the fourth term in (8.42)
over l'', changing the order of integration, and applying the Fundamental Theorem
of Integral Calculus on R we obtain
J2 ^±(-l)'-'/,-.±= f (r(d(^))(y)(eii,....k))dy. (8.44)

8.6. Differential forms and Stokes'Theorem 575
Here we have
//,±= / (((t>i,±)*(^)(yr)(ef)dy^, where
0/,± : yr^ <i>(yi,---,yi-u ^ ,yi+u ■■■,yk) ■■ i'"'^ -> u
is a parametrization of the smooth parts 9,-±0 of the z-th {n — l)-dimensional faces
of 9(0 (7*^)). Further, the direction of the outer normal to 9,_±0 is given by
±sgn(detD0)(-l)''~'Di0 x ••■ x D^ x •■■ x Dk<i>.
Indeed, the inner product of this vector with :^Di<p is positive (compare with
Exercise 7.59.(ii)). This enables us to recognize the left-hand side in (8.44) as /g. u),
and the right-hand side as /, doi; thus we have now obtained:
Theorem 8.6.10 (Stokes' Theorem). Let 0<k <n, let l'' = [0,lf C R^ let U
be open in R", and let <p € C^{I^, U); in addition, let m e Q''~^ (U) be a C^-form.
One then has
Jd<P J<p
Note that, moreover, a limit argument yields:
do).
dtp
Proposition 8.6.11. The definition of co \-^ dco : Q!^ '(f/) —>■ Q!^{U) is
independent of the choice of coordinates in R".
The following theorem is a generalization of Formula (8.43). It is yet another
expression of the fact that the exterior differentiation d is invariant under coordinate
transformations; but it is more general, because the mapping / below need not be
a diffeomorphism.
Theorem 8.6.12 {d and pullback commute). Assume U C R" and V C R'' are
open subsets, and f € C^(U, V). Let m e Q''~^(V) be a C^-form. Then we have
the following identity of elements in QJ^{U):
d(f*oj) = f*(doj).
Proof. It suffices to show that, for all (p e C^(l'', U),
f(dof*)a)= f(rod)a).
J<P J(i>
By means of Definition 8.6.8 and Formulae (8.43) and (8.39) we find

576 Chapter 8. Oriented integration
I (do f*) CO = I (p*o(do f*)a)= I do((P*o /•*) CO
J(l, Ji>' ' Ji>'
= I do(fo(j>)*CO= f (fo(j>)*odcO
= j^^r°{r°d)co = j{rod)co. □
The next lemma is a special case of Formula (8.50) below, which provides a
procedure for antisymmetrizing multilinear forms.
Lemma 8.6.13. Let rj e Lin*(V, R) be antisymmetric in the last k — \ variables.
\ V*, if we define
Ar){vi,...,Vk)= Y^{-iy~'^r)i'*^i,vi,...Si,...,Vk). (8.45)
Then Arj € /\ V*,ifwe define
\<l<k
Proof. We prove Ar){v\,..., t;/_i, v, vi+\,..., t;„,_i, v, v,„+i, ...,Vk) = 0, for
1 < I < m < k. In view of the antisymmetry with respect to the last k — \
variables, the summands in (8.45) with index i different from / or m vanish in this
case. The summands with indices / and m, respectively, are of the form
(-1)'"' r](v, vi,..., t;/_i, t;/+i, t;/+2, ■ ■., t^n.-i, v , v,„+i,..., Vk),
(8.46)
(-1)'" 7]iv, Vi,..., V,_i, V , V,+ i, ..., V,„_2, t^,«-l, t^m + 1, . ■ . , t^fc).
Ignoring sign, the first term in (8.46) can be obtained from the second by
transpositions of neighbors; to achieve this, the element v has to be carried from the l-th
position to the (m — l)-th position. This requires m — l—l transpositions of neighbors;
consequently, the sign of the second term becomes (— i)'"-'-('"-1 -0 = _(_!)'-'□
8.7 Properties of differential forms
We begin with a result in linear algebra. Let A e Mat(«, R). The computation
of det A using expansion by a row or a column of A is well-known. It is possible,
however, to expand det A by several rows, or columns, simultaneously.
Indeed, let 0 < A; < «. For / = (z'l,..., 4) e ik as in Definition 8.6.3,
introduce |/| = Ylispskh^ ^^ ^' = (i'l^ ■ ■ ■ J'n-k) ^ ^"-k by {m,...,4} U
{Zj,...,/^,_j(.} = {!,...,«}. Furthermore, for / and J e ik, write Ajj e
Mat(A;, R) for the matrix whose (p, ^)-th entry equals fl/pj,, for 1 < p,q < k;
then Ajiji e Mat(« — k,R). We now claim that there is the following expansion
by the rows of A labeled by / e 1^:
det A = (-1)1^1 Yli-iy-'^ det Aij det^ry-. (8.47)
7 sit

8.7. Properties of differential forms 577
In order to verify this, denote by aj e R" the y-th column vector of A. Next
define fl'- e R" byfl;- = Ujj (i.e. thez-thentryofaj-)if ^ ^ {M, • ■ •, ik }> and a-. = 0
otherwise. Finally, set a". = aj — a', e R". Observe that the linear subspace
spanned by the a', and the a"., for 1 < j < «, is of dimension < k and < n — k,
respectively. Because det e Lin"(R",R) is antisymmetric, the decomposition
aj = a'j +a'j leads to the decomposition det A = Xiysit ^^^ ^^^' where the entries
a'J' of a'-' e Mat(«, R) are given by
I(ip, jq), for some I < p,q <k\
{i'j., j^), for some 1 < r, s < n — k;
and a'^ = 0 otherwise. In order to compute det A''^, let us shuffle the rows and
columns of A''^ so as to place Ajj e Mat(A;, R) as defined above in the upper left
corner. To this end we have to perform
(M - 1) + ■ ■ • + 0, -k) + O", -l) + ... + (j,-k)^ \I\ + \J\ mod 2
permutations. The lower right corner of A''^ then is made up by Aj/j/, while the
other entries equal 0.
Lemma 8.7.1. Let V be a vector space of finite dimension. Then there exists a
unique bilinear mapping
k I k+l
/\V* X /\V* ^ /\V* with (CO, r])^^a)Ar],
which satisfies, for I € Ik and J € li,
e*Ae*=(elA---A e?) A (^ A • • ■ A e*) = e* A ■ • ■ A e* A ^ A • • ■ A e* .
The operation A is known as the exterior multiplication. It is associative and
anticommutative, which means that t] am = (—\)'''m a r).
Proof. In order to meet the conditions we define, for linear combinations of basis
vectors (see Lemma 8.6.4),
\y^M,e]\A[y^r]jey[= ^ M,r]je]Ae*.
The last assertion follows by successive application of e^ A ef = —ef A e'j. What
is left to prove is that the definition of exterior multiplication does not depend on
the choice of the basis in V. To this end we obtain Formula (8.48) below for mat],
where m = ef A • • • A e* and r] = e*- A- ■ ■ Ae*-. Consider t;i, ..., Vk+i € V. Then,
by Definition 8.6.2,
(MAr])(vi, ...,Vk+i) =deti4.

578 Chapter 8. Oriented integration
where the ^-th column vector of A e Mat(A; + /, R) is given by
(Viic,,..., Vi^q, Vj^q, ■■■ , vj,^y (I <q <k + l).
We compute det A using expansion by the top k rows as in Formula (8.47), thus we
find, with P = (\,... ,k), and 1 < ^i < • • • < qk < k +1 if Q € Ik,
(CO A n)(vu ..., Vk+i) = (-l)i^(^+') J2 (-1)'^' det ApQ dst Ap,Q,
Qsik
(8.48)
= (_l)^^(^+i) J2 (-l)ieia)(t;„,..., V) '/(V,' ■ • ■' \')-
esii.
This shows that the definition of to A ?; is in terms of m and ?; only, and does not
depend on the choice of the basis (ei,... ,e„) of V. □
Remark. The permutation
/I ... k k+\ ... k + l \ _
\qi ... qk qi ... q, J
where Sk+i is the permutation group onk + l elements, is said to be a shuffle. Such a
permutation describes a possible way of shuffling a deck of k cards through a deck
of / cards, placing the cards of the first deck in order in the positions qi,..., qk
and those of the second deck in order in the positions q[,.. .,q[. The set of all
shuffles in Sk+i is denoted by Sk,i and consists of ( j^ ) elements. Furthermore,
sgn(o-g) = (-l)i*^(*^+i)+iei. Accordingly Formula (8.48) takes the form
(ftj A ri){vx,..., Vk+i) = Y^ sgn(o-) (o) o a)(vi,..., t;^) (?? o a)(Vk+i,..., Vk+i).
(8.49)
Here cr(vi,. ..,Vk) = (Va{i), ■■■, v^^k)), etc.
The results above are used in the preliminaries for the proof of the next theorem.
Wewrite/\ ' V*forthe linear subspace of Lin'^"'"'(V, R) consisting of the mappings
that are antisymmetric with respect to the first k variables as well as the last /
variables, and similarly we introduce /\ ''"' V*. Next define (see Lemma 8.6.13
for the case ofA;= 1,1 = k — 1 and m = 0)
A,., € Lin (/\ V*, /\ V*) by A^.^oj = ^ sgn(or) oj o a, (8.50)
where cr(k + I + i) = k + I + i, for 1 < i < m. The associativity of the
exterior multiplication from Lemma 8.7.1 implies the commutativity of the following

8.7. Properties of differential forms 579
diagram:
k,l,m /^. k,l+ni
yy y*^^ yy y*
Au
At,,+„, . That is, Ak^,^„i o A^j = AkJ_^_,„ o Ai,„.
k+i"i At+,„, k+l+ni
yy y* ^ yy y*
(8.51)
Theorem 8.7.2 (d^ = 0). Z^f U be open in R". For d acting on C^ forms we have,
for A; e N,
0 = d^ : Q''-\U) ^ Q''+\U).
In particular, in the terminology of Definition 8.6.9 we have that every exact
differential form is also closed.
Proof. Consider a C^ form o) € ^*^-'(f/) andx € U. From (oix) € A*""' T*U
it follows that Da)(x)vi e /\''~^ T*U too, for a given ni e r^f/. According to
Definition 8.6.9 we find da)(x) e /\'' T*U by application of Aj ^t-i to
i,fc-i
Da)(xy € /\ r;f/, £>ftj(xnt;i, ...,Vk) = Dm(x)(vi)(v2, ..., t;^).
We may differentiate the resulting identity dcoix) = Ai^k-iDcoix)" at x in the
direction of a fixed vector t;i e T^U. In view of Lemma 2.4.7 we have that
D(dct))(x)v\ e A T^*^ is the result of application of Ai^k-i to the mapping in
A^''''"^7:;f/given by
(t;2, ■ ■., Vk+i) i-> (Z)^ftj(x)(t;i)t;2)(t^3, ■ • •, Vk+i).
On account of Lemma 2.7.4 we may write this mapping as
D'^coix)" : (V2,..., Vk+i) i-> D'^a)(x)(vi, V2)(v3,..., Vk+i),
where D^a)(xy €/\^'^''"^ T^U.
Now it follows that Ai^k-iD^(^(xy € /\' is the element which on evaluation on
(t;i, • • •) equals D(dct))(x)v\. Again by Definition 8.6.9, application of Ai^k to this
element gives d(dM)(x), that is
d'^ioix) = Aik o Ai^k-iD'^a)ixy = A2,k-i o AijD'^a)(xy,
where we used Formula (8.51). Finally, Theorem 2.7.9 implies
AijD^M(xy(vi,..., Vk+i)
= D^a)ix)ivi, V2)iv3,..., Vk+i) - D^a)(x)(v2, Vi)(vj,..., Vk+i) = 0. '^

580 Chapter 8. Oriented integration
Proposition 8.7.3. d is an antiderivation, that is, for C' forms co e Q''(U) and
t] € Q'(U) we have
d{u> AT]) = du> /\r] + {—\)'^(i> A drj.
Proof. As in Definition 8.6.9 and in the proof of Theorem 8.7.2 we have
\,k+l
dia)Ari)(x) = Aik_^_iD(a)Ari)(xy, where D(a) A7])(xy € A T*U
(8.52)
is given by D(a) A ri)(xy(vi,..., Vk+i+i) = D(a) A r])(x)(vi)(v2,..., Vk+i+i).
Accordingly we now consider D(m a v)(x)vi € A''^' T*U, for fixed t;, € T*U,
and note that, on account of Proposition 2.7.6,
D(ct) A ri)(x)vi = Dct)(x)vi A ri(x) + a)(x) A Dr](x)vi
= Da)(x)vi A r](x) + (-l)'''Dr](x)vi A m(x),
the exterior multiplication being anticommutative by Lemma 8.7.1. Thus, we obtain
D(a) A ri)(xy = Dccixy AYjix) + {-l)"Dr]{xy Au,{x). (8.53)
Because it is an exterior product of elements in /\ ' T*U and /\ T*U, we have
\,k,l
DM(xy A r](x) = Akj{Doi{xy ■ r](x)), with Dcoixy ■ ri(x) e /\ T*U.
Formula (8.51) now implies
Ai^k+i(Da)(xyAri(x)) = Ai^k+i o Ak,i(Da)(xy ■ r](x))
= Ak+i,i o Ai^k(Da)(xy ■ 7](x)) = Ak+i,i((da))(x) ■ r](x)) = (da))(x) A r](x).
Exactly the same arguments with the roles of k and / interchanged apply to the
second summand in (8.53); hence we get, using (8.52),
d(a) A ??) = (do)) A ?? + (-if'idTi) Aa)= (do)) A ?? + (-l)*^'+*^('+')&j a (dTj). □
Definition 8.7.4. We introduce a standard notation for differential A;-forms on an
opensubset f/ofR". Ifx,- : U —>■ R is the z-th coordinate function. Definition 8.6.9
implies that dxi(x) e T*U is constant, equaling the i-th standard basis vector
ej- e R"*,for 1 < i < wandx e U. Formula 8.36 therefore implies that to e Q''(U)
can uniquely be written as
0) = \2a)jdxj with (Oj e ^°(U), x i-> ct)(x)(ej) : f/ —>■ R,
dxj = dxji A • • • A dxi^ if I = (i\,..., ik).

8.8. Applications of differential forms 581
Furthermore, we define (compare with Definition 8.6.3)
e =(ei,...,e„)€(R")", ^= (e,,... ,1?,..., e„) € (R")""',
dx = dxi A • • • A dx„ e ^"(U),
dxj- = dxi A ■ ■ ■ A dxj A ■ ■ ■ A dx„ e ^""^(11). '^
Corollary 8.7.5. For U open in R" and f € Q°(U) a C^ differential form, one has
df=J2 Difdxi €^\U).
l<i<n
More generally, for m as in Definition 8.7.4 which is a C' differential form,
do) = Y^ dcoj A dxj = Y^ Y^ DjCDj dXj A dxj e ^'^^^{U).
Islk Islk l<i<n
Here doij € ^'(f/) is the exterior derivative ofoij e Q,°{U).
Proof. According to Definition 8.6.9 one gets df(x)(ei) = Dif(x), for !</<«,
and this implies the formula for df. Successively using mathematical induction over
k e N, Proposition 8.7.3 and Theorem 8.7.2, one derives/i(/ix/) = 0, for all/ e Ik.
Hence the first equality in the formula for do) follows from Proposition 8.7.3, while
the second equality follows from the formula for dcoi. □
8.8 Applications of differential forms
Example 8.8.1 (Exterior derivative of differential 1- and (n — l)-form). For a
C^form
CO = 2, (^jdxj ^Q}{U), one has dco = 2. DjCOj dxt Adxj.
In the notation of Definition 8.7.4 we have (—l)'~'rfx/ A dxf = dx. For m e
Q"''^(U) we define (—l)'~^ftj,- = o)(ef), which yields
CO = y^ (—iy''^cOi dxf, and so dco = i 2, DjCOjjdx. liV

582 Chapter 8. Oriented integration
Example 8.8.2 (Relation between vector fields and differential forms). We apply
the results from Example 8.6.5 in order to define b„_i and b i acting on vector fields.
Thus, to a C' vector field / : f/ —>■ R" we may, on the one hand, assign the C'
form
b„_if = ifdet€Q"-HU), 0>det)(t;,,...,t;„_,) = det(/t;, ••• t;„_,),
for arbitrary vector fields t;i, ..., t;„_i on U. As in Example 8.6.5, we find
b„_,/ = J2 (-!)''"'./;■ ^^ e ^"-'(f/);
SO
^(b„_,/) = b„(div/) := diy fdx € ^"(U),
by the preceding example. Moreover, for y e 9,-_±/" (see the proof of Stokes'
Theorem 8.6.10), we see that
r(K-if)(y)(ef) = (ifdst)i<P(y))iDi<P(y),..., oi^),..., D„<P(y))
= det(/ o <P(y) Di(j>{y) ■ ■ ■ oi^) ■ ■ ■ D„<P(y))
= {f o(j), Di4> X ■■■ X D;4> X ■■■ X D„(j)}(y).
Furthermore, for X € /",
0*(rf(b„_i/))(x)(e) = div/(0(x))rfx(£),0(x),..., D„,pix))
= (div/)o0(x)det(Z)i0, ...,Z)„0)(x) = (div f) o(j>(x)detD(j>(x).
Thus Gauss' Divergence Theorem 7.8.5 is seen to be a special case of Stokes'
Theorem 8.6.10.
On the other hand, to a C' vector field / : f/ —>■ R" we can assign the C' form
^i/= J2 fid^i^^\U).
\<i<n
Then, in the notation of Formula (8.4),
d(bif) = 2. DjfidxjAdxi= 2. (Djfi — Difj)dxjAdx!
In particular, for « = 3,
di\>lf) = (Dif2 - D2f\)dXi AdX2 + (1)2/3 - D^f2)dX2Adx^
+(D3fi - Difj) dx-i A dxi.

8.8. Applications of differential forms 583
That is
^(h/) = b2(curl/),
because ^zC^) = ^i '^•^2 ^ ^^3 + 82 ^^3 ^ ^^i + ^3 '^•^1 ^ dx2.
Consequently, Stokes' Integral Theorem 8.4.4 also is a special case of Stokes'
Theorem 8.6.10.
We summarize the situation in R^ in the following table:
function g vector field gradg bi(gradg) = dg;
vector field/ vector field curl/ bzCcurl/) = d(bify, (8.54)
vector field/ function div/ b3(div/) = d(b2f).
These identifications are valid only in the standard coordinates on R^ and use the
orientation of R^. We conclude that
b2(curlgradg) = d'^g = 0, bsCdiv curl/2) = d'^ibih) = 0;
therefore the identities from Formula (8.17) follow from the identity d'^ = 0 (see
Theorem 8.7.2).
The classical notation for a vector field is t; = t;' t^, and for a differential form
3.V'
0) = Mjdx', where according to the Einstein summation convention the summation
is carried out over those indices that occur as subscript and superscript. Upon the
transition from v to m the indices ' of the coefficients are lowered to /; hence the
notation b (a character indicating a half step drop in pitch) for these "musical"
isomorphisms. ik
Example 8.8.3 (Special case of Brouwer's Fixed-point Theorem). Assume that
U is open in R", that (p e C^{I", U), and that K := (j)(I") has a nonempty interior.
Let g : U ^ dK be a C^ mapping. Then the restriction gl^i^ of g to dK cannot
equal the identity ondK. This result implies, among other things, that a membrane
cannot be retracted onto its boundary without being punctured somewhere.
Indeed, let x = (xi,..., x„) be the coordinate mapping on R" and let g =
iSi' ■ ■ ■' 8n)- Consider the integrals
/ Xidx2 A ■ ■ ■ Adx„ and / gidg2 A ■ ■ ■ Adg„.
Ja(t> Jd<p
If g(x) = X, for all X e dK, then the two integrals are equal. By Stokes'
Theorem 8.6.10 and Corollary 8.7.5 it then follows that
vol(ii:)
= / dx = I dX\ A • • • A dx,i = I dg\ A • • • A dg„.
J tp J tp J tp

584 Chapter 8. Oriented integration
This results in a contradiction, because the left-hand side does not vanish, whereas
the right-hand side does. Indeed, we have gj = x; o g = g*Xj, for 1 < z < «. By
Theorem 8.6.12 we find dg; = d(g*Xi) = g*(dxi). Thus, for x € U and for any
«-tuple of vectors Vi,..., v„ € T^U — R",
(dgi A • • • A dg„)(x)(Vi, ...,V„) = {g*(dXi) A • • • A g*(dx„))(x)(Vi, ..., V„)
= (dxi A • • • A dx„){g(x)){Dg(x)vi,..., Dg(x)v„)
= det(Dg(x)vi ■ ■ ■ Dg(x)v„) = 0,
since the n vectors Dg(x)vi,..., Dg(x)v„ are elements of the (n — l)-dimensional
linear subspace Tg^^^(dK), and therefore linearly dependent.
Now assume A!' to be a convex open set, and let B be its closure. Every C^
mapping f : U ^ R" which maps B into itself has a fixed point in B, in other
words, there exists anx € B with f(x) = x. Indeed, if x ^^^ f(x) for all x € B,
one can assign to x the unique point of intersection g(x) with dK = dB of the
half-line from f(x) to x. The mapping g : B ^ dK thus defined can be extended
to a C^ mapping g : U ^ dK for an open neighborhood U of B, but this leads to
a contradiction with the foregoing. ik
Example 8.8.4 (Jacobi's identity for minors). Let U and V be open in R", let
<p : U ^ V be a C^ mapping and denote the (z, i)-fh minor of the matrix of D<p {x)
by <pij{x). Then we have Jacobi's identity for minors:
^ (-iy£>;0,; = 0 (!</<«).
Indeed, consider the standard volume form dy = dy\ A • • • A £?}'„_i € ^"""^ (R"""^).
Fix I <i <n and write f,- = (0i,..., 0,-,..., 0„) : U —>■ R"""^ In the notation of
Example 8.8.1 we have
^:idy)= Y, ^;(dy)(ej)dxj€Q"-\U).
By Definitions 8.6.7 and 8.6.2,
^*(dy)(ej) = dyiD^ieu ..., D^j,..., D^ie„)
= dy(Di^i,..., D~li,..., A,f,) = det(£>,f,- ■ ■ ■ SJlt ■ ■ ■ D„^i) = (j>ij.
Now d(dy) = 0, since it belongs to ^"(R"~'), whence
0 = ^ndidy)) = dmdy)) € Q"(U)
by Theorem 8.6.12. Thus the assertion follows from Corollary 8.7.5 and the identity
dxj A dxj = (—l)^""'/ix. Note that, for « = 2, Jacobi's identity takes the well-
known form DiD2(t>i - I>2^i0/ = 0, for 1 < z < 2. ik

8.9. Homotopy Lemma 585
8.9 Homotopy Lemma
The formulation of Lemma 8.9.1 below requires a certain amount of preparation.
Let U C R" be open. Assume (<l>')/sR to be a one-parameter group of dif-
feomorphisms on U with C^ tangent vector field t; : f/ —>■ R", that is (see
Formula (5.21))
d
<D'(x) (x € U).
'^""^ dt
Then we have the induced mapping L„, the Lie derivative in the direction of the
vector field v, acting on the C' forms co e Q!^{U) (see Definition 8.6.7)
dt
(<D')*to e Ql'iU). (8.55)
r=o
It follows, by Theorem 8.6.12, that for all C' forms co e Ql'iU) and r) e ^'(f/),
and C^ forms u) e Q!^{U), respectively,
Lvico Arf) = {Lycci) AYj + CO ALyYj, and Ly{dcS) = d^L^cS). (8.56)
We say that L„ is a derivation.
We further introduce /„, the contraction with the vector field v, acting on ^'^ (U)
as follows. If / e ^°(f/), then i^f = 0, and if 1 < Jfc < «,
/„ : ^*^(f/) ^ ^*^""'(f/) with {iyO)){v2,...,Vk)=CL,{v,V2,...,Vk).
(8.57)
By expansion according to the upper row of the determinant {df Arj){v, v\,..., Vk),
for all / e C (f/) and ?; e Q''(U), where t;i,..., t;t are vector fields on U, we see
i^(df A ??) = (/„rf/)?? - rf/ A /„??. (8.58)
We claim that z„ is an antiderivation (see Proposition 8.7.3), that is, for C° forms
0) e ^'^(U) and t] e ^'(f/) we have
iy(a) At]) = i^o) Ari + (-l)''o) A i^r). (8.59)
Using Formula (8.58) we can prove this by mathematical induction over A; e N as
follows, for/ €C\U):
iv((df Aco) Arj) = iv(df A (co A rj)) = iv(df)ci) Arj — df A iy(a) A rj)
= iy(df)ct) Arj — df A ivO) Ari + {— 1)*^+'^/ Aco A i^rj
= iv(df am) ATI + (-l)^+^(fi?/ AO)) A i^T).

586 Chapter 8. Oriented integration
Lemma 8.9.1. We have the following homotopy formula as identity of linear
operators on C^ forms in Q''(U):
Ly = d O iy -\- iy O d.
In particular, the operators Ly and d commute, becauseL^od = doL^ = doi^od.
Note that for o) e ^'^(U) one has iyO) e ^^■"' (U), and hence divO) € ^'^(U),
while do) € Q''+^(U), from which iydco € Q''(U).
Proof, d is an antiderivation according to Proposition 8.7.3 and iy is an antideriva-
tion too according to Formula (8.59). But then Dy := diy + i^d is a derivation,
since for C' forms o) e ^'^(U) and t] e ^'(U),
D^(m At]) = d{i^M At] + (—lyo) A iyt]) + iy{dM At] + (—\)''m a drj)
= diyO) A t] + (—l)'^~^iya) Adt] + (—l)'^da) A i^rj + {—1)'^'^cl> Adi^r]
+iyda) At] + (—l)'^iya) Adt] + (—l)*^+'rfftj A iyt] + (—l)'^'^a) A i^dt]
= DyO) At] + 0) A D^r).
Set k = <d\ application of the chain rale to f i-> / o O' then results in the homotopy
formula, for / e C^{U). Furthermore, for m = df € Q^(U) one then shows, by
(8.56) and using d'^ = 0 (see Theorem 8.7.2),
Lyco = d(L,f) = d(d(iyf) + iy(df)) = d{iycc) = D,a). (8.60)
Hence we know that Ly and Z)„ are derivations agreeing on the elements of Q°(U)
and Q^(U), therefore they agree on those belonging to Q''(U), for 0 < A; < «. □
Example 8.9.2. Suppose a e Q" (R") is a C' form that vanishes nowhere and let v
be a C^ vector field on R". Then there exists a uniquely determined function diva v
on R", the divergence of v with respect to the volume form a, such that
L^a = div„t;Q! e ^"(R").
The homotopy formula from Lemma 8.9.1 then implies d(ivOi) = (div + ivd)(x =
LyO. = diva va. Further, suppose that v is the tangent vector field of the one-
parameter group of diffeomorphisms (<l>')/sR oil R"> ^iid consider arbitrary (p e
C^([ 0,1 ] ", R"). Application of Stokes' Theorem 8.6.10 to /„« € ^"-'(R") now
leads to the following analog of Formula (7.52):
— I / a = I diVa va = I lyOi.
Here the first identity follows from
d\ f d\ f , f d\ , f

8.9. Homotopy Lemma 587
Example 8.9.3. The theory of differential forms sheds another light on Step IV on
differentiation in the third proof of the Change of Variables Theorem in Section 6.13.
As in Example 8.8.2, the function g o S" det Z)H" on V corresponds to the form
{'Ei"y{gdy) € ^"(V). On account of the homotopy formula we then find, using
ih?iid{gdy) €^"+HV) = {0},
d
du
(3"r(gdy) = L^igdy) = (d o i^ + i^ o d)(gdy) = d o i^(gdy)
11=0
= d(gi^dy) = d(^ J2 (-iy-'g^idyr)=div(g^)dy.
(8.61)
Here we applied the formula from (8.37) that i^ dy = X!i<i<n(~^)'~'^''^^ ^
^"-^(V), and Example 8.8.1.
It is tempting to terminate the computation in (8.61) as soon as one encounters
d applied to an « — 1 form and to invoke Stokes' Theorem; this, however, would
lead to a circular argument, because the proof of that theorem uses the Change of
Variables Theorem. ik
Definition 8.9.4. Let U and V be open in R" and let Oq and Oi : f/ ->■ V be
two C^ mappings. Then Oq and Oi are said to be C^ homotopic mappings if there
exists a C' homotopy between Oq and Oi, that is, a C' mapping F : / x f/ —>■ V
satisfying
r(o,-) = <i>o, r(i,-) = <i>i.
Now assume that Oq and Oi are C^-diffeomorphisms. A C^ homotopy T between
Oo and Oi is said to be a C' isotopy if T{t, •) : f/ —>■ V is a C^ diffeomorphism,
for every f e /. O
Lemma 8.9.5 (Homotopy Lemma). Let U and V be open in R" and let <l>o and
Oi : f/ —>■ V be two homotopic C' mappings. Then, for I <k <n, there exists an
operator H = Hk, the homotopy operator, satisfying
Hk&Un{^{V),a''-\U)), ^-% = doHk+Hk+iod on Q''(V).
Proof. Suppose r:/xf/—>■ VisaC^ homotopy between Oq and Oj. Further,
let i, : f/ -> / X f/ be given by l,(x) = (t, x). Then F o j, = F(r, •) e C (f/, V),
for f e /. Let 0) e Q^(V). Application of the Fundamental Theorem of Integral
Calculus on R to f i-> (F o it)*(^ € Q''(U) (evaluated on k arbitrary vector fields
on f/, if necessary, to convert it to a scalar function) now gives
Jo dt'
^0) - %ay = (F o LyYay - (F o jo)*ftj = / —(F o L,Yaydt € Ol'iU). (8.62)

588 Chapter 8. Oriented integration
Next we introduce O'' : / x f/ ->- / x f/ by (i>''(t, x) = (t + h, x). Then
<D\r, x) = (1, 0,..., 0) =: eo € R X R".
d_
dh.
/i=0
Additionally we have i,_|_/, = O'' o j, : f/ —>- / x f/, and so
d_^^d_
dt''' ~ dh
h+h = h °
h=0
dh
C*")* = l; o L,„ : ^'^(7 xU)^ Q.'{U),
h=a
where Le^ is the Lie derivative in the direction of the vector field eo on 7 x f/. Using
the homotopy formula from Lemma 8.9.1 in Q!^{V) and Theorem 8.6.12, we find
^(r o L,T = r^i* j oT* = L*o Le, o r* = L* o(do 2,„ + /,„ od)or*
= doii*oie,or*) + ii*oie,or*)od : ^^(V)^ ^^(f/).
(8.63)
Formulae (8.62) and (8.63) then show that 77 is as desired, if we define 77^^ ^
Lin(^*^(V), Q''-^(U))hy
Jo
oi,. oF*) dt.
(8.64)
□
For application later on, in the proof of Theorem 8.11.2, we note a consequence
of the Homotopy Lemma (compare this result with Step IV on differentiation in the
third proof of the Change of Variables Theorem in Section 6.13).
Proposition 8.9.6. Let U and V be open in R" and let <l>o and <1) i : f/ —>■ V be
two homotopic C^ mappings, by means of F e C^(I x U, V). Consider a C^ form
0) e Q"(V) and a compact set K C U such that supp(r(f, -yo)) C K, for all
t € I. Then we have
Ju Ju
Proof. In view of the Homotopy Lemma 8.9.5 we have ^*m — OqCO = d{HcS),
with Ho) e Q.''~'^{1}). Hence Stokes' Theorem 8.6.10 gives
/ ^\ay - I %ay = I d(Ha)) = f Ha) = 0,
Ju Ju Ju Jdu
since supp(77ftj) Pi dU = 0. Indeed, the condition supp((r o l,)*o)) C K and
Formula (8.64) imply supp(77ftj) C K, while K DdU =Si. □

8.10. Poincare's Lemma 589
8.10 Poincare's Lemma
Next we prove the following generalization of Lemma 8.2.6: on contractible open
sets (see Definition 8.10.1 below), closed and exact differential forms coincide. In
addition, this result yields a different proof for the formulae from Lemma 8.2.6.
Definition 8.10.1. An open set U C R" is said to be contractible if there exists a
point x° e f/ such that the mapping x i-> x° on f/ and the identity mapping on U
are C' homotopic. In other words, there is a C' mapping F : I x U ^ U such
that r(0, x) = x° and r(l, x) = x, for all x e f/. O
Note that a star-shaped open set is contractible. A bounded open subset in
R^ bounded by two concentric spheres of different radii is simply connected (see
Definition 8.2.7) but not contractible.
Tlieorem 8.10.2 (Poincare's Lemma for differential forms). Assume U C R" to
be a contractible open set, ando) e Q''(U) to be a closed C^ form, that is, do) = 0.
Then m is exact on U, that is, there exists a C^ form t] e ^'^~' (f/) with m = dr).
Assume in particular that U is star-shaped and that x° = 0, in the definition of
being star-shaped. If
U> = 2_. ^i\,...4k dXji A • • • A dXi^, (8.65)
i<ii<"-<'t<"
then t] = Ho) is an example of a C' form having this property, where, for x € U,
.1
l<ii <-"<h<n
ii<—<it
■ 2, (~^y ^^i; dXji A • • • A dxij A • • • A dx;^.
Proof. We apply the Homotopy Lemma 8.9.5. In the case under discussion F o j, e
C^(U, U), for t € /; in particular, F o li is the identity on U, while F o jq is the
mapping on U with constant value x°. Consequently,
CO = (Fo LiYco - (F o LoYa) = d(Ha)) + H(da)) = d(Ha)),
as do) = 0. Hence we see that t] = Hm satisfies dt] = m.
If U is star-shaped with x° = 0, we can choose F(f,x) = tx. For m as in
Formula (8.65) we have
(r*<w/i,...,,j)(f,x) = ftj,-,,...,,-,(rx),
F*(rfx,-.) = rf(F*x,-p = ditxi.) = tdxi. +x,-. dt.

590 Chapter 8. Oriented integration
It follows that
iie,or*)a)it,x)
= ieo ^ (^ii,..Jk (f^) X] ^'' '^■^'■'^ A • • • A (Xij dt) A- ■■ A(t dxQ
= 2~1 ^n,...,ik(^^) Z_\ (—iy~^t''''^X;. dXj^ A • • • A dXi- A • • • A dXii..
l<i]<-"<ik <" 1 <j <k
Thus we find the desired formula for Hoi. □
Example 8.10.3 (De Rham cohomology). Differential forms are tools in the study
of topological properties of a submanifold in R". Here we take no more than a first
step towards elaborating this observation.
Let / e C^(V), with V a submanifold in R". If the values of / do not change
under small variations of the point at which the function is evaluated, then / is
constant on the connected components of V, see Proposition 1.9.8.(iv); and this
is the case if and only if rf/ = 0 on V. Consequently, the number of connected
components of V equals the dimension of the vector space
H\V):={f€Q.\V)\df = Q}.
One dimension higher, an analog of a function / as described above is a function
acting on curves, with the property that the values of the function do not change
under small variations of the curve along which the function is evaluated. In other
words, this now involves a C' form u) e ^^{V) whose integral along a mapping
y does not change under small variations of y. Natural differential 1-forms u) are
those of the form df with / e ^°(V), that is, u) is exact; and for these
fai=fdf =
J y J y
/(beginning y) - /(end y).
This value does not change under variations of y that leave the end points of im(y)
invariant. Henceforth we shall, for that reason, consider variations with fixed end
points. Or, rephrasing, we require f m = 0 for "small" closed curves y. And if
we choose y = 90, Stokes' Theorem 8.6.10 yields 0 = /g, <w = / do) for "small"
surfaces (p; and this is true of m if and only if do) = 0, that is, if m is closed. In
view of Example 8.8.1, exact forms m are always closed. As a consequence, the
vector space that codifies the information of interest with respect to this problem is
the quotient space
H^V) :={m€QHV) \dM = 0}/{M€QHV) \ m = df for f € Q°(V)}.
Continuing in this way, we define H''(V), the A;-th de Rham cohomology of V, for
k e No, as the quotient vector space
H''(V) :={m€Q''(V) \dM = 0}/{M€Q''(V) \M = dr]foTr] €^*^-'(V)}.

8.11. Degree of mapping 591
The H''(V) give a measure of the topological complexity of V. Thus, Poincare's
Lemma 8.10.2 asserts that H''(V) = (0) if V is a contractible open set. On the
other hand, dim H''{V) > 0 if V is a compact orientable submanifold of dimension
d without boundary. Indeed, if m = dt] € Q''(V), then
0)= ??=/?? = 0;
Jv JBV JVi
but we also have volj(V) > 0. liV
8.11 Degree of mapping
Definition 8.11.1. Let U and V be open in R". A mapping <1) : f/ —>■ V is said
to be proper if the inverse image of every compact set in V is also compact in U
(compare with Definition 1.8.5). O
We want to prove the following theorem.
Tlieorem 8.11.2 (Degree of mapping). Let U and V be open in R" and V be
connected, let ^ : U ^ V be a proper C^ mapping. Then there exists an integer
deg(<l)) e Z, the degree of the mapping O, with the property that for every C^ form
0) e Q"(V) with compact support
f <D*ftj = deg(<D) f 0). (8.66)
Ju Jv
We say that y e V is a regular value for <1) if <1) is regular at every point
X e O'^ly}) C U, that is, if Z)<l)(x) e Aut(R") (see Definition 3.2.6). A local
version of Formula (8.66) in a neighborhood of a regular value is easy to prove; this
proof also makes it clear why deg(<l)) is an integer. Note that y isa regular value
if y ^ ^(U), and that, according to Sard's Theorem (see Exercise 6.36), the set of
singular values for <1) is a negligible subset of V.
Lemma 8.113. Let the notation be as in Theorem 8.11.2. Let y € V be a regular
value for O. Then there exists a neighborhood Vq ofy in V such that Formula (8.66)
holds for every m with compact support contained in Vq.
Proof. On account of the Submersion Theorem 4.5.2, the set <!>""'({>'}) is a sub-
manifold of dimension < 0, and therefore a discrete set in U; and this set is finite
(say m e No elements) owing to <1) being proper. Because of the Local Inverse

592 Chapter 8. Oriented integration
Function Theorem 3.2.4 there exist disjoint connected open sets f/,- C U and an
open set Vq C V such that
(D-i(Vo) = [J Ui, ^\^_ : U; -^ Vq C^ diffeomorphism (1 < / < m).
l<i<m
(8.67)
If the support of m is contained in Vq, the support of ^*m is contained in <1)~^ (Vq),
thus
JU ,^; JU
But in view of (8.67), the Remark following Definition 8.6.8 yields f^((i>\^)*ct) =
CTj fy 0), where cr,- = ±1, according as det D{^\ ) ^ 0. Hence we find
deg(<D) = ^ a/ € Z. (8.68)
l</<m
In particular, therefore, deg(<l)) = 0 if j* ^ <l*(f/). □
Next we show, by deformation arguments and a partition of unity, that the general
case of Formula (8.66) can be reduced to the special case from Lemma 8.11.3. To
show this we first derive the Isotopy Lemma; here we recall the notion of isotopy
from Definition 8.9.4.
Theorem 8.11.4 (Isotopy Lemma). Assume U to be a connected open set in R",
and let x° and x' e f/. Then there exists a C^ isotopy F : / x f/ —>■ f/ with
r(0,x) = X, for all X e U, andr(l,x°) = x^, while outside a fixed compact
subset of U the F (t, ■), for all t € I, equal the identity.
Proof. If the assertion holds for a pair x° and x^ € f/, we say these are isotopic. This
defines an equivalence relation between the elements of U. We shall demonstrate
that every equivalence class is an open set. Then f/ is a disjoint union of open sets,
and in view of the connectedness of U there can only be one such class. As a result,
it suffices to show that the assertion of the lemma holds for x ^ lying in a sufficiently
small ball B C U about x°.
We may assume that x° = Oandx' = (x,\ 0,..., 0), which may require a prior
translation and a rotation in R". Let x '■ U ^ / be a C^ function which is 1 at x°
and which vanishes outside ^B. Define, for x e U and t € I,
F(f,x) =x + tx(x)(x^ -x°) = i-xi +tx(x)xl,X2,...,x„) e U.
That is, F(0, •) is the identity on U; in addition, F(f, •) is the identity on U outside
B; and F(l, x°) = x'. We now have
£>iF,(r,x) = l+r£>,x(x)x;,

8.11. Degree of mapping 593
while X i-> Dix{,x) is a bounded function on B. It follows that we can arrange, by
taking \x\\ small enough, that Xj i-> Ti{t, x) is monotonically strictly increasing.
for all f e / and X2,... ,x„. Consequently, sX\T{t, •) are C diffeomorphisms. □
Proof, (of Theorem 8.11.2.) Assume y° & V to be a regular value for the mapping
<1> : f/ —>■ V and Vq to be an open neighborhood of y°, as in Lemma 8.11.3.
Application of the Isotopy Lemma 8.11.4 to the connected set V yields, for every
y € V, a C^ diffeomorphism (p,, : V ->■ V such that ^PyCy") = y and (p,, is C^
isotopic to the identity on V. The collection {^y{Va) | y € V } forms an open
covering of supp(ftj). In view of the compactness of supp(ftj) there exists a C^
partition of unity {xj I 1 < y < /} subordinate to this covering. By changing over
to xj<w we may assume the support of u) to be contained in an open set ^y{Va),
for some y e V. Because the identity on V is C^ isotopic to ^y, the mappings
<1) and ^y o ^ : U ^ V are C^ homotopic. Furthermore, it follows from the
Isotopy Lemma and the properness of <1) that the condition on the supports in
Proposition 8.9.6 is satisfied. Hence, this proposition yields
Ju Ju Ju
The support of >I'*ftj is contained in Vq, and so we find, by Lemma 8.11.3,
f <D*(«P» = deg(<D) f «P>.
Ju Jv
For the diffeomorphism ^P,, : V —>■ V one observes that det D'i^y > 0, because ^P,,
is C^ isotopic to the identity on V; hence, by the Remark following Definition 8.6.8,
/■«P>= /■
Jv Jv
CO. □
Example 8.11.5 (Degree of polynomial and Fmidamental Theorem of Algebra).
Let p : C —>■ C be a complex polynomial function of degree n, that is
pit) = ^ Ckz'', Ck € C, Cn 7^ 0.
Then the degree of p as a polynomial equals the degree of p considered as a mapping
R^ —>■ R^. This is obvious for Pq(z) = c„z", on account of Formula (8.68) and
detZ)po(x) = \p'q(z)\^ > 0(usetheCauchy-Riemannequation),forz = x 1+1x2 ^
0; and
gives a C°° homotopy between po and p. An immediate consequence of this is the
Fundamental Theorem of Algebra. According to this theorem there exists, for every
polynomial function p : C —>■ C with positive degree, a ^ e C with p(z) = 0.

594 Chapter 8. Oriented integration
Indeed, p not surjective implies that the degree of the mapping p equals 0, but
then the degree of the polynomial p also equals 0. In fact, p possesses precisely n
different roots if 0 is a regular value for p. See Exercises 3.48 and 8.13 for other
proofs. ik
Definition 8.11.6. Assume X and Y are C'' submanifolds in R" of dimension d,
and let <1) : Z —>■ F be a mapping of manifolds. We say that <1) is a C*^ mapping if,
for every x € X, there exist C'' embeddings 0 : f/ —>■ R" and i/f : V —>■ R" with U
and V open sets in R'^, for which
X e im(0), <l)(x) e im(T/f), <l> := T/f~'o<l)o0 : f/-> V a C*^ mapping.
Analogously we say that a bijective mapping <1) : Z —>■ F is a C* diffeomorphism
if, for every x e X, the mapping <t : f/ —>■ V is a C*^ diffeomorphism. Let
cl> e ^'^(R") be a C' form with supp(ftj) n F C im(T/f), then oi^, the restriction
of cl> to F, is defined as the differential form fjr*CL> e Q,^{V). Finally we define
Jy My '■= fy i^*o), the integral of m over the submanifold F. O
Using Lemma 4.3.3 one readily verifies that the choices of 0 and ■\j/ are irrelevant
in the first two definitions above. Furthermore, fy My = fy fjr*CL> is independent of
the choice of fjr, provided det Difjr"^ o t/c) > 0 for a fjr with the same properties as
f.
From the foregoing one immediately derives:
Tlieorem 8.11.7 (Degree of mapping). Let W be open in R", let X and Y be
compact oriented C^ submanifolds of dimension k, and assume Y C W to be
connected. Let <1) : Z —>■ Y be a C^ mapping. Then there exists an integer
deg(<l)) e Z, the degree of the mapping O, with the property that, for every C' form
I <D*ftjy = deg(<D) I COY. (8.69)
In particular, deg(<l)) ^ 0 implies that <1) is surjective. Furthermore, deg(<l)) is
invariant under C^ homotopy of^.
In the notation of Definition 8.11.6, the compactness of X and F implies that
<5 : f/ —>■ V is proper. Formula (8.69) now means that, in fact, for r] = ■\j/*M e
I ^*ri = deg(<l)) /
Ju Jv

8.11. Degree of mapping 595
Example 8.11.8 (Hairy sphere of even dimension has a cowlick). There exists a
C^ tangent vector field to S""' without any zeros if and only if « e N is even (see
Exercise 6.22 for another proof).
Indeed, assume that / : S"~^ —>■ R" is a C^ mapping with f{x) e T^S""^ and
f(x) ^ 0, for all X e S""'. Then let g(x) = ,. ,1 ^../(x), and define
r : / X S"~^ -> S""' by r(f,x) = icosnt)x + (sinnt)gix).
One sees that indeed we have r(t,x) e S""', since ||x|| = ||g(x)|| = 1 and
{x, g(x)} = 0, for X e S"-'. Furthermore, r(0,x) = x and r(l,x) = —x.
Thus Id\^„_, is C^ homotopic with —/rf|^„_i, which implies that the two mappings
are of the same degree. The degree of —/rf|^„_i : x i-> —x is (—1)", because
(—Id)*ct) = (—l)"ct), with CO as in (8.71). From (—1)" = 1 it follows that n is even.
An example of a tangent vector field / to S^""' without any zeros is the
following:
/(X) = (-X2, Xi, -X4, X3, . . . , -X2„, X2„_l) (x € S^""^). Ik
Example 8.11.9 (Winding nmnber and Kronecker's integral). Let X c R"
denote a connected compact and oriented C^ submanifold of codimension 1 (it
can be shown that orientability of Z is a consequence of the other conditions), let
a e R" and let 0 : Z —>■ U := R" \ {a} be a C^ mapping. Note that the compact set
Y := <p{X) need not be a manifold. The number w{Y, a) e Z, the winding number
of Y with respect to a in R", that is, the number of times the set Y winds around
a in R", is defined as deg(<l)) with <1) = jt o 0 : Z -> S"~^ Here n : U ^ S"~'
denotes the radial projection with respect to a given by 7r(x) = , ^^. (x — a). For
computing w{Y, a) we have the following generalization of Example 7.9.4, known
as Kronecker's integral:
1 f 1
w{Y,a) = / i^^adx e Z. (8.70)
Here the integrand is a closed differential (« — l)-form on U.
Indeed, we may assume a = Q and we will apply Theorem 8.11.7 with u) e
^"■~' (f/) equal to a C°° differential form whose restriction to S""' determines the
hypersurface area on S""'. Formula (7.15) implies that we can take u) = ixdx, the
contraction of dx € Q"(U) with the vector field x i-> x on f/, which satisfies
L
a)=\S"-'\. (8.71)
The properties above of m also can be seen as follows. In the notation of
Example 8.8.1,
0)= J2 (-l)'"'^/^^e ^""'(f/); and da) = ndx € ^"(U).
l<i<n

596 Chapter 8. Oriented integration
(The second identity also follows from the homotopy formula in Lemma 8.9.1.)
Because S""^ = dB", we find by means of Stokes' Theorem and Example 7.9.1
/ 0)= I da) = n vol„(B") = hyperarea(S"~') = |S"~^|.
Js"-^ Jb"
Now, on the strength of Example 2.4.8,
D7r(x)v = V + some multiple of x (x € U, v € R")- (8.72)
lUII
Using the definition of m. Formula (8.72) and the antisymmetry of dx, we obtain
forx e U and any (n — l)-tuple of vectors t;i,..., t;„_i e Tj^U — R"
(7r*M)(x)(vu ..., v„_i) = M{7r(x)){D7r(x)vi,..., D7r(x)v„_i)
In other words,
or := n*a) = j^/.^x = |S"-'|b„_,/ € ^""'(R" \ {0}), (8.73)
where b„_i / is the differential form associated as in Example 8.8.2 with the Newton
vector field f(x) = |y,_i||, ,,„ x from Example 7.8.4. A direct computation shows
that cr is a closed differential form, contrary to m. We may prove this also by means
of the homotopy formula from Lemma 8.9.1, using that / : R" —>■ R" is the tangent
vector field of (OOrsR with <!)' : R" -> R" given by <l)'(x) = tx. Formula (8.70)
now follows from (8.73) since we have, on account of (8.71) and Theorem 8.11.7,
deg(<D) = ^^^^ f 0) = —^ f <p*in*a)) = —^ f <p*a. (8.74)
For Y fixed the function a i-> w(Y, a) is a continuous function R" \ F —>■ Z,
and therefore it is locally constant. So it is constant on the connected components
of R" \ Y; and since limQ_^oo w(Y, a) = 0, we have w(Y, a) = 0 for all a in the
unbounded component of R" \ F. ik
Example 8.11.10 (Special case of Jordan-Brouwer Separation Theorem). Let
V C R" be a connected compact C^ submanifold of codimension 1 which is ori-
entable. (It can be shown that the last condition is a consequence of the preceding
ones.) Then the complement R" \ V consists of two nonempty disjoint open
connected sets, and V is the boundary of both these sets. We will now prove this
result.
Let a ^ V and consider w(V,a), the winding number of V with respect to
a, that is, in the notation of the preceding example we take X = Y = V and

8.11. Degree of mapping 597
<p = I. More precisely, we study what happens to Kronecker's integral in (8.70)
when a crosses V. Write a± = ^€e\, for e > 0 arbitrary but fixed. On account
of Theorem 4.7.1.(iv) we then may assume that V, locally near the fl±, is given by
{(0> y) \ y ^ R"""' near 0}. The main contribution to the difference
comes from the x e V closest to the fl±, that is the y near 0. In fact, for x far away
from the a±, the integrand, being a difference, is small and the integral is small
too, the integration being over the compact submanifold V. Hence we may as well
assume that V = {0} x R"""', manifestly ignoring the compactness of V. Now, for
X = (0, y) € V, we have x — a± = (±e, y2, ■.., ynY € R" and so
||x-fl±|| = (e'^+Wyf)'-, h-a.j^dx(e2,..., e„) = det(x-a± e2 ■ ■ ■ e„) = ±e.
We obtain, with the substitution y = ex in the third term,
x:±-(v,..)=E]|^X lu^^'^-"^
2e r 1.2/- 1
L-'Ce^+llyf)?''^ \S"-'\L
■dx = I,
15"-'I yR«-i (e2 + II>-II2)5 \S"-'\ V-. (1 + ||xf)^
on the strength of Exercise 7.23. In this case, of V being a manifold, we obtain that
the function a i-> w(V, a) jumps by ±1 when a crosses V. (These approximate
calculations are rigorous because we are dealing with a Z-valued function.) In
turn, this implies that R" \ V has at least two connected components. Furthermore,
suppose / is a ray emanating from a ^ V, and let a' e / \ V. Suppose that
k is the number of times that / intersects V between a and a'. Conclude that
w(V,a) — w(V,a') = k mod 2.
Next we prove that there are exactly two connected components. To this end
note, analogously to Exercise 7.35.(i), that there exists a number S > 0 such that
<1) :]-S,S[ X V ^^ R" with ^(t,x) = x + t v(x),
is a C^ diffeomorphism onto an open neighborhood V^ of V in R". Here v is a
continuous choice for a normal to V. Then it follows from Theorem 1.9.4 that V^
is a connected subset of R", and this implies that V^ \ V = V+ U V_, where the
V± both are connected subsets of R". It now suffices to show that every connected
component C of R" \ V intersects either V+ or V_. To this end, considers € dC.
If X ^ V, then x e R" \ V, which is open; so x cannot be a boundary point of any
connected component of R" \ V. Therefore 9C C V, which implies that C must
intersect a V±.
Observe that of the two connected components of R" \ V, precisely one is
unbounded, call it Co; and the other is bounded, call it Ci. If V is given the outward
orientation it gets as 9Ci, then
a e Ci ^!^^^ w(V, a) = i (0 < i < 1).

598 Chapter 8. Oriented integration
Because V is differentiable, infinitesimally (via the normal to V) one has a ready
criterion for deciding at which side of V a given point lies. The winding number
of V with respect to a point makes this criterion into a global one. The proof given
above can be adapted to the case of V being closed but not compact. The general
formulation of the Jordan-Brouwer Separation Theorem is valid for a set V C R"
that is a homeomorphic image of S"~'. The proof must take into account that V
may then be much more irregular. ik
Example 8.11.11 (Number of solutions of an equation). As a further application
of Example 8.11.9, we show how the number of solutions of an equation (j)(x) =0,
counted with signs, within an open set can be computed by means of Kronecker's
integral taken over its boundary.
Let Q C R" be a connected bounded open set for which 9^ is a compact
submanifold of dimension n — I. Suppose that ^o is an open neighborhood in R"
of the closure Q and that 0 : ^o —>■ R" is a C^ mapping. Assume that (p (x) ^ 0,
forx e 9^, and that 0 is a regular value for 0. Then 0~^({O})n^ is a finite set, say
[at I 1 < / < m }; and Z)0(fl,) € Aut(R"), for l<i <m.'ifn : R" \ {0} -^ S""'
denotes the radial projection and cr e Q""^ (R" \ {0}) is as in Formula (8.73), we
have
—^ f <P*a= T sgn(det£>0(fl,)). (8.75)
In fact, <j>~^({0}) n ^ is a discrete set in Q. If this set were infinite, it would
have a cluster point in 9^; by continuity, in this case (p would vanish at such a point,
contrary to the assumptions. Select an open ball Vq about 0 of radius S > 0 such that
Vo n 0 (9 ^) =0, and further disjoint connected open neighborhoods f/,- C ^ of a,,
with i < i < m, for which the conditions in (8.67) are satisfied. Next, introduce
Wj = Ui n 0~^(j Vq); then the restriction of 0 to each dW; is a C^ diffeomorphism
from this manifold onto {x e R" | \\x || = |}. The set U := Q\ Ui<,<„, Wj is open
in R" and its boundary is the disjoint union of the (n — l)-dimensional submanifolds
dQ and 9 Wj, for 1 < i <m. We may assume that (p has no zeros on ^o \ ^. Hence
the differential (n — l)-form (j)*cr on the open neighborhood ^o \ {a,- | 1 < i <m}
of U is closed, on account of cr being closed and Theorem 8.6.12. Therefore we
may apply Stokes' Theorem to f^ <p*a and Formula (8.74) in order to obtain
/ 0V = deg(77: o (0l9^)) = - V A&g{7i o (^Ig^y.)).
ICh-
But the orientation of 91V/ is the outward one with respect to Q,, while <P\^w. '■
IV/ —>■ {x e R" I ||x|| = I} is a diffeomorphism where the sphere is oriented by
the outward normal. Therefore there is an extra minus sign when applying
Formula (8.68), and this proves Formula (8.75). Note the similarity with the arguments
in Example 7.9.4. ik

Exercises
Exercises for Chapter 6
Exercise 6.1 (Not Jordan measurable, compact set). Let {r„ \ n e N} be an
enumeration of Q n ] 0, 1 [, and let 0 < e < 1 be chosen arbitrarily. For each
« € N we select an open interval /„ C ] 0, 1 [ such that r„ e /„ and length(/„) = ^.
Now define A = U,,^!^ I„ and K = [0, l]\A.
(i) Show that the inner measure of A cannot exceed e.
(ii) Prove that A is a dense subset of [ 0, 1 ], that is, A = [ 0, 1 ]. Conclude that
the outer measure of A equals 1.
(iii) Prove that A is a not Jordan measurable, open set in R.
(iv) Prove that A!' is a compact set in R which is not Jordan measurable.
Exercise 6.2. Let B = [ 0, 1 ] x [ 1, 2 ]. Prove
/ (J^i+J^2) ^dx = log(^~y
Exercise 6.3. Demonstrate that | is the area of the bounded set in R^ bounded by
the line {x e R^ | xi = X2} and the parabola {x e R^ | x| = 2xi}.
Exercise 6.4. Show that the volume of the solid in R^ under the paraboloid {x e
R^ I x^ + x| — X3 = 0} and above the square A!' = [ 0, 1 ] x [ 0, 1 ] equals |.
Exercise 6.5. Verify that the volume of the bounded solid in R^ bounded by the
parabolic cylinder {x e R^ | x^ + X3 = 4} and the planes {x e R^ | xi = 0},
{x e R^ I X2 = 0}, {x e R^ I X2 = 6}, {x e R^ I X3 = 0} equals 32.
599

600
Exercises for Chapter 6: Integration
Illustration for Exercise 6.6
Exercise 6.6. Prove that the volume of the bounded solid in R^ bounded by the
paraboloid {x e R^ | xf + x| — X3 = 0}, the cylinder {x e R^ | x^ + x| = a^ },
for a > 0, and the plane {x e R^ | X3 = 0} equals ^na'^.
Exercise 6.7. Let B be the unit disk in R^. Prove /^ x^x| dx = ^.
Exercise 6.8. Let B be the ball of radius R about the origin in R^. Calculate
fgXiX2X3\\x\\^dx.
Hint: Do not plunge into the calculation straight away.
Exercise 6.9. Let B = {x e R^ I |xi| < 1, |x2| < 1}. Prove /^ \\x\\-'^ dx
41og(l + V2).
Exercise 6.10. Define B = {x e R^ I 1 < xi < e''\ X2 > X3, x| + x| < 4]
Prove
1 , 8-4V2
/.
— dx =
B Xi 3
Exercise 6.11. Let / e C([a, b]), and define V = {x e R" | ^^ < Xi < X2 <
■ ■ ■ < x„ < Z>}. Show
lllnxj)dx = ^_{l\mdt)".
i<j<"
Exercise 6.12. Let B be a rectangle or a ball in R" and suppose / e C(B). Prove
that there exists a point Xq e B such that f^ f(x) dx = /(xq) vol,, (B).

Exercises for Chapter 6: Integration 601
Exercise 6.13. Let U C R^ be the open disk about 0 and of radius a > 0. Prove
f \\x\\ dx = -^■, f e-ll^ll' dx = nil - e'"').
Exercise 6.14 (Needed for Exercise 6.28). Let a and Z> > 0. Using polar
coordinates prove that Tzab is the area of the ellipse {xeR^I -h + li '^ ^
Exercise 6.15 (Sequel to Exercise 2.73). Now we can give the background for that
exercise.
(i) Prove that, for all a e R,
/ /■" _ 2 \2 fi f^ _2 n fi £_
{ I e ^ dx) =2 I I e '' rdrda = / e t^"'^" da.
Wo ^ Jo Jo 4 Jo
(ii) Conclude by means of the substitution a = arctan t that, for all a e R,
.1 ^-a\\+l-)
Exercise 6.16. f/ C R^ denotes the interior of the triangle having vertices (0, 0),
(1, 0) and (0, 1). Prove
Ju 2
Hint: Consider the C°° diffeomorphism <1) : x i-> (xi — X2, xi + X2), and find the
triangle V for which V = (L>(U).
Exercise 6.17. Assume 0 < a < b and letU = {x e R \ a < ||x|| < b}. Prove
j \\xr^dx=47Tlog(^-y
Exercise 6.18. Prove that 167r is the volume of the open bounded set in R^ bounded
by the paraboloids {x e R^ | x^+x| —X3 =0} and {x e R^ \ xf+x^+x^ = 8}.
Exercise 6.19. Let C = {x e R^ 11-^11 < 2, xf + x| > I, X3 > 0}.
Calculate /^ a/3x3 — x| dx, by means of substitution of cylindrical coordinates x =
(r cosa, r sina, X3).

602 Exercises for Chapter 6: Integration
:/fsJ
Illustration for Exercise 6.18
Exercise 6.20. Let f/ C R^ be the open unit ball. Prove, for all y € R^,
{ I \A 4^ /sin II y II \
y^eos(x,,)^x = —(^^-cosll.ll).
Conclude that vol3(f/) = \ti.
Hint: Use the fact that the integral is invariant under rotations acting on y.
Exercise 6.21. Let f/ = {x e R^ | xi < 1, X2 < 1}. Prove
I
Jl , ..2 ,.2..2
^^r+.v,—^f.vj /^_ 7^^^
n{e - 1)
Hint
: Consider <I>(x) = {x\, xiJ\ — x\).
Exercise 6.22 (No C' field of unit tangent vectors on even-dimensional sphere
- sequel to Exercise 3.27). We use the notation from that exercise. Deduce from
the definition of <I>, that t i-> det <I>/(x), for x e R", is a polynomial function,
which is strictly positive if x e A and |f | is sufficiently small. More precisely, show
det <I>/(x) = 1 + X!i<;<n ^'Q!i(^)> where the a,- e C(R"). Conclude by integration
over A that
vol„(<I>,(yl)) = vol„(yl)+ V f' / a.i(x)dx.
On the other hand, deduce from Exercise 3.27.(iv) that
vol„(<I>,(yl)) = (l+r2)"2vol„(yl).
For « e N odd, conclude that a mapping / satisfying all the conditions in
Exercise 3.27 does not exist.
Background. We have proved the result from Example 8.11.8 that a hairy sphere
of even dimension has a cowlick, that is, on the unit sphere in R" there exists a C'
field of unit tangent vectors if and only if n is even.

Exercises for Chapter 6: Integration 603
Exercise 6.23 (Sequel to Exercise 1.15 - needed for Exercise 8.37). Let B" =
{y € R" I llyll < 1}, and define ^ : B" ^ R" by ^(y) = (1 - Ibll^)"'^^)'.
(i) Using Exercise 1.15 prove that 'P : B" —>■ R" is a C°° diffeomorphism, with
inverse <I> := ^'^ : R" -^ B" given by <I>(x) = (1 + \\xf)-^^^x.
(ii) Prove the following equality of functions on B":
D,.«P,. = (l-||-||2)-V2(3.. + «p,«p.).
If we regard x e R" as an element of Mat(« x 1, R), then xx' = (^/JCy) i<,j<„ €
Mat(«, R).
(iii) Let A e 0(«, R) (hence A'A = I), let x e R", and write z = Ax. Then
prove det(/ + xx') = det(/ + zz'), and conclude, making a suitable choice
for A, that
det(I + xx') = I + \\xf (x€R").
(iv) Show that, for every continuous function / or g with compact support on R"
or B", respectively.
Jr" Jb" V(l-||3'||2)i /
(l-||).f)iV (l-||ylP)'+''
dx
Exercise 6.24. Let n>2,letK€ $(R") and let <I> : R" -> R" be the C' mapping
defined by
<I>i(x) =xi+ai (ai e R),
<^i(x) = Xj + fl,(xi,... ,x,_i) (I <i < n, tti e C'(R'~')).
Prove that ^{K) € |(R"), and that vol,, {^{K)) = vol„(K).
Exercise 6.25 (Solids of revolution - sequel to Exercise 4.6). Let / e C([a, b ])
be nonnegative and let y : [a,b] —>■ R^ be the curve given by y (t) = (f(t), 0, t).
Let V be the surface of revolution in R^ obtained by revolving im(y) in R^ about
the X3-axis, as in Exercise 4.6. Let L be the compact set in R^ bounded by V and
the planes {x e R^ | X3 = a } and {x e R^ | X3 = Z>}.
(i) Prove that L is Jordan measurable in R^, and that vol3(L) = ^ f f{t)^dt.

604 Exercises for Chapter 6: Integration
Assume K C R^ to be a compact subset in the half-plane {x e R^ | X2 = 0, xi >
0} and which is Jordan measurable in R^, if the plane {x e R^ | X2 = 0} is
identified with R^. Construct, in a way analogous to that in Exercise 4.6, the solid
of revolution L in R^ obtained by revolving K in R^ about the X3-axis.
(ii) Prove that L is Jordan measurable in R^ and that V0I3 (L) = In fj^xi dxi dx^.
Exercise 6.26 (Archimedes' Theorem). Prove that the volumes of an inscribed
cone, a half ball, and a circumscribed cylinder, all having the same base plane and
radius, are in the ratios 1 : 2 : 3.
Exercise 6.27. Two cylinders are inscribed inside a half ball in the following fashion.
The lower face of one of the cylinders is in the plane face of the half ball, and the
circumference of the top face of that cylinder lies on the round surface of the half
ball. The lower face of the other cylinder lies in the upper face of the first cylinder,
and the circumference of its top face lies on the round surface of the half ball. How
should the heights of the two cylinders be chosen for the sum of their volumes to
be maximal ?
Exercise 6.28 (From Newton to Kepler - sequel to Exercises 5.24 and 6.14). Let
the notation be as in Example 6.6.8 on Kepler's second law. In particular, suppose
t i-> x(t) is a C^ curve in R^ such that the position x(t) and the acceleration x"(t)
are linearly dependent vectors, and more precisely, that there exists 0 ^ k €R such
that
x"(t) = -k\\xm-'x(t) = -grad ( - —^) (t € R).
\ ||x(OI|/
This corresponds with an inverse-square law (for the magnitude ofx"(t)) as studied
by Newton. The acceleration is centripetal for k > 0, and centrifugal fork < 0. In
the following we often write x instead ofx(t) to simplify the notation, and similarly
x' and x". We will determine the geometric properties of the orbit {x(t) \ t €R}
and we begin by computing the velocity of the normalized position vector.
(i) Assuming x ^0, use Example 2.4.8 to prove
(||x||-'x)' =-\\x\\-^(x,x'}x+\\x\\-'x'
= \\x\\-H-{x,x')x + {x,x)x')=: \\x\\-^v.
Show (v, x) = 0 in order to obtain v = XJx, with A. e R and J e S0(2, R)
as in Lemma 8.1.5.
(ii) On account of Example 6.6.8 there exists / e R with / = det(x x') =
(Jx,x') = —(Jx',x). With A. as in part (i), verify A. (x,x) = det(x XJx) =
det(x v) = (x, x) det(x x') = l(x, x), and conclude —(x, x'}x + (x, x}x' =
V = IJx.

Exercises for Chapter 6: Integration 605
(iii) Combine parts (i) and (ii) to find
(\\x\\-^xy = l\\x\\-^Jx = -yJx".
k
Integrating once with respect to t, obtain a constant vector e e R^ such that
(*) \\x(t)\\-'x(t)=€-yjx'(t) (t€R).
k
Verify that ||x' + j7e|| = |j|. In other words, the hodograph (f) 686c = way),
the curve traced by the velocity vector x', is the circle of center —yJe and of
radius \j\. Next, take the inner product of the equality (•) with x and apply
(ii) once more to conclude that
/ /2
||x|| = (e,x) - -(Jx',x} = (e,x) + —.
Deduce from Exercise 5.24.(iii) that the orbit of x is a (branch of a) conic
,2
section with eccentricity vector e and d = j, the sign of which depends on
that ofk. This is Kepler's first law.
(iv) By a translation in R we may assume ||x(0)|| = min{ ||x(f)|| | f e R}.
Deduce from part (iii) and (ii), respectively,
Ik(0)11 = TT^—T and ||x(0)|| ||x'(0)|| = /.
kil +e)
(v) According to Exercise 3.47.(i) the total energy per unit mass H = ^f— tAt
is constant along orbits. Starting from (•) in part (iii) prove that this constant
value equals i!/ = ^(e^ — l). Conclude that the orbit is an ellipse, a parabola,
or a hyperbola, as i!/ < 0, // = 0, or // > 0, respectively.
(vi) Assume that k > 0, and that the orbit is an ellipse, with semimajor axis a
and semiminor axis b. This implies that there exists minimal T e R+ with
x(t) = x(t + T), for all t € R. After a time T the area swept out by the
radius vector equals the area of the ellipse. Deduce from Example 6.6.8,
Exercise 6.14, Exercise 5.24.(ii), and part (iii), successively, that T satisfies
/ 3 T^ 471-2
-T = nab = nd'^(l-e'^)~^, or —- = ——,
2 a"^ k
where the constant at the right-hand side is independent of the particular
orbit. This is the assertion of Kepler's third law. Again on account of
Exercise 5.24.(ii), prove / = {ka(l — e^))^. Furthermore, show
k
2a
M\x\\ a/

606 Exercises for Chapter 6: Integration
Demonstrate ||x'(0)|| = max{||x'(OI| | r e R} as well as ||x'(|)|| =
min{ ||x'(f)|| U e R}, and also
T /cl+e ,/^\t Ic I — e
\W(0)f = -- , \\x'(-)f = --^,
a 1 — e \2/ a I + e
IU'(0)||||x'(|)|| = ^ = -2H.
Exercise 6.29 (Volumes, Bernoulli and Euler numbers - sequel to Exercise 0.16
- needed for Exercises 6.30 and 6.40). Define 0" = {x e R'| | x^ +Xj+i <
1 (1 <;■<«)}.
(i) Prove
9„ = vol„(0") = / / • ■ ■ / dx„ ■ ■ ■ dx2dx\.
Jo Jo Jo
For the calculation of this integral we introduce polynomial functions p^- on R by
po(x) = l, pk{x)=\ pk-iit)dt (k €N).
Jo
(ii) Prove
p„(0)=e„ («€N), po(l) = l, Pk(l) = 0,
p'k(x) = -Pk-i(l-x) (k€N).
Next, we introduce the formal power series in y (that is, without considering
convergence)
f(x,y)=J2PkMy'-
(iii) Prove that / satisfies the differential equation
W ^(x,y) + yf(l-x,y) = 0, and (**) /(l,3') = l.
6x
(iv) Prove that j4:(-^^ y) + y^fi^, y) = 0, and conclude
fix, y) = a(y) cos(x>') + b(y) sm(xy).
Substitute x = 0 into (•) and prove b(y) = — 1, and, using (**), show that
a(y) = tany + ^. Conclude
tan y + sccy = f(0, >-) = 1 + ^ p„(0)y" = 1 + ^e,y.

Exercises for Chapter 6: Integration 607
For all « e N we find that 9„ equals the coefficient of y" in the power series
expansion of tan y + sec y. On account of Exercise 0.16.(xiii) we have
tany = J2^„y"'-' = ^(-l)«-'2^"(2^" - 1)-|^ y^'-^ [\y\ < |).
Here the B2« are the Bernoulli numbers. Moreover, we have
Here the E„ are known as the Euler numbers. One has
£0 = 1, £1 = 1, £2 = 5, £3=61, £4 = 1385,
(v) Deduce, for all« e N,
02„_:=/„ = (-ir-'2^''(2^'-l)-^, 02„= ^"
(2«)! ' (2«)!
Exercise 6.30 (Sequel to Exercise 6.29). We use the notation from that exercise.
The formulae from part (v) of that exercise can be proved in a somewhat more direct
way.
(i) Prove, for i e No and x e R,
Pi+iix) = I pi+i{t)dt-\ pi+i{t)dt
Jo Ji-x
= Pi+2(0) - / p,+i(1 -t)dt = p,+2(0) - / / Piis) ds dt.
Jo Jo Jo
(ii) Use part (i) and mathematical induction on « € Nq to show, for x e R,
-ly 2,- (-1)"
(20! "" (2n + 1)!^
p..,w= E '^--('»yT-"-7^-""-
0<!<n
Conclude with P2n+iW = 0 that

608 Exercises for Chapter 6: Integration
(iii) Now derive the following identity of formal power series:
That is.
(-1)\.2,A ^ X- (-1)" ^2n+l
(iv) Now discuss the case of the function sec.
Exercise 6.31. Let ('POrsR be the one-parameter group of C°° diffeomorphisms of
R given by 'i'(x) = e'^'x, set L = [0, 1 ] and suppose / e C'(R). Prove
^ /" f(x)dx=2e^'f(e^') (t €R)
"t J^il'{L)
by means of the chain rule, the substitution x = e^'y, and the transport equation
from Example 6.6.9, respectively.
Exercise 6.32 (Special case of Urysohn's Lemma - needed for Exercise 6.33).
Let K C R" be a compact set, let U C R" be an open set, and assume K C U.
(i) Prove that a continuous function / : R" —>■ [ 0, 1 ] exists such that f(x) = 1,
forx e K, and supp(/) C U.
Hint: Consider the open covering {U} of K.
(ii) Verify that, for given a, b € R with a < b, the function a + (b — a) f :
R" —>■ [a, b] has the value b on K, and a onR" \U.
Exercise 6.33 (Special case of Tietze's Extension Theorem - sequel to
Exercise 6.32). Let K C R" be a compact set, assume a, b € R with a < b, and let
f : K ^ [a,b]hea continuous function. Then there exists a continuous function
F -.R" ^ [a,b] such that F|^ = /.
(i) Replace / by (/ — a)/(b — a) and conclude that one may assume [a,b] =
[0,1].
Let t = ^. We assert that there exists a sequence (fk)ksfi of continuous functions
on R" such that
fk : R" ^ [ 0, f (20'-' ]; 0 < fix) - J2 M^) < (20' (^ e K).
l<i<k

Exercises for Chapter 6: Integration 609
(ii) Let A = /"'([ 0, t ]) and B = /"'([ 2t, 1 ]). Prove by using Exercise 6.32
that there exists a continuous function fi : R" —>■ [0,t] such that /i (x) = 0
for X € A, and /, (x) = f for x e B. Verify that 0 < /(x) - fi (x) < 2t, for
x€K.
(iii) Now prove, by induction on A; e N, that one can find fk as above, such that
fk = Oonthesetwhere/-X;i<,<fc fi(x) < t (20*^"^ and that ./> = t {Itf'^
on the set where / — X!i<!<fc fii^) — i^O''-
(iv) Prove that F = XiteN fk satisfies the requirements.
Exercise 6.34 (Every closed set is zero-set - needed for Exercise 6.37).
(i) Let g € C(R"). Prove that C = N(0) = {x € R" | g(x) = 0} is a closed
subset of R".
We shall now prove the converse of the assertion in part (i) in five steps. Let C be
an arbitrary closed subset of R".
(ii) Prove that there exists a countable collection {^^tltsN of open balls B^ C
R" \ C with
R" \ C = IJ Bk.
(iii) Show that for every A; e N there exists gk € C°°(R") with the properties (see
the proof of Theorem 6.7.4) gk > 0, and gk(x) > 0 if and only if x e Bk-
(iv) Verify that for every A; e No the number nik = sup{ \D"gk(x)\ e R | x e
R", a. e Ng, la I < Jfc} is well-defined.
(v) Prove that, for every a e Ng, the series JZfcsN ~^^"§k converges uniformly
on R"; and use termwise differentiation to conclude that g e C°°(R"), if
^ = E;;ri^^^-
nikli''
(vi) Verify that C = {x € R" | g(x) = 0 ]
Exercise 6.35. Let K and Ks, for S > 0, be as in Lemma 6.8.L Let / e C(R")
and assume N(f) n AT = 0. Prove that a number S > 0 exists with N(f) DKs = id.

610 Exercises for Chapter 6: Integration
Illustration for Exercise 6.36: Sard's Theorem
Exercise 6.36 (Sard's Theorem - sequel to Exercise 2.27 - needed for
Exercise 6.37). The following is an (easy) version of this theorem. Let U C R" be open
and let <I> : f/ —>■ R" be a C' mapping, then
vol„({ <I>(x) I X e f/ is singular point for <I>}) = 0.
We will prove this in a number of steps.
(i) Verify that it is sufficient to prove the assertion for every rectangle B in R"
with B CU.
Let B be a fixed rectangle thus chosen.
(ii) Prove that a number m > 0 exists such that ||<I>(x) — <!>(«)|| < m\\x — a\\,
for all X, a € B.
Now define
K = {a € B \ ^ is singular at a }.
Leta e A!'be fixed for the moment. Because D<I>(a)(R") is aproper linear subspace
of R", there exists an affine submanifold L of codimension 1 in R" going through
^(a), such that, for all x e R",
<^(a) + D<^(a)(x -a) € L.
Now let e > 0 be chosen arbitrarily.
(iii) For every x € B with ||x — a\\ < e, prove that <I>(x) belongs to
the ball in R" about <!>(«) of radius me;
the tubular neighborhood in R" about L of half thickness eA.(e),
where A. is as in Exercise 2.27.(ii).

Exercises for Chapter 6: Integration 611
(iv) Conclude for such x that <I>(x) is contained in a rectangular parallelepiped
in R" which can be written as a product of a cube in L of dimension n — \
having edges of length 2me, and of an interval in R of length 2eA.(e). Then
prove
vol„(<I>{x €B \\\x-a\\<e})< (Ime)""' 2eA(e).
Note that the constants on the right-hand side of this inequality are
independent of a e K.
(v) Verify that the number of balls in R" of radius e about a € K (where the
point a now is considered to be variable), required to cover K is of the order
e)(e~"), e I 0; and conclude that
Yol„ i^(K)) = OiX(€)), e|0.
(vi) Why has Sard's Theorem now been proved ?
Exercise 6.37 (Functional dependence - sequel to Exercises 6.34 and 6.36). (See
also Exercise 4.33.) Let U C R" be open and let <I> e C (f/, R")- We assume that
the rank of D^(x) is lower than n, for all x e U.
(i) Prove by Exercise 6.36 that vol,, (im(<I>)) = 0.
(ii) LetC C f/be a compact subset. Conclude by Exercise 6.34 that a submersion
g € C°°(R") exists such that g o^(x) = 0, for all x e C.
(iii) Next, take « = 2 and assume D<I>(x) has constant rank 1, for all x e U. Prove
that, locally on U, one of the two component functions of <I> can always be
written as a C' function of the other one.
Exercise 63S (Volume of a neighborhood of a parallelepiped). Define A' =
{1, 2,...,« }. Consider linearly independent vectors vj e R", for j e A', and
let P = {X!i<i<H h '^j I 0 < fj < 1} be the parallelepiped spanned by the vj.
Suppose S > 0 and write
P^ = {}) € R" I there exists x € P with H}) - x|| < 8}
as in Lemma 6.8.1. Then, with the notation #K for the number of elements in
K CN,
vol„(P,)= J2 ( H vol„_„,(P^)V,„S"'.
0<m<n ^ K(zN,#K=n-m '
Here P^ denotes the (« — m)-dimensional parallelepiped spanned by the n—m of
the Vk satisfying k & K,
(*) vol„_,„(P^) = (det((t;t, Vk'))k,k'^Ky'^,

612
Exercises for Chapter 6: Integration
while c„, is the m-dimensional volume of the unit ball in R'" (see, for instance.
Exercise 6.50.(viii)), which makes the factor c,„ S'" equal to the volume of an m-
dimensional ball of radius S. The details of the following proof are left for the
reader to check.
Proof. Let x e R". Because P is a closed subset of R", there exists p € P such
that ||x — pll < ||x — p'll, for every p' € P. The convexity of P implies that we
have a strict inequality here if p' ^ p. In other words, the element p = 7i{x) is
unique, hence we have a mapping jt : R" —>■ P, the projection to the nearest point
in P.
The relative interiors of the faces of P are the sets Fjj^k of the form
{ Yl ^J ^j\ J ^^ ^ h = 0; j €J ^tj = U j €K ^0<tj <l},
where (/, J, K) denotes any partition of A' into three disjoint subsets. Now suppose
p e Fjj^K, then y := x — p €R" has the following properties.
(i) i € /
(ii) ;■ e J
(iii) k€K
-{y, Vi) =
dt
{y^ ^j) =
iy' ^^) = i
\\\x
-(p+
tvdf
-{p-tv^W
-(p +
Vk)f
>0;
>0;
= 0.
Conversely, a convexity argument yields that (i), (ii) and (iii) imply that p = 7i{x).
Condition (iii) means that y belongs to Ck, the orthogonal complement of the Vk,
fork e A!', whichisalinearsubspaceof R" of dimension equal torn = n—#K. The
conditions (i) and (ii) then describe a polyhedral cone Cjj^k in Cr- The condition
X e P^ now corresponds to the condition \\y\\ < S. Write Bk (S) for the open ball in
Ck with center at the origin and radius equal to S. The fact that y is perpendicular
to the face Fj j ^ then implies that the ^-dimensional volume of jt"' (F/ y /^) n P^
is equal to the #Ar-dimensional volume of Fj jj- times the m-dimensional volume
of Ci,j,K n Bk(S). Furthermore, the #A!'-dimensional volume of Fjj^k is equal to
(•), which is independent of the partition (I, J) of N \ K. If, for given K C N,
we let (/, J) run over all partitions of N \ K into two disjoint subsets, then the
union of the C/_ y, k is equal to Ck- On the other hand, the intersection of Cjj^k
and Crj\K, if i}, J) 7^ (i', J'), is contained in a linear subspace of Ca: of positive
codimension, and therefore has m-dimensional volume equal to 0. It follows that
the sum of the m-dimensional volumes of the sets Cjj^k H Bk (S), over all partitions
(/, J) of N \ K, is equal to the m-dimensional volume of Bk(S), which in turn is
c,„S'". □
Exercise 6.39 (zeta function and dilogarithm - sequel to Exercise 3.20 - needed
for Exercises 6.40 and 7.6). As in Exercise 0.25 Riemann 's zeta function is defined
by >^{s) = XifcsN ^~^' for J e C with Res > 1.

Exercises for Chapter 6: Integration 613
(i) Demonstrate that one has, with f/ = ] 0, 1 [ ^ C R^,
3r(2)
- Z] „2 Zl (2n)2 -Z2(2n + 1)2 ~ ^ [i ""'" '^V
= y^ / / (XiX2)^"rfXirfX2 = / ^r^dx.
.~r' ^0 yo y^ 1 - Ji:r-^7
nsNo "
2
(ii) Deduce by means of parts (i) and (ii) of Exercise 3.20 that ^(2) = -^ =
1.644934066848 •••.
(iii) Perform one of the integrations in the double integral in part (i) and
demonstrate by partial fraction decomposition that
• 1
/ ^°Kt^)^" = t-
This result can also be obtained by series expansion of the integrand according
to powers of x, leading to
riiog(i±£)^.=2y:-i
Jq X ^\\-x) ^-' (2n-{
And series expansion also gives
I
Mog(l-x)^ n^
ax = .
0 X 6
Background. The function Li2 : ] —oo, 1 [ —>■ R with
r' log(l - 0
Li2(x) = - /
Jo t
dt
is called the dilogarithm. Thus Li2(l) = ^, and one has Li2(x) = JZnsN ^' ^°^
\x\ < 1. By means of differentiation one obtains the following/wwcZzowa/e^Mflriow
for the dilogarithm:
n^
Li2(x)+Li2(l-x) + (logx)(log(l-x)) = — (0<x < 1).
6
(iv) Prove in a similar way (recall that arctanx = JZnsN (~^)" It+T^
^ (-1)" f ^ /"' arctanx
= 0.915 965 594 177-•• .
The integral on the right-hand side cannot be calculated by antidifferentiation;
its value is known as Catalan's constant G.

614 Exercises for Chapter 6: Integration
Background. We have the function Ti2 ; R —>■ R with
^. ^ ^ /■*' arctanr ,
Ti2(x)= / dt.
Jo t
So Ti2(l) = G, and Ti2(x) = l]„gN„(-l)" (f;^, for \x\ < 1. Furthermore, the
Clausen function CI2 ; R —>■ R, and the Lobachevsky function JI : R —>■ R are
defined by
Chix) = - log 2 sin -t dt = y^ —T—, JI(x) = - / log |2sinf| dt.
Therefore JI(x) = | Cl2(2x). Substitution ofx=7z and x = ^ in Cl2(x) gives
f' Iog(sint)dt = -- log2 and CI2 (-) = G.
/■ 1 G n^
(v) Conclude that / ^--r rfx = 1 .
if/l-xfx^ 2 16
(vi) Prove as in part (i)
/ dx = I Y](X1X2)" dx=y^(l x"dxV
Jul-XiX2 Ju^^^^^ ^^^^yo >
Furthermore, show by means of integration by parts in the third equality
/ dx = 2 / V^ x'lx'] logxi dx
= 2 2~^ / -^'i' logxi dxi I x'2 dx2 = —2^(3).
«sNo '
Exercise 6.40 (^(2«) and Euler's series - sequel to Exercises 3.20,6.29 and 6.39
- needed for Exercise 6.52).
(i) Prove, as in Exercise 6.39.(i), with D " = ] 0, 1 [ " C R",
Y^ (-1)' ^f 1 ,

Exercises for Chapter 6: Integration 615
Define
v„ = vol„(Y"), Y" = {y € r:; I yi+y2 < 1, y2+y3 < 1, • ■ ■, yn+yi < 1}.
(ii) Conclude by Exercise 3.20.(iii) that
^^" V^ (—1)* /^N.2«+l
Let a = |(1, 1,..., 1) e T". The 2«-tope T" has «-fold symmetry about the
axis Ra; and T" is the union of n congruent pyramids, each having a as its apex,
and as basis the intersection of T" with the y-th coordinate plane for 1 < j < n,
respectively. Indeed, let F" be the collection of the x e T" satisfying x„ < xj, for
1 < y < «. In view of the cyclic symmetry in our problem we have
u„=«vol„(r").
Note that
r" = {x € R'l I Xn < Xj (1 < ;■ < n), xj+xj+i < 1 (1 < ; < « - 2)}.
Indeed, the two "missing" equations for x e F" are a consequence of the other
X„_i + X„ < X„_i + X„_2 < 1, X„ + Xi < X2 + Xi < 1.
Define p : R" —>■ R"~' as the projection onto the first n — \ coordinates, and
^ :R" ^ R" by
^(y) = (l-yn)p(y) + y„a.
Now introduce the (2n - l)-tope 0" = {y e R^ | yj + yj+i < 1 (1 < y < «)}.
(iii) Verify (p : 0""' x D ^^ F" is a C°° diffeomorphism and detD'i^iy) =
5(1->-»)""'■ Conclude that
1
vol„(F") = -^vol„_,(0"-').
2n
(iv) Now prove by Exercise 6.29.(v)
N«-lo2/i-l/o2'' <^ ^' "
^' ^ ^ ^ \2ny. ^"+' 2(2«)!
and derive from this (cf. Exercise 0.20)
U2n) =i-iy-^b2nf"^
2 (2«)!
y^ (-1)' ^ l/jrx2"+i E„
^^ (2Jfc+l)2"+' 2\2/ (2«)!'
The series on the left-hand side of the second identity are known as Euler's
series.

616 Exercises for Chapter 6: Integration
Example. For 1 < « < 4 and 0 < « < 3 the sums of the series above are,
respectively,
71^ 71^ 71^ 71^
6 ' 90' 945' 9450'
71 71^ 571^ 6\7l^
4 32 1536 184320
(v) Verify that the results above can be summarized as follows:
T ^-'^"' =.„(-)" («€N).
Here each u„ e Q is found as the volume of an ^-dimensional convex polytope
T" with rational vertices. We have found by direct computation of vol,, T"
^u„f"-'= -(secf + tanO {\t\ <^.
(vi) From part (iv), derive the following estimates;
1 \B2„\ ^ n^ 1 l/i\^"+' _^ 2('-V"^'
(277:)2" '^ (2n)\ " 3 (271^"' 3\7i) '^ (2n)\ '^ \7i)
(vii) Verify that the solid generated by symmetrizing T^ with respect to the origin
is a rhombic dodecahedron (SwSsxa = twelve). That is, it is a dodecahedron
whose faces are congruent rhombi having diagonals of lengths v2 and 1,
respectively. Verify that its volume equals 2.
Exercise 6.41 (Another proof for J^ e ^' dx = ^). (Compare with
Exercises 2.73 and 6.50.(i).) From Example 6.10.8 we know
(/" e-''dxY=[ I e-^^'+y'^dydx.
^Jr+ ' Jr+ Jr+
(i) Introduce the new variable t via y = xt, and conclude that
(f e-''dxf=l f e-''^'+''hdtdx= f f e-'^'^'+''hdxdt.
^Jr+ ' Jr+ ^r.,. Jr^ Jr+
(ii) Now finish the proof, using f^ e~^"('+'"^x dx = ^7[t^-

Exercises for Chapter 6: Integration 617
Exercise 6.42 (Probability density, expectation vector and covariance matrix
of distribution - needed for Exercises 6.43,6.44,6.51,6.96,6.97). In stochastics
a function / : R" —>■ R is said to be the probability density of a distribution on R"
if
fix) > 0 (x € R"), /" fix) dx = 1.
Jr"
In the case of convergence of the following integrals, the vector // e R" and the
matrix C € Mat(«, R) given by
= / Xjfix
Jr
l^j = / Xjfix)dx (1 < ;■ <«),
Cij = I ixi - fii)ixj - fij)fix)dx (1 < i, J < n)
Jr"
are said to be the expectation vector and the covariance matrix of that distribution,
respectively.
(i) Prove that C € Mat"'"(«, R) is a positive semidefinite matrix.
In the case where n = \, the vector // e R is said to be the expectation and the
number C > 0 is known as the variance of the distribution; the usual notation then
is cr^ instead of C. The number cr >0 itself is known as the standard deviation of
the distribution.
(ii) Now set n = \, and verify cr^ = f^x^fix) dx — ji^.
Exercise 6.43 (Sequel to Exercise 6.42). Let // e R and a > 0. Let / =
fill, a) € C°°(R) be defined by
1
ay/In
Demonstrate
/ fix) dx = 1, I ix — li)fix) dx = 0, I ix — ii)^fix) dx = a^.
Jr Jr Jr
Hint: The last identity follows by differentiation with respect to cr of the first
identity.
Background. In the terminology of Exercise 6.42 the function /(/x, cr) is said to
be the probability density of the normal distribution on R with expectation ji and
variance a^.
Exercise 6.44 (Sequel to Exercise 6.42 - needed for Exercises 6.92 and 6.93).
Let <2 € Mat"'"(«, R) be positive definite. Let E be the ellipsoid, and B the unit
ball in R",
E = {x^^" \{Qx,x)<\], and B = {x € R" | Ikll < 1 }■

618 Exercises for Chapter 6: Integration
(i) Prove that E is Jordan measurable in R", and that vol„(E) = ^^'"^^\
Hint: See Formula (2.29).
(ii) Prove
/.
,-^(2...)^^= (2:r).
X l-> — e ^{C-x,x)
R« VdetQ
Background. Now replace <2 by C~' in the expression above, for a symmetric
positive-definite matrix C € Mat(«, R). In the terminology of Exercise 6.42 the
function on R"
1
(27r)S-v/detC
is said to be the probability density of the normal distribution on R" with the origin
as expectation vector and C as covariance matrix (the latter property will be proved
in Exercise 6.93).
Exercise 6.45. Let m e R be arbitrary, define f/ C R^ and (p : R^ ->- R^ by
f/= {x eR^ I xi+X2 < m} and "i^iy) = iyi, yi-y\).
(i) Find V C R^ such that >I':V->.f/isaC°° diffeomorphism.
Assume / and g belong to Cc(R).
(ii) Prove
/ f{xi)g{x2)dx = \ \ f(yi)g(y2-yi)dyidy2.
Ju J-ooJr
2
(iii) Now assume f(x) = e~^ forx e R. Prove
/. /.»/^/2
/ f{xi)f{x2)dx = ^j f(x)dx.
Hint: The present function / is absolutely Riemann integrable over R, but
does not have a compact support. Verify that the result from (ii) is valid for
this function /.
Exercise 6.46. Define the C°° function /;, : R"* \ {0} ^ R by f,,ix) = \\x\\p, for
p € R. Calculate the values of p for which fp is absolutely Riemann integrable
over f/ = {x € R"* I 0 < ||x|| < 1} and V = {x € R"* | ||x|| > 1}, respectively;
for these values of p, calculate the integrals of fp over U and V, respectively.

Exercises for Chapter 6: Integration 619
Exercise 6.47. Define in R^ the sets V = {x €R^ \ x^ > a}, where a e R+, and
B(R) = {x € R^ I ||x|| < /?}, where R > a.
(i) Calculate /^^^^^v pf '^^ ^^ fyj^dx.
The set IV C RMs defined hy W = {x €R^ \ X1+X2+X3 > I}.
(ii) Calculate f^y j^ dx.
Hint: The answer can be very easily found by using the result of J j^ dx;
why?
Exercise 6.48 (Sequel to Exercise 5.51). Let V c R^ be the pseudosphere from
Exercise 5.51. Prove that the volume of the set in R^ which contains the X3-axis
and which is bounded by V equals '^n.
Exercise 6.49 (Sequel to Exercise 3.8 - needed for Exercise 8.35). Suppose
f € C^ (R^) has compact support, and let / = V--T. In four steps we now prove
that
(*) [ -^(x)dx:=[ }-^^(Dif(x) + iD2f(x))dx = -7T,
Jr2 Z dz Jr2 X1+IX22
/(O).
(i) Prove the convergence of the integral in (•) by means of = rrr for
Let >I'(r, a) = r(cosa, sina), for (r, a) e [0, 00 [ x R, and let / = / o ^P.
(ii) Prove by Exercise 3.8.(iii) that the integral in (•) equals
'iLji^i^-r'iy''^'''''-
(iii) Prove
r r df r r df
f f —(r,a)dadr= f f -^(r,a)dr da = —In f (0,0).
Jr.,. J-71 dr y_jr ^R^ dr
(iv) Prove (•) by means of the Ijt-periodicity of a i-> f(r,a).

620
Exercises for Chapter 6: Integration
25
15
5
12 3 4 5
Illustration for Exercise 6.50: Euler's Gamma and Beta functions
Gamma function, and Beta function on ] 0, 1 [ x ] 0, 1 [
Exercise 6.50 (Euler's Gamma and Beta functions - sequel to Exercise 2.79
- needed for Exercises 6.51, 6.52, 6.53, 6.55, 6.56, 6.57, 6.58, 6.59, 6.60, 6.63,
6.64,6.65,6.69,6.89,6.96,6.98,6.104,7.5,7.21,7.23,7.24 and 7.28). Define the
Gamma function F : R_|_ —>■ R by
dt.
r(p)= f e-'t"-
Jr+
(i) Prove
r(p + i) = pr(p) (p€R+), r(«+ !) = «! («€No).
Further, compare with Example 6.10.8 and Exercises 2.73, 6.15 and 6.41,
r(p) = 2 / e""' u^''~^ du; in particular ^{-) = V^-
(ii) Show, for p\ and p2 € R+,
r(Pi)r(p2) = r(pi + P2) 2 /"' cos^'"-' « sln^''^-' a da.
Jo
Now define B : R^ —>■ R, the Beta function, by
B(pi,p2) =
r(pi)r(p2)
r(pi + P2)'
(iii) Prove
r
■ada='-B(^^±^,^^±^)
2 \ 2 2 J
(pi, p2 eR+).

Exercises for Chapter 6: Integration 621
(iv) Show, using the substitution t = ■^, for pi and p2 € R+,
B(p,, p2) = [ t"'-' (1 - tr--' dt= [
Jo Jr
= 2/ dv.
m'"-'
R.,. (1 + U)"^+P2
(v) Setting p, = ™ and p2 = |, and substituting f = u", prove
:^ + |)
/ du = ^ (m, « e N).
7o yr^^ «r(^ + i)
(vi) Prove
B(p,p) = 2'-2''b(p, i),
and in addition Legendre's duplication formula (see Exercise 6.53 for a
different proof)
r(2p) = 2^"-' n-'i^ r(p)r(p + ^) (p € r+).
Conclude, using part (v) (see Exercise 3.44), that
ZZ7 /•' 1 1 /1\2
T ■" io VT^ ' ~ 4V2^ U J ■
(vii) Prove
/ K (2«)!v^
Conclude that
,^ (-1)*^ (n\ /•' , „ («!)222"
o^„2^+^V^^ yo (2« + l)!
Prove
/ cos^" ada = I sin^" ada =
Jo Jo
l-3---(2«-l) (2«-l)!!
JT =: JT.
2-4---2« (2«)!!
Show by expanding the logarithm in Exercise 2.79 in a power series in x cos a
and integrating term-by-term
,^(2n- m x^"+'
arcsmx = x + y (\x\ < 1).
^ (2«)!! 2« + l

622
Exercises for Chapter 6: Integration
(viii) Let B" = {x € R" | ||x|| < 1}. Prove
vol„(B") =
n-
r(f + i)
IT'
(2k)\
n = 2k;
, n = 2k — 1.
Hint: Apply mathematical induction and the formula
vol„+i (£"+')= j vol,, (ball of radiusVl - h^)dh.
Exercise 6.51 (Sequel to Exercises 6.42 and 6.50 - needed for Exercises 6.83
and 7.30). In statistical mechanics the following integrals play a role. Let a e R+,
then
I |! ?= 7, « odd;
V 2 ; 2(v^)''+''
f x"e-"'''dx= i , r-
(f)!(2»"+i
Prove these formulae in the following two ways.
(i) Using parts (i) and (vii) of Exercise 6.50.
(ii) Consider the functions on R+
1
n even.
a i-> / X
Jtu
e "^ dx = ^- and
a \-^ I e ""^ dx = -J — ,
respectively, and apply the Differentiation Theorem 2.10.13.
Let« e N. In the terminology of Exercise 6.42 the function /„ : R —>■ R with
/«w =
f 0,
2
x"-'e-
2ir(f)
X <0;
-, X > 0,
is said to be the probability density of the x -distribution with n degrees of freedom.
(iii) Verify that /j^ /„ (x)dx = 1, for « e N.
In particular, the function f^ix) = J-x e 5^ , for x e R+, occurs in kinetic gas
theory.
Exercise 6.52 (Sequel to Exercises 6.40 and 6.50). Riemann's zeta function is
defined by l^{s) = JZfcsN ^~^' ^°^ ^ € C with Re j > 1, as in Exercise 0.25.

Exercises for Chapter 6: Integration 623
(i) Prove, for Res > 1,
Jr., e^ - 1 Jk., e^ + 1
l-s
)r(s)ns).
Hint: Use ^^^ = JZfcsN ^ *^ ^"^^ interchange the order of summation and
integration. Then prove by Exercise 6.50 that f^ x^~^e~''^ dx = -^.
(ii) Using Exercise 0.20 or 6.40.(iv), show that
h+ e^ - 1 Ja x-\ An
Verify that for « = 1 the answer is in agreement with that in Exercise 6.39.(iii).
(iii) Replace x by ax, for a > 0, in the first integral in part (i), and deduce by
means of differentiation with respect to a of the resulting formula
(iv) Imitate the proof in part (ii) to show
f X^" /•' (logx)2" /77:\2'. + l
/ ——dx =2 ° ; dx = (-) E„ («€No),
yR, coshx Jo x^+1 \2/
/»
■I-
2/1-1
rfx = (-1)"-' (2^" - 1)77:^"-^ (« € N).
R_,_ sinhx 2«
Background. The integral f^^ ^^ dx = j^ occurs in the theory of the energy
intensity of black body radiation, and the one in part (iii) in the quantum theory of
transport effects.
Exercise 6.53 (Another proof of Legendre's duplication formula - sequel to
Exercises 2.87 and 6.50). Multiply both sides of the identity in Exercise 2.87.(v)
by x''~^ for p e R_|_, integrate with respect to x over R_j_, and change the order of
integration on the right-hand side. Conclude that Legendre's duplication formula
from Exercise 6.50.(vi) follows.
Exercise 6.54 (Asymptotic behavior of some integrals and Stirhng's formula).
We make the following assumptions: (i)a <0 < b, and g and h e C([a, b]); (ii)
there exists c > 0 such that h(t) >iCt^, for t € \a,b\, (iii) there exists H > 0 such
that lim,_^o -^ = y. Then we have
277:
lim v^ / g(f)e-""('^ dt = g(0)J —
A->0O J^ \ H

624 Exercises for Chapter 6: Integration
Indeed, in the following we may assume x e R+. The substitution of variables
t = ^ implies
I(x) := v^ /" g{t)e-''''^'Ut = f fix, s)ds,
where / : R+ x R ^^ R is given by f(x, s) := l[^a,^b]g(s/V^)e~-"'^'^^'>-
Set m = max{ \g(t)\ \ t €[a,b]} and deduce from (ii)
\f(x, s)\ < me-'''^'^^^' = me-''' (x € R+, ^ € R).
2
Note that the function s i-> me~" is absolutely Riemann integrable over R.
Furthermore, / (x) is uniformly convergent for x e R+ on account of De la Vallee-
Poussin's test from Lemma 2.10.10. Conclude from (i) and (iii) that we have
lim;c_^oo f(x, ^) = ^(0)e~2^'", for j e R. Use the Continuity Theorem 2.10.12
and Example 6.10.8 to verify
lim I(x) = g(0) f e-^-"'" ds = g(0)J^.
For the asymptotic behavior of the Gamma function r(x) = /j^ e~^ y^~^ dy, for
X —>■ oo, from Exercise 6.50 substitute y = xe' and then apply the preceding result
with g(t) = 1 and h(t) = e' — \ — t to obtain Stirling's formula
lim j- = v27r, that is r(x) ~ w27TX^~^e~^, x —>■ oo.
A'—>0O -,—.Vy^~-
Background. For more precise results, see Exercise 6.55.
Exercise 6.55 (Asymptotic expansion of Gamma function - sequel to
Exercise 6.50). For the complete asymptotic behavior of the Gamma function r(x) =
/jj e~^' y-''~^ dy, for x —>■ oo, from Exercise 6.50 introduce a new variable t by
means of y = xe'. This gives
r(x) = xV-^' f e-'^^'-^-'Ut.
Jr
The function /?. : R —>■ R with h(t) = e' — 1—t attains its absolute minimum 0 at 0
and has quadratic approximation jf^ near 0, that is,/z (f) = ^t^(l + 0(t)), t —>■ 0.
Therefore apply the change of variable u = u(t) given by h(t) = |m^, in other
words, u(t) = sgn(t)-\/2h(t). Show that u is a bijection from R into itself. Further,
deduce
lie' - 1) t + 0it'^)
limu (t) = limsgn(r)—, = limsgn(r) r—tt-t = 1.
This implies that u'(0) = 1. Prove that u'(t) ^ 0, for all t ^ 0. Deduce from the
Global Inverse Function Theorem 3.2.8 that the inverse mapping t = t(u) is a C°°

Exercises for Chapter 6: Integration 625
function on R with t(0) = 0. Differentiate the equality /z(f) = ^u^ with respect to
the variable u to find
1
t'(u)(e' — I) =u, hence (•) t'(u)(-u'^+ t(u)) = u.
Next consider the MacLaurin expansion of t in powers of u; that is, write, without
assuming convergence
t (u) = u + aii^ + h a^^u^ + ■ • • .
Verify that this implies
1 1
t'(u){-u^ ->rt(u)) = M + (- + 3fl2)w^+ (a^ + la\ + Aa■i)w' ^
\
+(-(A; — l)at_i + A;flt + (A; — l)fl2«fc-i H 1" 1ai,_iai + afc)M* H .
Hence, equating like powers of M in (•) conclude that ^2 = ~\->o-'i = ^''^4 = ~270'
and in general, fori; > 3, obtain the following recursion relation for the coefficients
1
(k + l)ak = --(k - l)ak_i -(k- l)a2«fc-i - (k - 2)a^ak_2 2at_,a2-
Show that accordingly
r(x) ~ x^'e-" j e-5^"' Yl(k + 1K+im*du
Prove that the integrals on the right-hand side vanish fork odd, and use Exercise 6.51
to conclude that, for x —>■ oo,
r(x) ~ Ix'^e-' ^(2jfc + 1)^2^+1- (^^^'^
fcsNo
jfc! (V2x)2«^+i
/^ ,_i _, ^-^ (2k + l)\ 1
~ '^ \ "^ 12x "*" 288x2 51840x3 2488320x4 "*"*"/'
In the computation above we ignored the fact that the MacLaurin expansion for t
actually terminates and then contains an error term. It is therefore that we use the
sign ~ to indicate that we have only equality in the following sense. Let there be

626 Exercises for Chapter 6: Integration
given a function / : K+ —>■ R. A formal power series in -:, for x e R+, with
coefficients b^- e R,
ksNo
is said to be an asymptotic expansion for / if the following condition is met. Write
s„ (x) for the sum of the first n + \ terms of the series, and r„ (x) = x" (/ (x) — s„ (x)).
Then one must have, for each « e N,
lim r„(x) = 0.
.V—>0O
Note that no condition is imposed on Iim„_^oo /"n (^), for x fixed. When the foregoing
definition is satisfied, we write
f(x) ~ Yl ^'^~' -^ ^ °°-
frsNo ^
Background. See Exercise 0.24 for related results.
Finally an application. Using Exercise 6.50.(viii), verify
1 /277:e\*
voI„(B") ~ ^= , « ^ oo,
^Tzn \ n J
and deduce Iim„_^oo voI„(B") = 0.
Exercise 6.56 (Product formula for Gamma function and Wallis' product -
sequel to Exercise 6.50). The well-known formula Iim„_^oo (1 — -)" = e~', for all
t € R, makes it plausible that, for all p > 0,
(•) r(p)= / e-'t"-^dt= lim / t"-^(l--)" dt.
In order to prove (•) we define, for « e N and p > 0,
E„ =U t"-'e-'dt-U = \ t"-'e-'(l-e'(l--jjdt\.
(i) Prove that e^ > I + x, for all x e R. Use this result to show that, for
0 < t <n,
hence
£.s/%-V'(l-(l-^)>,.

Exercises for Chapter 6: Integration 627
(ii) By mathematical induction over « e N verify that (1—x)" > 1—nxifx < 1.
Use this to show, for « e N and p > 0,
1 f" 1
E„<- tP+^e-' dt < -Tip + 2).
n Jo n
Conclude that Iim„_^oo E,, = 0, and that (•) is valid.
Background. Write Xn for the characteristic function of the interval [ 0, «] and
Mt) = Xn(0(l - ^T- Part (i) asserts 0 < f„(t) < e^', while Iim„^^/„(0 =
e~', for all t € R. Arzela's Dominated Convergence Theorem 6.12.3 therefore
immediately implies the identity in (•).
(iii) Prove, by means of the substitution t = nu and Exercise 6.50.(iv),
„. . ,. nPn\ ,. n''n\
r(p) = hm = hm (p > 0),
n->oo p(p -I- 1) . . . (p + „) n->oo (p)„+i
in the notation of Exercise 0.11. Use this to derive the following product
formulae:
where y = lim„_^oo {X!i<fc<n i ~ ^ogn) is Euler's constant.
(iv) Conclude that
1
«2r! iJiiTnX ^
hm -.—; ■. = hm = Jn.
»-°°i(i + l)---(i+«) "-°ol.3-5---(2«-l)(2« + l)
From this, show
2"«!V2« + 1 /jT
"^ l-3-5---(2«-l)(2« + l) " V 2"'
and verify that in this way we have obtained Wallis' product from
Exercise 0.13.(iii)
11 A Ad d In In n 1 t-t / 1\
hm =-, or -= (1 -).
»->°ol 3 3 557 2«-12« + l 2 n ^ t\ An^J
Background. Part (iii), and therefore Wallis's product also, amounts to the
following property, which is an asymptotic variant of the functional equation for the
Gamma function. For p > 0,
«'T(«) ~ r(« + p), «-> oo.

628 Exercises for Chapter 6: Integration
Exercise 6.57 (Analytic continuation of Gamma function - sequel to
Exercise 6.50 - needed for Exercises 6.58, 6.62, 6.89 and 8.19). We prove that the
Gamma function can be extended to a complex-differentiable function on the
subset ^ = C \ (—No) of the complex plane C.
(i) Prove thatthe integral r(z) = /jj e~' f'~' dt is well-defined, for every z € C
with Re z > 0.
(ii) Now let z €Q, then there exists « e N with Re(z + n) > 0. Define
TM = r(, + »)
^^^ (z + n-l)---(z + l)z'
Verify that this definition is independent of the choice of « e N.
(iii) Prove r(z) i^ 0, for all ^ e ^.
(iv) Verify that F is a complex-differentiable function on ^.
Exercise 6.58 (Reflection formula for Gamma function - sequel to Exercises
6.50 and 6.57 - needed for Exercises 6.60, 6.61, 6.74, 6.89 and 6.108). Define
/ : R \ Z ^ R by (see Exercise 6.57)
fix) = r(x) r(l — x) sinTTX.
(i) Define /(O) = n. Prove by f(x) = r(l + x) r(l - x) ^ that / is
continuous at 0, and also that / is infinitely differentiable at 0.
(ii) Demonstrate that f(x + 1) = xr(x) \ sin(—jtx) = f(x). Conclude
that / : R —>■ R is continuous and periodic, and therefore bounded on R.
(iii) By means of Legendre's duplication formula from Exercise 6.50.(vi), prove
/(f)/(^) = ./«.
(iv) Let g = (log of)^\ Verify that g is a bounded function on R satisfying
^x + 1-
^©+KV)=^^(^>-
Show that this implies g = 0 on R. Prove that log of is a linear periodic
function on R, and that therefore / is constant and equals tt. Thus we find
the following reflection formula for the Gamma function (see Exercise 6.59
for another proof):
r(x)r(l-x) = ^^ (x€R\Z).
sinjTX

Exercises for Chapter 6: Integration 629
(v) Using Exercise 6.50.(iv), prove (see Exercise 0.15.(iv) for a different proof)
/■' f x''~^ n
/ t"~\l - ty" dt = / dx = (0 < p < 1).
Jq Jvl,^+x smnp
Show that
dx n
in € N).
JVL.,
/r , 1 + x" n sin -
(vi) Conclude by means of Exercise 6.50.(v) and (vi), and by using part (iv), that
/ dt = ^-^r(-) = 1.402182105325 ■■• .
Background. The reflection formula can also be proved via F (1 — x) = —xF (—x)
using the product formulae for the Gamma function from Exercise 6.56.(iii) and
for the sine function from Exercise 0.13.(ii). Combination of Exercise 6.57 and the
formula ^^ = ^ F (1 — z) sin jt^ shows that ^ is complex-differentiable on C. This
also follows from the product formula for ^ in Exercise 6.56.(iii), which is valid on
all of C.
Exercise 6.59 (Another proof of reflection formula for Gamma function - sequel
to Exercises 0.15, 3.1 and 6.50). Using Corollary 6.4.3 verify, for 0 < x < 1,
F(x)F(l-x)= / e-^"+'^U-X-dt.
jRl ^h^ h
By means of Exercises 3.1 and 0.15.(iv) deduce (see Exercise 6.58.(iv) for another
proof)
F(x)F(l -X) = /" e-^> ^^ ds= i ^^ ds2 = -^^^- (0 < X < 1).
A^ ^^2 + 1 yR... ^2 + 1 SmJTX
Exercise 6.60 (Fresnel's integral - sequel to Exercises 0.8,6.50 and 6.58 - needed
for Exercises 6.61 and 6.62). One has
M I dx = — ^ = ^ ^ (0 < p < 2).
Jr., xp 2F(p) 2F(p)sin(pf)
In particular,
/■ sinx n f sinx [tF f sinx ,—
/ dx = -, / —^dx=-, / —■=dx = \'27T,
Jr., X 2 ^R ^x V 2 ^R x^

630 Exercises for Chapter 6: Integration
and Fresnel's integral
I sin(x
')^^ = Jt-
Similarly,
1^^" = - (o<P<i).
2r(p)cos(pf)
We now prove (•) in two steps.
(i) Prove
1 1
xP Tip) _.,.
and show
/" t"-^e-'''dt (x €R+),
Jr+
/ dx = / f'' W e "" sinxdxdt.
Jr., xp Tip) h_^ Jr.,.
(ii) Using Exercise 0.8, prove
/• sinx 1 /• t''-'^
/ d.x = -—-\ -dt.
Jr., xp T(p)Jt,., l+f2
Now apply Exercise 6.58.(v).
(iii) Show (see Exercise 8.19 for a different proof)
Jr.,
j_i I sin J r./_N I sin ^ JT^ — 1 < j < 1;
xdx = T(s) \ (s—) „ .
cos cos 2 0 < J < 1.
Exercise 6.61 (Values of Airy function - sequel to Exercises 2.90,6.58 and 6.60).
We use the notation from Exercise 2.90. Prove directly using Exercises 6.60 or 8.19,
or deduce from Exercise 2.90.(vi) and the reflection formula from Exercise 6.58.(iv),
Ai(0) = -A—, Ai'(O) = - ^
3tr(|) • ' 3ir(i)
Exercise 6.62 (Functional equation for zeta function - sequel to Exercises 0.16,
0.18,0.25,6.57 and 6.60 - needed for Exercise 7.31). The notation is as in
Exercise 0.25. Deduce from (•) in that exercise
.t^ ^ ^^^' .(. + 1) rb2(x-[x]) ^ , , p ^

Exercises for Chapter 6: Integration 631
Using integration by parts and Exercise 0.16.(i) and (iii), show that the integral also
converges for —2 < Re j, and prove
Conclude that
s(s + l) f b2(x- [x])
r(^) = -^^/ ' dx (-2<Re.<-l).
Next use the identity Z>2(x-[x]) =^^km ^g^^fromExercise0.18.(ii) and then
Exercises 6.60 and 6.57 to show, for —2 < Re j < —1,
C cos t -c—V
m = -2s(s + 1) / -^dtj^i^kn)
s-l
fcsN
r(^ + 2) cos(f(^+2))^'
Deduce the following/mwc/zowa/ equation for the zeta function (see Exercise 6.89
for another proof):
T-^n'i;{\-s) = zos(Ui)Y{s)Kis) (-2<Rej < -1).
Background. It follows from complex analysis that the functional equation actually
is valid on the common domain of the left- and the right-hand side. It is seemingly
easier to work with h\ instead of Z>2, but in that case the interchange of integration
and summation is a more delicate matter. One can prove that the partial sums
2Zi<fc<Af ^'"fc^"^^ ^^ bounded on R and then use Arzela's Dominated Convergence
Theorem 6.12.3.
Exercise 6.63 (Sequel to Exercises 0.4 and 6.50). Let / € No and let P/ be the
Legendre polynomial as in Exercise 0.4. Use repeated integration by parts to show
l^\l^f f Piixfdx = (-1)' /" (x^ - \)'(—^\x'^ - l^dx
= (-l)'(2/)! /" {x'^-lidx.
Now apply Exercise 6.50.(vii) to deduce
2
h"'
Ydx =
2/ + 1

632 Exercises for Chapter 6: Integration
Exercise 6.64 (Dirichlet's formula - sequel to Exercises 3.2 and 6.50 - needed
for Exercises 7.26 and 7.37). Let p e R^, and let A^ c R^ be the triangle with
A^ = {x e R^ I xi +X2 < 1}.
(i) Prove the following formula:
L
xf'-'xf-'rfx= ^(^'>^(^2>
Ia^ ' ' r(i + p,+ P2)
Hint: Apply Theorem 6.4.5, and subsequently Exercise 6.50.(iv).
(ii) Let p3 e R+. Prove
r(p,)r(p2)r(P3)
/.
Xf ~'x(-~'(l -Xi -X2)''^-'rfx =
Ui'^ "' '* '"' '"■' "" r(p,+ P2 + P3)
Hint: Begin as in part (i); substitute X2 = (1 — xi)t in the inner integral.
(iii) Let B^ = {y € R^ | \\y\\ < 1}. Prove that we then have the following
analog of the formula from Exercise 6.50.(iii):
/.
yf'yfCvT^W)''^
dy r(£j±i)r(^)r(£fi)
(iv) Suppose / e C([ 0,1 ]). Prove the following, known as Dirichlet's formula:
f xf-'xf^-'V(^i +^2)dx = ^[P'^'^^P^] f'x"^+"--'f(x)dx.
J A'- r(pi +P2) Jo
Show the formula from part (i) to be a special case.
Hint: Use Exercise 3.2.
Exercise 6.65 (Generalization of DiricWet's formula - sequel to Exercises 3.19,
6.50 and 6.64 - needed for Exercise 7.52). Let A" be as in Exercise 3.19.
(i) Prove the following generalization of Exercise 6.64.(i):
l<J<n
Hint: Use Exercises 3.19.(i) and 6.50.(iv).
(ii) In particular, conclude that vol,,(A") = —,.
(iii) Consider B'^ = [x € R" | J2i<j<n l-^jl* < 1}, for jfc € N. Prove
vol,
2\" ni)"
Verify that, on geometrical grounds, limt_^oo vol„(Bj^') = 2".

Exercises for Chapter 6: Integration 633
(iv) Suppose / e C([0, 1]). Prove the following generalization of Dirichlet's
formula, for p e R!|.:
/..n/''-'^£-->-=Kli''
-^+T.i<j<nP.if(^x)dx.
('Ll\''j
(v) Let aandd € R![ and define 0" = {x € R:[ | T.i<j<n {^) < 1 }• Using
part (i) prove
Js^il ' r{i + i:,,^.,,4)
Hint: Consider (p : R^ ->■ R;|_ given by x,- = (p,- (y) = aty^', for 1 < / < «,
and note that 0" = >I'(A").
In particular, the ^-dimensional volume of 0" and of the ellipsoid {x e R" |
2Zi<;<n (j^) < 1 }> respectively, are equal to, see Exercise 6.50.(viii)
—-^ -(- and ^ ^ n a,=vol„(B") rT a.-.
(vi) Suppose that pj e 2No,for 1 < j <n. Deduce from part (v) that the average
of the monomial function x \-^ x^ = ni<;<n'*^;^ '^^^^ ^" satisfies, in the
notation of Exercise 6.50.(vii) and with (—1)!! = 1
J_/" ^■»^^^,_n.^,<„(Pr
in(B")jB" n,<,<M(«
n,^,<„(p.-i)!!
vol„(B")V '" n,<,.M(« + 2/) ^^
Exercise 6.66 (Bessel function - sequel to Exercise 6.50 - needed for Exer-
i
'2-
cises 6.67, 6.68, 6.70, 6.98, 6.102, 6.107, 7.22, 7.30 and 8.20). Let X > '
Define the function
/a : R ^ C by fi(x) = f e"'-""'" sin^ a da.
Jo
(i) Prove by the substitution of variables a h> tt — a, for x e R,
fx{x) = 2 / cos(x cos a) sin^ a da =2 I cos(x sin a) cos^ a da.
Jo Jo
Conclude that fx is real-valued.

634 Exercises for Chapter 6: Integration
(ii) Show that
IQ, g-'-v COS<. gjjjZ^ ^ ^^ (« e N, X € R),
A(">(x) = (-0" Tcos",
^0
and conclude that f^ = fx+i — fx on R.
(iii) Verify by integration by parts that (2A. + l)fl(x) = —x fx+i (x), for x e R.
(iv) Prove that fx satisfies the following differential equation for u e C^(R):
X u"(x) + (IX + 1) u'{x) + X m(x) = 0 (x e R).
The Bessel function of order A. is defined as the function
A:R,^R wi,h A(„ = _j_L_^(|;^'.-".,„"„.„.
(v) Demonstrate by integration that J\_{x) = J^^, for x e R+.
(vi) Show by part (iii) that Jx+i(x) = —J[(x) + j Jx(x), for x e R+. Conclude
that
(vii) Using part (iv), prove that on R+ the function Ji satisfies the following,
known as Bessel's equation:
x^ m"(x) + X u'(x) + (x^ - X^)u(x) = 0 (x e R+).
(viii) Proveby means of Exercise 6.50.(iii) that lim_v 4,0 (t) Jx(x) =
1
r(A+i)-
(ix) For fixed X e R prove that the following power series expansion is convergent
uniformly in a e R:
-/^cos<. ^ y^(_n*^_cos*^Q!.
Next apply the theorem on the termwise integration of a uniformly convergent
series, then Exercise 6.50.(iii) and (vii), to conclude that
^2J /-fk\r(X + k + l)\2J ^'

Exercises for Chapter 6: Integration 635
Now let A. < — i. Then the defining integral for 7^ diverges. Nonetheless it follows
by Exercise 6.57 that the right-hand side in part (ix) is well-defined if A. ^^^ —n, for
« e N. Therefore we define the Bessel function J^ for those values of A. by means
of the series in (ix).
2 cosji:
(x) Using Exercise 6.50.(vii), prove J_i_(x) = J j^^^, forx e R_|_.
(xi) Verify that Jx and J_x both satisfy Bessel's differential equation from (vii).
(xii) Prove by part (viii) that Jx and J_x are linearly independent functions over
R.
Background. In texts on ordinary differential equations it is proved that the
dimension over R of the solution space of Bessel's differential equation equals 2.
Bessel's differential equation often occurs when spherical coordinates are used, see
for example Exercise 6.68.
(xiii) For « e No and x > 0, show that
J„{x) = — / e'-"'""-'""rfa = - / cosina-xsma)da.
In y_jr n Jo
ffint: Write
and prove by means of Newton's Binomial Theorem, for m € No,
— /" (e'" - e--)»+2'V-'>- da = (-1)'" ("« + 2'" Y
In y_^ \ m J
Exercise 6.67 (Kepler's equation - sequel to Exercises 0.19,3.32 and 6.66). This
equation reads, for unknown x e R with y e R^ as a parameter,
X = y\ +>'2sinx.
Geometrically this comes down to a point of intersection of the graph of the affine
functions \-^ x — y\ withthegraphof the sinex i-> )'2sinx. Ifx = fjr{y)Wiihfjr :
R^ —>■ R is a solution, then /,,, : R —>■ R with fy^iyi) = i^iy) — yi = yi sin 1/^(3')
is an odd Ijt-periodic function. Therefore yi i-> D\fjr{y) is even and Ijt-periodic.
Prove by implicit differentiation
Difiy) =
1 — y2Cosx
Prove by Fourier series expansion according to the variable y\, compare with
Exercise 0.19, with coefficients depending on )'2,
\ [^ 1 X—V coswvi [^ coswvi
D,f{y) = —\ dy,+T -i ^-^dy,.

636 Exercises for Chapter 6: Integration
By means of the substitution yi = yi(x) = x — y2 sinx and of Exercise 6.66.(xiii),
show that
1 r ^.r-^cosnyi r
= 1 + 2 ^ 7„ (ny2) cos nyi.
Hence conclude that
X = i/(y) = y, + 2 2_^ .
Here we do not consider the problem for which values of y the series actually
converges.
Exercise 6.68 (Helmholtz' equation and Bessel function - sequel to Exercises
3.9, 3.17 and 6.66 - needed for Exercise 8.35). We use the notations from those
exercises. We look for / e C^ (R^) that satisfy the following, known as Helmholtz'
equation:
(A + ii^)f(x)=0 (x€R^At>0),
which occurs, for example, in the eigenvalue problem for the Laplacian, see
Exercise 7.64.(iii). Here we try to find a solution / of the form
fo^:(r,a,e)^g(r)Y;"(a,ey,
where (r, a, 9) € R+ x ] —n, jt [ x ] —|, | [, while g e C^(R+) and / € No and
m e Z with \m\ < I. By means of Exercises 3.9.(vi) and 3.17.(i), prove that g must
satisfy the differential equation
W r^ 8"(>-) + 2r g'(r) + (ii^r^ - 1(1 + 1)) g(r) = 0.
Verify that h : r h> -Jr g{r) must then satisfy the differential equation
1n2>
2>
Show that M : X i-> /?(-) must be a solution of Bessel's differential equation
1\2n
r'^h"{r) + r h'{r) + (u?r'^ - (l + -V) hir) = 0.
x^ u"(x) + X u'(x) + (x'^ ~ (^ + ~) ) "C-^) = 0-
Conclude by Exercise 6.66 that, with a and Z> e C, every solution g of (•) is given
by
1
In particular, /±(x) = — ^^ '^"' , for x e R^ \ {0}, satisfy (A + //^)/(x) = 0.

Exercises for Chapter 6: Integration 637
Exercise 6.69 (Hypergeometric differential equation - sequel to Exercises 0.11
and 6.50 - needed for Exercises 6.70, 6.106 and 7.5). Assume a, b and c in R.
Let Z C Che the complex plane excluding the interval [ 1, oo [ along the real axis.
For every z € Z the function [ 0, 1 ] —>■ C with t i-> (1 — zt)"" is well-defined, and
uniquely determined by the requirement that its value be 1, for z = 0. Consequently,
for a, b and c in R, ^ e Z, the following function is well-defined on [ 0, 1 ] by
t h^ t''-^(i - ty-''-^(i - zty.
Moreover, the integral from 0 to 1 of this function converges, for a, b, c e R,
0 < b < c, z € Z; thus one defines the hypergeometric function 2F\{a, b;c:-) =
F(a, Z>; c : •) : Z —>■ C by
W F(a,b-c:z)= ^.,^^^f ■■ f t'-\l-tr-'-\l-ztydt,
r(b)r(c-b) Jo
with r as in Exercise 6.50.
(i) In the notation from Exercise 0.11 prove
r(a+n)
(a)„ = (n € N),
T{a)
and conclude from that exercise that, if ^ & Z,t e ] 0, 1 [, with \zt\ < 1,
■^ ria)n\
nsNo
(ii) Now use integration term-by-term of the power series in (i), then
Exercise 6.50.(iv) to conclude that F(a,b; c : z), for all ^ e C with \z\ < 1,
is given by the following hypergeometric series:
r(c) ^r-^ r(a+n)r(b + n) „
F(a, b;c:z) = 7
r(a)r(b) ^f-^^ r(c + n)nl ^
_ ab a(a+ l)b(b + 1) 2
^71^^ c(c+l)l-2 ^
a(a + l)(fl+2)Z>(Z>+l)(Z> + 2)_3
c(c +l)(c + 2) 1-2-3 ^
The terminology "hypergeometric" is explained by, inter alia, the relation with the
geometric series
'—r. i- — z
«sNo

638 Exercises for Chapter 6: Integration
(iii) Assume 0 < a < c and z ^ Z. Prove F{a, b\ c : z) = F(b, a; c : z), that is,
F(a,b;c:z)= ^. ^Jf r f t"-'(I - tf-"-'(I - zty'dt.
r(a)r(c-a) Jo
(iv) Demonstrate (compare with Exercise 6.50.(vii)), for \z\ < 1,
(l+z)" = F(-«, 1; 1 :-2); log(l + z) = zF(I, V, 2 :-z);
-113 2\ ^r-^(2n-\)n ^2"+'
arcsin z
1 • 3 • • • (2« - 1) ^2"+!
^^ (2«)!! 2n + l
^ + 2-. 2-4---2« 2« + r
/I 3 \ 7^"+'
arctan. = ^^2'^= 2^ "^^ = I^(-^>"^^
«sNo
r(a)r(b)
For the moment, write /(z) = p', F(a, Z>; c : z).
(v) Provethat, for |z| < 1,
\ dz ^ ^ T{c + n)n\
and conclude that the function / satisfies
(vi) Verify, for every p e R,
d / d \ d^ d
Tz\'Tz^n='ir-^^'^'^Tz
and conclude from (v) that u = F(a,b; c : •) satisfies the following hyper-
geometric differential equation:
(**) z{l-z)^{z)+{c-{a+b+l)z)^{z)-abu{z) = Q (z € C).
dz^ dz
(vii) Let / e No. Assume the function u satisfies the hypergeometric differential
equation with
a = —I, b = I + \, c = I.
Define ^ : R ->■ Rhy z = ^(t) = ^. Verify that v = u o^ satisfies
Legendre's differential equation from Exercise 0.4.(i). Note that in this case
the parameters a, b and c do not satisfy the restrictions mentioned above.

Exercises for Chapter 6: Integration 639
We now want to prove directly from its integral representation (•) that F(a,b; c : ■)
satisfies the differential equation in (••). To do this we define
g(z,t):=t\l-tr-\l-ztr'-';
then
^ -iz, t) = t"-'(i - ty-"-'(i - zty-'hiz, t),
at
with
hiz, t) = Z> (1 - 0(1 - Zt) + (Z> - c) r(l - zt) + (a + 1) ztii - t)
= b{\- ztf + {{a + b + l)z -c)t{\- zt) + {a + \) z{z - l)f^
So now we have
^(2,0 =bt''-\\-tY-''-\\-ztr''
at
+((fl + b + l)z-c)t\l - 0^-'^-'(1 - zty-'
+{^ -z){a + 1) t''+'(1 - 0'~*~'(1 - zt)-''-^.
In addition we write for the integrand in (•)
k{z,t) := t''-\\ - ty-''-\\ - zt)-".
Note that then
"—iz,t) =at\l-tY-'-\l-ztr''-'\
az
\^{Z, t) = aia + l)t'+\l - rr-'-'(l - Zt)-"-^;
az^
9? T 9^A; 9A;
^^(^' ^) = (^ - ^)t^(^' 0 + ((« + ^ + 1)Z - c) — iz. t) + aZ>/:(z, 0-
at az^ az
(viii) Prove that the results above do indeed imply that F{a, b;c:-) satisfies (••).
Exercise 6.70 (Confluent hypergeometric differential equation - sequel to
Exercises 6.66 and 6.69 - needed for Exercise 6.71). Assume a and c in R are such
that 0 < a < c. We define the confluent hypergeometric function or the Kummer
function \F\{a\ c : ■) = K(a; c : •) : C —>■ C by
r(c)
K(a; c:z) = z—-f^ r f t"-\l - tf-'-'e''dt.
r(a)r(c-a) Jo

640 Exercises for Chapter 6: Integration
(i) Let /x be as in Exercise 6.66. Prove, by the substitution of variables sin^ | =
s,
u(t) := e^-fj—) = l^^^^^^^-* K{X+-- 2X+l:t) (t € R).
Next, use Legendre's duplication formula from Exercise 6.50.(vi) to show
that the following formula holds for Ji, the Bessel function of order X:
Jx(x) = p^^^^^^g-''(|)V(A + ^; 2A + 1 : 2ix) (x € R+).
(ii) Use termwise integration of a power series, then Exercise 6.50.(iv), to
conclude that, for all z € C,
nsNo
_ . a_ a{a +1) 2 , a{a + l){a + T) 3
c 1 ^ c(c + 1) 1 • 2 ^ c(c + l)(c + 2)l-2-3^
Hint: Use the ratio test to prove the convergence of the series for ^ e C.
(iii) Show that part (ii) implies K{a;a:z) = e', for all a > 0 and ^ e C.
(iv) Let a, b, c, z and F(a, b;c:-) be as in Exercise 6.69. Note that
lim (1 - ^)-* = e-^',
and use Exercise 6.69.(iii) to prove the following equality for the integral
representations:
(•) lim F(fl, b;c:-) = K(a; c : z).
b-*oo b
With a, c and z fixed, show the series for F(a, Z>; c : |) from Exercise 6.69.(ii)
to be uniformly convergent for Z> e [ 2|z |, 00 [, and conclude that the identity
(•) also holds for the series representations.
For the moment, write f(z) = j^K(a; c : z)-
(v) Prove, for z €C,
\ dz I ^--' r(c + «)«!
to conclude that the function / satisfies

Exercises for Chapter 6: Integration 641
(vi) Verify that part (v), in the same manner as in Exercise 6.69.(vi), implies that
u = K(a; c : •) satisfies the following confluent hypergeometric differential
equation:
W Z ^(Z) + (c-z) ^(Z) -au(z) = 0 (z€ C).
dz'^ dz
(vii) Let F(a,b;c:-)he: as in Exercise 6.69. Verify that m : ^ i-> F{a, b; c : |)
satisfies the differential equation
.d'^u ^ ^ ( a + \ \du
( Z\d"u ( a + 1 \du
0.
Conclude that (•) is obtained from this equation by taking the limit for b —>■
oo, as expected on the basis of part (iv).
(viii) Assume u satisfies (•) and write w(z) = e~^^z^''u(z), k = ^c — a and
m = |c — |. Verify that w then satisfies the following, known as Whittaker's
equation:
<Z)-(--- + ^U(2) = 0.
G
dz"-'' \A z z} ,
(ix) Prove by Exercise 6.66.(iv) that the function u from part (i) satisfies the
following confluent hypergeometric differential equation:
t u'\i) + (2A + 1 - 0 u\t) - (A + -) m(0 = 0 (r € R).
Background. One observes that we have now proved that the confluent
hypergeometric differential equation is a limiting case of the hypergeometric differential
equation, and that Whittaker's and Bessel's differential equations transform, by a
substitution of variables among other things, into a confluent hypergeometric
differential equation (see also Exercise 6.71).
Exercise 6.71 (Further confluence - sequel to Exercise 6.70).
(i) As in Exercise 6.70.(vii), determine a differential equation for m : ^ i->
K{a\ c : ^^, and by taking the limit for a —>■ oo derive the following
differential equation:
d^u du
(*) Z^^iz)^C^iz)-uiz)=^ (2€C).
dz'- dz
(ii) Verify that (•) is satisfied by oFi(c : z) := r(c) J^nsNo r(c+H)»! ^"-

642 Exercises for Chapter 6: Integration
Assume u satisfies (
differential equation
_2
(iii) Assume u satisfies (•), and write v(z) = u{ — ^). Show that v satisfies the
-c
v"iz) + i2c-l)v'iz) + zviz)=0.
Note that, with c = A. + 1, this is the differential equation from
Exercise 6.66.(iv) associated with the Bessel function.
(iv) Finally, examine the effect of the substitution w(z) = u(cz), and thus derive
the differential equation w'(z) — w(z) = 0, for w = qFq : ^ i-> e' =
Z^nsNo ii\^ ■
Exercise 6.72 (Cauchy-Schwarz inequality - needed for Exercise 6.76). Here
we give two proofs of this integral inequality. It is assumed that / and g are
contained in Cc(R")-
(i) Prove that the following numbers are well-defined:
\\f\\2={f \f(x)\^dxY\ {f,g)=l f{x)g{x)dx.
J R" J R"
(ii) Verify that Proposition 1.1.6 as well as its proof apply and directly yield
l(/. ^)l ^ Il/ll2ll^ll2' known as the Cauchy-Schwarz inequality.
(iii) Prove ab < \{a^ + Z>^), for all a, Z> e R. Substitute a = -^ and b = ^
and integrate, to conclude again that the Cauchy-Schwarz inequality holds.
Exercise 6.73 (Holder's and Minkowski's inequalities - sequel to Exercise 5.41
- needed for Exercises 6.74, 6.75 and 6.79). For p > 1 and / in Cc(R"), we
define
MP
^ = {fjm\"dx)'
J- + -L
Prove the following, known as Holder's inequality:
(i) Assume p > 1 and ^ > 1 satisfy - + - = 1, and consider f,g€ Cc(R")-
L
\f(x)g(x)\dx<\\f\\Jgl
R„ P"-"1
Hint: Compare with Exercise 5.41.(v).
(ii) Next, assume p^ > 1, for i < k < r, and JZ!</:<»■ ~ = 1' ^^^ l^t f^- e
Cc(R"), for 1 < jfc < r. Prove
f n iAwi^^< n 11/^
kWp,-
l<k<r

Exercises for Chapter 6: Integration 643
(iii) Let p > 1 and f,g€ Cc(R")- Prove the following, known as Minkowski's
inequality:
\\f + 8\\„<\\f\\„ + \\8\\,-
Hint: The inequality readily follows for p = 1. Therefore assume p > 1.
One has
f \f(x) + g(x)\Pdx <f \f(x)\\f(x)+g(x)r'dx
Jr" Jr"
+ 1 \g{x)\\f{x)+g{x)r'dx.
JR"
Now apply part (i), use (p — 1)^ = p, divide by {/jj„ |/(x) + g{x)\'' dx) ^,
and make use of 1 ' — '
Exercise 6.74 (Hilbert's inequality - sequel to Exercises 6.58 and 6.73). Let
/ = R_(_ and let A; : /^ —>■ / be homogeneous of degree —1. Let p > 1 and ^ > 1
with - + - = 1, and define c > 0 by
= j kix,\)x-^l"dx.
(i) Prove that c = Jj k(\, y)y~'^l'' dy.
(ii) Prove that, for all / and g e Cdl) one has, in the notation of Exercise 6.73,
/ Hx,y)f{x)g{y)dxdy
<': 11/11,11^11,-
Hint: We have
/ f(.x) I k(x, y)g(y)dydx = I f(x) I xk(x, xw)g(xw) dw dx
= I fM I k(l, w)g(xw) dw dx
= I k(l,w) I f(x)g(xw)dxdw.
Now apply Holder's inequality from Exercise 6.73.(i) to the inner integral
and use the fact that fj \g(xw)\'' dx = ^ fj \g(y)\'' dy.
(iii) Consider in particular k(x, y) = j^. By means of Exercise 6.58.(v), prove
the following, known as Hilbert's inequality:
Jr- x + y sm(f) p 1

644 Exercises for Chapter 6: Integration
Exercise 6.75 (Sequel to Exercises 5.41 and 6.73). Let k e C(R"+'') and assume
that Ci > 0 and C2 > 0 satisfy
f \k(x,y)\dx<ci (yeR"), f \k(x, y)\ dy < C2 (x € R").
i 1 1
P
the notation of Exercise 6.73, the following inequality:
Let / e Cc(R'0> g e Cc(R"), p > 1 and ^ > 1 with i + ■!- = L One then has, in
/ Hx,y)f{y)g{x)dxdy
I Jvu'-^p
<':y"4^'11/11,11^11,-
Hint: On account of Young's inequality from Exercise 5.41 one has, for all t > Q,
\f{y)g{x)\ < -tnf(yW' + -r''\g(x)\''.
p 1
Now minimize the resulting upper bound for the integral with respect to all t > 0.
Exercise 6.76 (Poincare's inequality - sequel to Exercise 6.72 - needed for
Exercise 6.77). Let / e Cc(R") be a C' function.
(i) Prove /jj„ f(x)'^dx = —ij^n Xj f(x) Djf(x) dx, for 1 < y < «.
(ii) Use (i) and the Cauchy-Schwarz inequality from Exercise 6.72 to show that
j nxfdx<in {xjf{x)fdx)''\j Djf(x)^dxy'\
(iii) Let U C R" be a bounded open set. Prove the existence of a constant c =
c(U) > 0 such that, for all / having the additional property that supp(/) C
U, we have the following, known as Poincare's inequality:
f f(x)^dx<c f ||grad/(x)f^x.
Ju Ju
Exercise 6.77 (Heisenberg's uncertainty relations - sequel to Exercise 6.76). In
quantum physics the state of a particle is described by means of a wave function;
this will be understood to be a C^ function with compact support / : R" x R —>■ C,
forwhich/jj„ \f(x,t)\^dx = l,forallf e R. Here \f(x, t)\^ gives the probability
density (see also Exercise 6.42) of the particle being found at the point x e R" at
time f e R. The collection of these wave functions is acted upon by the position
operators Qj and the momentum operators Pj, for 1 < j < «, as follows:
Qjfix, t) = xjfix, t), Pjfix, t) = -L=^(x, t).

Exercises for Chapter 6: Integration 645
(i) Verify that the Qj and Pj are self-adjoint operators with respect to the Her-
mitian inner product on the collection of wave functions
\g) = / f(x,t)g(x,t)dx.
if:
Then the expectation vectors x° e C" for the position and p° e C" for the
momentum, respectively, of the particle described by the wave function / are given
by
x° = { Qjf, f) and p° = ( Pjf, f) (1 < ; < n).
(ii) Check that in fact we have x° e R" and p° e R".
The uncertainties or standard deviations Axj > 0 and Apj > 0, in the y'-th
coordinate of the position x and of the momentum p, respectively, of the particle
described by the wave function / are defined by
(Axj)^ = \\(Qj - x°)ff = f x] fix, t)J{^dx - (x°j)\
{Apjf=\\{Pj - p°)/ip=I -^-^{x, t)Wj)dx - (p°)\
■' Jr" 9^7 •'
(iii) Check that one may assume x° = 0 and p° = 0, using the substitutions where
f(x) is replaced by f(x + x°) or by e~^~^ ^'' ■ ''^ f(x), respectively.
(iv) Prove the following, known as Heisenberg's uncertainty relations:
Ax J Apj > - (i < j <n).
That is, the smaller the uncertainty in the position x of the particle, the larger
the uncertainty in its momentum p, and vice versa.
Hint: See Exercise 6.76.(ii).
Exercise 6.78 (Sobolev's inequaUty). Let C^(R") = QCR") n C'(R"). In
general, the constant occurring on the right-hand side of Poincare's inequality from
Exercise 6.76.(iii) depends on the open set U C R". In contrast with this, Sobolev 's
inequality asserts the following. Let p e R with 1 < p < n and define p* e R by
111 * «
— = (that is, p = p > p).
p* p n n — p
Then there exists a constant c = c{n, p) > Q such that in the notation from
Exercise 6.73, for all/€ C^^R"),
(*) 11/11/,* <c Y. w^jfWp-

646 Exercises for Chapter 6: Integration
This is an estimate for a function / in terms of its partial derivatives which is valid
regardless of the support of /. This generality is possible owing to the occurrence,
on the left-hand side, of an integral expression containing p* instead of p. In
estimates of this kind the rate at which / converges to 0 is of importance, and |/|''*
converges to 0 more rapidly than |/|''. We now prove (•) in five steps.
(i) Apply the Fundamental Theorem of Integral Calculus 2.10.1 twice, to prove,
for i < j < n
2|/(x)| < f \Djf(x)\ dxj =: fjix)"-' (X € R").
Conclude that
|2/(x)|"/(''-')<( Yl fj(x))( f \D„f(x)\dxy^"~'\
(ii) Now integrate with respect to the variable x„, and apply Exercise 6.73.(ii),
with
1 1 1
— = ••• = = -, and fj (1 < ;■ < «).
Pi Pn-i n - 1
Conclude that
/'|2/(x)|"/('-')rfx„
^R
W n \Djf(x)\dxjdx„j n \D„f(x)\dx„j
<
l<j<n—l
(iii) Integrate with respect to the variable x„_i, and again apply Exercise 6.73.(ii),
this time with the functions, for 1 < y < « — 2,
\Djf(x)\dxjdx„j , n \D„f(x)\dx„j
Repeat these operations to prove
^J2/(x)|"/('-')rfx< fj (^j^JDjf(x)\dxY^"~'\
1)
l<J<n
that is.
^^ |/(x)|"/("-') dx) < 2 ( n J^„ \DjfM\ dx) .
iSjS"'

Exercises for Chapter 6: Integration 647
(iv) In the case p = 1, the inequality (•) immediately follows from (••) by means
of the elementary inequality
(***) n Sj ^ ^( E Sj)" (gj > 0, 1 < ;• < n).
l<J<n l<J<n
(v) Now let p > 1. Then
n — 1 ^ n — 1
p = p > 1;
n n — p
and as a consequence we have/"^''* e C^(R"). Apply (••), with/replaced
by /~ii~'' . Then use the fact that, for q obeying ^ + ^ = 1,
f \f(x)\"-^"*-'\Djf(x)\dx
{j^ \f{x)f-^"*-'^^dx)'^\j^ \Djf{x)\Pdx)'^".
<
One has
,n — \ ^ s ^ n — \ 1 1 «—1^ n — \
( P-^M=P\ =—> P= P-
n n q p* n n — p
Therefore, (••) leads to
Now use (• • •) once again to prove the validity of (•), for p > 1.
Exercise 6.79 (Hardy's inequality - sequel to Exercise 6.73). On the left-hand
side in Sobolev's inequality from Exercise 6.78 an exponent p* occurs which differs
from p on the right-hand side. In that exercise the point was made that this is related
to the rate at which the integrands converge to 0. On R the desired increase in the
rate of convergence is obtained if on the left-hand side one writes - f{x) instead
of f{x); in this context, see also Exercise 7.72. That is, for functions / e Cl(I)
where / = R_|_, one has Hardy's inequality, which reads
-f(x) dxj <-^(^j \f'(x)\!>dxj .
We prove (•) in three steps.
(i) Write F(x) = ^f(x), and prove fj Fix)" dx = -p fj F(xy'-^ x F'(x) dx.

648 Exercises for Chapter 6: Integration
(ii) Note that x F'{x) = -F(x) + f'(x), and deduce (p - 1) fj F(x)p dx =
pf^F(xr-'f'(x)dx.
(iii) Check that there is no loss of generality in assuming that F(x) > 0. Then use
Exercise 6.73.(i) and arguments such as those in Exercise 6.73.(iii) to prove
Exercise 6.80 (Inequality by Fourier transformation). For / e -5(R^0> write
(compare with Exercise 6.73)
||/||^ = sup{|/(x)||x€R"}, \\f\\i=f \f(x)\dx.
Show that there exists a constant c = c„ > 0 such that, for all / e Siti"^),
\\f\\oo<C„ J2 ll^"-^ll'-
|q:|<«+1
Hint: Prove
l<j<n |Q:|<n+l
and deduce
\m\<c'„ii + \\^fr"'^ J2 \^"f(^)\
\a\<n+l
ti ^-^ ^r~-~ m2n-^
|q:|<«+1 |Q:|<n+l
Verify ||/||oo < (27r)~"||/||i by means of the Fourier Inversion Theorem, and
conclude that
ii/iioo < < /" (1 + mY"-^ d^ Yl ii^"/iii-
" |Q:|<n+l
Exercise 6.81 (Eigenfunction for translation). Suppose/ e C(R")isanontrivial
eigenfunction for translation, that is, for every y e R" there exists X(y) e R with
(*) f(x + y)=X(y)f(x) (x€R"), /(O) = 1.
Show X(y) = f(y), for y € R". Select g € Cc(R") n C°°(R") satisfying
/r" /(>')^(>')^>' = 1- Deduce
ix)= f fix + y)giy)dy= f
Jr- Jr-
fix) = f(x + y)g(y) dy = g(y - x)f(y) dy,
JR" JR"
and prove/ e C°°(R") by differentiationunderthe integral sign. Now differentiate
(•) with respect to the variable y and show f{x) = e^/W^, for all x e R" (compare
with Exercise 0.2.(iii)).

Exercises for Chapter 6: Integration 649
Exercise 6.82 (Another proof of Theorem 6.11.3.(iii)). The assertion can be
proved without recourse to the Differentiation Theorem 2.10.13.
(i) Using Taylor expansion of t h> e~" show that there exists a constant c > 0
such that for all r e R we have \r(t)\ < ct'^ if e"'' = 1 —it + r{t).
(ii) Deduce, for f, ?; e R",
m + r))- m = f e-'<-.?>(e-'<-."> - l)f(x)dx
= f e--^''^K-i{x,n}+r({x,n}))f(x)dx
JR"
IR"
=:-i J2 rij(:~f)(^) + R(^,V).
Conclude by part (i) and the Cauchy-Schwarz inequality that
\R(^,m <f \r((x,rjmf(x)\dx<cM^ f \\xf\f(x)\dx
Jr" Jr"
= 0(\\ri\\\ IhlUO.
Deduce that / is differentiable in f; apply this argument repeatedly to
conclude that /"e C°°(R"), and that Theorem 6.11.3.(iii) holds.
Exercise 6.83 (Another proof of Example 6.11.4 - sequel to Exercises 2.84 and
6.51).
(i) By taking the real and imaginary parts of ^, for n = I, show that the result
from Example 6.11.4 is equivalent to the formulae, valid for f € R,
—e~2?" = / e~2-*" cos(fx)rfx, 0= / e~^^ sin(fx)/ix.
(ii) Demonstrate that the first identity follows from Exercise 2.84 or by series
expansion of the cosine and by Exercise 6.51, and that the second one is
trivial.
Exercise 6.84. Assume 0 ^ g € -5(R") and that g vanishes in a neighborhood in
R" of 0. Let f ='g e -5(R"). Prove that all moments of / vanish, that is.
f x"fix)dx = 0 (a €N2).
Jr"

650 Exercises for Chapter 6: Integration
Exercise 6.85. Prove that the convolution f*g€ -5(R"), if f and g € -5(R").
Hint: Use that |x^| < I'^l T,o<j<\p\ W^ " >'ll-' ll>'ll'^'~-''' ^r all x, y € R" and
Exercise 6.86 (Parseval-Plancherel's identity - needed for Exercises 6.94,6.100
and 731). Let /, h € -5(R'0 and y € R".
(i) Prove /^„ /(f)/7,(f)e'(v.?>^| = /^„ f(x)h(x - y)dx.
(ii) Conclude that /^„ f(x)h(x)dx = /^„ f^(^)h(^)d^.
Now let g e -5(R"), and introduce h = (Inyf.
(iii) Verify that h € -5(R") and 'h = g.
(iv) Conclude that we have the following, known as Parseval-Plancherel's
identity:
f f(x)g(x)dx = (2nr" f mm)d^-
Exercise 6.87 (Poisson's summation formula for R - sequel to Exercise 0.18).
Let/ e -5(R). Multiplybothsidesof the identity in ExerciseO.18.(ii)for« = 2by
the second derivative f"(x), integrate over R, and use integration by parts twice,
keeping in mind that limj:_^±oo/(^) = lim_v->±oo/'(x) = 0. Verify Poisson's
summation formula (see Exercise 6.88.(ii) for a different proof)
^/(/:) = ^/(;27r/:),
fcsZ ksZ
and derive
J2f(x+k)=J2 T(?-^k) e^""'"'"* (.X € R").
fcsZ" fcsZ"
Exercise 6.88 (Poisson's summation formula and Jacobi's functional equation
- sequel to Exercise 0.19 - needed for Exercises 6.89, 6.90, 6.91 and 6.96). For
/ € S{W) the function F : R" ^ R is well-defined by F(x) = J^j^Z" f(^ + /)•
Check that this series, as indeed the series of the a-th derivatives also, is uniformly
convergent on R", for a e Z". Thus one defines a C°° function F on R" which
is invariant under translation with elements from Z". Moreover, (fk)ks'Z," with
fic(x) = e-^'^'^'^Hsa maximal orthonormal system on [ 0, 1 ]"; expansion according
to Exercise 0.19 of F in the corresponding Fourier series therefore yields F =

Exercises for Chapter 6: Integration 651
(i) Demonstrate, for k e Z",
{F,fk) =1 V/(x + ;>-^'<*^''''rfx= / e-'<^*^'^>/(x)^x
= Tilnk).
(ii) Prove the following two formulae, the latter of which is known as Poisson 's
summation formula, for x e R":
fcsZ" fcsZ" fcsZ" fcsZ"
Define ■\j/ and 0 : R_(. —>■ R by
fix) = ^£—''^ 0(x) = ^{f(x) -1) = ^£—''\
nsZ HSN
(iii) Using part (ii) and Example 6.11.4, prove the following, known as Jacobi's
functional equation:
f(-j=x^-f(x) (xeR+).
(iv) Conclude that 0(^) = x^ <p(x) + ^(x^ — 1) for x e R+.
(v) Demonstrate 0 < 0(x) < J^nsN^'""'' = T^^ = Oie"'"'), x -> oo.
(vi) Conclude by parts (iv) and (v) that 0(x) = 0{x~^), x \Q.
Exercise 6.89 (Functional equation for zeta function - sequel to Exercises 0.25,
6.50,6.57, 6.58 and 6.88). For j e C with Re j > 0 one has
n-iT{-)—=i e-'^'-'^xi-'^x (« € N).
^^'n' Jr.,
Sum over « e N and interchange the order of summation and integration. Let
(j) : R+ —>■ R be as in Exercise 6.88. Using parts (v) and (vi) from that exercise one
obtains, for Res > 1,
A(s):=7T-2r(-)^(s)= I y'e-'"''-^x'-'^x= / 0(x)x*-'rfx.

652 Exercises for Chapter 6: Integration
Prove with Exercise 6.88.(iv)
/ <p(x)xi~^dx= I <p(x)x^~^dx (Rej > 1).
Jo J\ l-s s
Therefore
A{s) = I (j>(x)x^~^ dx h / (j>(x)x^~^ dx (Rej > 1).
Ji i-s Ji s
The integral
(j>(x)x^~^ dx
converges for all s € C, and moreover it defines a complex-differentiable function
of s € C. This implies that A(s) can be defined for all j e C with the exception of
s = I and J = 0. Because r(|) ^ 0 for all s € C, the preceding result defines ^(s)
for those values of s also. In this connection, see Exercise 6.57 for the definition
of r(s), with s € C Furthermore, we find that s i-> A(s) is invariant under the
substitution s h> 1 — ^; that is,
A(.) = A(1-.); and r{^) C(s) = n'-^-r{^) C(l - s).
Multiply both sides by r(l — ^s), then apply the reflection formula from
Exercise 6.58.(iv), and Legendre's duplication formula from Exercise 6.50.(vi); this
yields \he functional equation for the zeta function
I'-^Tz'^il -s) = cos {-7r)r(s)^(s) (s € C).
In particular we find, by Exercise 0.25, the result from Exercise 0.20 or 6.40.(iv)
r(2n) = (-l)"-'i(2:r)^"-f?^.
2 (2«)'
Exercise 6.90 (Poisson's summation formula and Fourier Inversion Theorem
- sequel to Exercise 6.88). For / e -5(R") and x e [0, 2??: ]" define g e SiR")
hyg(t) = e-'^''-^^f(t).
(i) Verify J"(f) = /(f + x). Next apply Exercise 6.88.(ii) to g and deduce
Yl f(k)e-'^^'-'^ = Yl filnk+x).
(ii) Integrate this equality termwise with respect to the variable x over [ 0, In ]"
in order to obtain
(In)" f(0) = T f m^k+x)dx= I m dt
Obtain the Fourier Inversion Theorem 6.11.6 for / e -5(R").

Exercises for Chapter 6: Integration 653
Exercise 6.91 (Partial-fraction decomposition of hyperbolic functions - sequel
to Exercise 6.88). For r > 0 we define / : R -> R by /(x) = e^'l-'l. Then we
have (see also Exercise 6.99.(ii))
m = 2 ^ e-" cos fX dx = ^^^ (f € R).
2t_
t^T^^
(i) Show by summation of the geometric series, for f > 0 and 0 < x < 1,
y^ g-'|A+fc| ^ g-n _^ y^ g-'{^+k) _,_ y^ g/(A-fc) ^ coshf(x-^)
■^ i-j i-j sinh ^
(ii) Prove by using Exercise 6.88.(ii), part (i), and the fact that / is an even
function
coshf(x - t) 2 x-^ It
, ^ =-+2V-- —• cos(277:;cx) (r > 0, 0 < x < 1).
sinh^ t ^t^ + 47r^k^ v / v > _
(iii) Substitute x = 0 and x = |, in the identity in (ii) and derive the following
partial-fraction decomposition of the hyperbolic cotangent and the hyperbolic
cosecant, respectively, for x € R \ {0},
n cotanh(7rx) = —|-2x > — —, -^-——- = —|-2x > -
X r^-^ +«: smh(7rx) x r~:<-^
i-iy
1-2 1 1-2
(iv) Use the equality cotanh(7rx) — ^^^L^, = tanh(7r|) to obtain from (iii)
1
x2 + (2k - 1)2
TT TT X—\ 1
— tanh(—x) = 2x > (x e R).
(v) Differentiate the identity in (ii) with respect to x, take x = ^^, and use the
equality sinh 2=2 sinh j cosh j to find
fcsN
Prove
b = '^"I^;^Tl^ ""(!')•
cosh ^ ^-^ t'^ + 47r2jfc2
""-—= 2y(-l)^-' , ^^ ^ , (x€R).
2cosh(|x) ■p x2 + (2Jfc-l)2

654 Exercises for Chapter 6: Integration
Background. By replacing x by ix and using the identities cosh ix = cos x and
sinhzx = i sinx, we see that the identities in parts (iii) - (v) go over in their
counterparts in Exercise 0.13.(i).
(vi) From parts (iii) - (v) deduce the following formulae, by expanding the
denominators of the integrands in a geometric series and integrating termwise,
for f € R:
/■ sin f X , 71 / , ^ 1 \
/ ax = — I cotanh tt^ ),
yR... e" -I 2\ n^J
L
L
L
sinfx , n , ,7t .
ax =—tanhl—f),
R sinhx 2 2
cos f X 71
-dx =
R.i. cosh X 2 cosh (I f)'
e-'^? . 71
■ dx =
Rcoshx cosh(|f)'
Exercise 6.92 (Solution of heat equation - sequel to Exercise 6.43 - needed for
Exercises 6.103 and 8.35). Let jfc > 0 and define g : R"+' \ {0} ^ R by (see
Formula (6.48))
g(x,t) = gt(x) =
{A7ikt) 2 e 4i/ ,
0,
t > 0;
t <0.
(i) Prove that g : R"+' \ {0} -> R is a C°° function (see the proof of
Theorem 6.7.4 for the points (x, 0) € R"+' with x ^ 0).
(ii) Using Exercise 6.43, check that /jj„ gi(x) dx = I, for t e R+.
(iii) Prove, for x e R" \ {0}, r e R+ and 1 < ; < n.
Conclude that g on R"+' \ {0} satisfies the heat equation (6.46).
(iv) Conclude by application of the Differentiation Theorem 2.10.13 that the
function u from Formula (6.49) satisfies the heat equation; so that one has
u(x, t) = if * g,)ix) = / f(y)g,(x - y) dy
Jr"
= 7T-"^^ I fix - 2V^3')e-"-^'"' dy.
Jr"
Further show that m(x, 0) = /(x).

Exercises for Chapter 6: Integration 655
Exercise 6.93 (Fourier transform of probability density of normal distribution
and covariance matrix - sequel to Exercise 6.44 - needed for Exercise 6.94).
Let C € Mat+(«, R) be positive definite, and define / : R" ->■ R by
f(x)= -i_=e-i(c-'-v,.v>
(27r)2VdetC
(i) (Compare with Example 6.11.4.) Verify (Dkf)(x) = -(C-^x)kf(x) for
\ < k < n. Now write D = (Di,..., D„). Fourier transformation then
yields
-^km = ((C-'£>),,f)(f) (l<k< n).
Next, multiply by Cj^ and sum over 1 < A; < « to obtain — (Cf)j/(f) =
(Djfx^), for 1 < ;■ < «. Now conclude that /(f) = e'^-^^^'^K
(ii) Expand both sides of the identity
in a power series in f e R", and compare the coefficients. This leads to
/ f(x)dx = 1; / xjf(x)dx = 0, / xjXkf(x)dx = Cjk,
Jr" Jr" Jr"
for 1 < i,k <n.
Now let ji e R" and define
f{x) = „\ ^-i(C-'(.v-^),(.-^))_
(277:)2VdetC
(iii) Prove, for i < j,k ■<n,
I fix) dx = \, I (Xj - fij)fix) dx = 0,
^R" ^R"
(Xj - iJ,j)(Xk - lJ^k)f(x) dx = Cjk-
L
Background. In the terminology of Exercise 6.42 the function / = /(/x, C) is
said to be the probability density of the normal distribution on R" with expectation
vector ji and covariance matrix C.
Exercise 6.94 (Asymptotic expansion for oscillatory integral - sequel to
Exercises 6.86 and 6.93). Let V € Mat+(«, R) be positive definite.

656 Exercises for Chapter 6: Integration
(i) Prove by Exercises 6.86.(ii) and 6.93.(i) that, for all / e -5(R") and t e R+,
f f(x)e-^-'^^'''''^ dx = (2nt)-i(detVr'^ f /(f)e-i<^"'?'?> rff.
With the use of the theory of functions of one complex variable it can be shown
that a similar identity holds for the following oscillatory integral with amplitude
function f and quadratic phase function, which is obtained by replacing t with —io),
where i = V—1 and co > 0:
f /(x)e?'"(^^'^>^x = (277:ftj)-2|detVr2e'?^g"^ f /(f)^-^*^"'?'? > ^f.
^R" ^R"
Here sgn V equals the signature of V, the difference between the number of
positive and the number of negative eigenvalues of V, all counted with multiplicities.
Introduce the differential operator
(V-'D,D)= J2 (V-')jkDjDk.
i<j,k<i!
(ii) Prove by means of Theorem 6.11.3.(ii) and Theorem 6.11.6 that, for k e Nq,
(iii) Prove that we have the following asymptotic expansion (see Exercise 6.55)
for the oscillatory integral:
^R"
(^)'|detVrie'f='g"^^^((^(ajV)-'£>, £>)V)(0),
0) —>■ OO.
fcsNo
Exercise 6.95 (Principle of Stationary phase). Let0 e C°°(R")and/ e Q°°(R").
Define the oscillatory integral I : R+ —>■ C with phase function <p and amplitude
function f by
I{ay)= / e""^^-'">f(x)dx.
Jr"
We want to study the asymptotic behavior of / (m), for m —>■ oo, under the
assumption D(j) (x) ^ 0, for all x e supp(/), that is, (p has no stationary points in supp(/).
We define the differential operator L on R" by L = \\D(j>\\-^Y,i^j^,XDj(j>) Dj.
(i) Prove 1(0)) = ^ 4„ L{e"^1>)(x)f(x)dx.

Exercises for Chapter 6: Integration 657
(ii) Show that there exists a differential operator L^ on R" with
1(0)) = — f e"^^-'\LU)ix)dx.
io) Jj^„
(iii) Demonstrate that 1(a)) = 0{cl>~'^), (W ->■ oo, for all A; e N.
Background. The result in part (iii) shows that the function m h> I (m) may not
decrease arbitrarily rapidly, for m —>■ oo, only if the phase function (p has stationary
points in supp(/). In Exercise 6.94 we have a quadratic phase function (p, with
a stationary point at 0. In that case, I(o)) ~ cm~^, for m —>■ oo, with c ^^^ 0 if
/(O) 7^ 0.
Exercise 6.96 (Gamma distribution, Lipschitz' formula and Eisenstein series
- sequel to the Exercises 0.20, 6.42, 6.50 and 6.88). Let a and A. > 0. In the
terminology of Exercise 6.42, the function /q.^ ;, : R —>■ R with
fa.xix) =
0, X <0-
I r(«)
x"-v-^\ x>o.
is said to be the probability density of the Gamma distribution with parameters a
and A.. In particular, for « e N, the function
f„ 1 (x) = — x5-'e-5^ (x e R+)
9 ' 7
is said to be the probability density of Pearson's y} distribution with n degrees of
freedom.
(i) Prove
/ fci,xix)dx = 1, / xfa,xix)dx = -,
^R ^R A.
/ {x- iiff^^^{x)dx = —.
Thus I is the expectation and ^ the variance of this Gamma distribution. In many
cases the convolution and the Fourier transform are well-defined for functions not
contained in -5(R), and this is also true in the particular case of the /q.jl-
(ii) By means of Exercise 6.50.(iv) prove/ajj^ */Q,,jt = /Q,|_,_Q,2,A.>fora!i, 0:2 and
A > 0.
(iii) Show by Exercise 6.69.(i) that .^(f) = [j^f, for |f | < A.

658 Exercises for Chapter 6: Integration
In fact, this formula is true for all f e R. Moreover, the Fourier Inversion Theorem
is valid in this case. Let J{ be the upper half-plane {z € C {hnz > 0}.
(iv) For any z € J{, show that the Fourier transform of x i-> (x — z)~" equals
0, f > 0;
^^
r(«)
(v) Now use Poisson's summation formula from Exercise 6.88.(ii) to derive Lip-
schitz' formula
f^^iz + nf ik-l)\^
Furthermore, prove the following identity for the Lerch function A, valid for
Z € ^, X e R and a > 0:
(« + x)"-'^^--' = -^-^ V —.
(vi) Lipschitz' formula also can be obtained by expanding the right-hand side of
the following formula from Exercise 0.13.(i):
I 1 e^^'^
= Ti co\.{tiz) = —Tii — 2ni r—:-
Z + n 1 - e^jnz
nsZ
as a geometric series in e^^'^ (this is where the condition ^ e ^ is
necessary) and differentiating k — \ times with respect to z. Conversely, the
partial-fraction decomposition of the cotangent can be derived from
Lipschitz' formula.
The Eisenstein series Gk of index A; > 1 is the function Gk '■ 3^ ^ C given by
(0,0)5i(m,«)sZxZ "- ^ ' ^ mgN hsZ "- ^ ' ^
(vii) Apply Lipschitz' formula with z replaced by mz, to get the so-called Fourier
expansion of Gk at infinity
Gkiz) =2^(2k) + ^^^^ E ^n^^-^.^-"-

Exercises for Chapter 6: Integration 659
Here cr2k-i(j) denotes the sum of the (2k — l)-th powers of positive divisors
of j, while in the last equality we used Exercise 0.20.
Exercise 6.97 (One-sided stable distribution - sequel to Exercises 2.87 and
6.42). In the terminology of Exercise 6.42, the function / : R —>■ R with
fM =
0, X < 0;
—=x 2e -I, X > 0,
is said to be the probability density of the one-sided stable distribution of order i.
Define, for f > 0,
(i) Using Exercise 2.87.(vii), prove that /j^ fix) dx = I, for all t > 0.
(ii) The convolution is well-defined for the functions/,. Show/,2*/,2 = /(„_|_,,)2,
for ^1 and t2 > 0.
Hint: Verify that in x > 0 the left-hand side equals
^ Jo
y)y) '-e ^-y y dy,
and introduce the new variable ^ > 0 via z = 7^. Finally, use
Exercise 2.87.(vii).
Exercise 6.98 (Bessel function as Fourier transform - sequel to Exercises 6.50
and 6.66). The Fourier transform/is often well-defined for functions/ : R" —>■ R
which are not contained in -5(R"); in particular this applies to the characteristic
function / of the unit ball B" in R"; we therefore define
W m= f e-'^'-^Ux (f €R").
Jb"
(i) Check that / : R" —>■ C is a continuous function, and that /(O) = f^rpr-
(ii) Note that the expression on the right-hand side of (•) is independent of the
direction of f e R". Therefore, set f = (0,..., 0, ||f ||), and use the result
and the hint from Exercise 6.50.(viii) to prove that
/c
^(f) = ^^ I £-'■""?"(1 - h"-)"-^ dh.

660 Exercises for Chapter 6: Integration
(iii) Substitute h = cos a and conclude that
^ rc-^) Jo
In particular, for ^ ^0,
(**) /(f) = (|^)V|(||f||).
(iv) Prove by means of Exercise 6.66.(viii) that the expression on the right in (••)
does indeed approach the value of /(O) for f —>■ 0.
Exercise 6.99 (Fourier transform of x i-> e~ll^ll and Poisson's integral - sequel
to Exercise 2.87 - needed for Exercise 7.30). The Fourier transform / is often
well-defined for functions / : R" —>■ R not contained in -5(R"); iii particular this
applies to the function / : R" —>■ R given by f(x) = e-H^H.
(i) Prove
m = 2V^r(^)(i + III 11^)-^ (f € R").
Hint: Prove by Exercise 2.87.(v) that
interchange the order of integration, and use Example 6.11.4.
(ii) Deduce, for all f e R" and t > 0,
L
e-'i-'i)-<\\-n dx = 2"7r"-?'r(^^)
« + 1 \ t
Uf + t^)^
Let R'^+' = R"+' \ {0}. We define Poisson's kernel P : R;'+' ^^ R by (see also
Exercise 7.70.(ii))
rm U\
P(x,t) =
.'^ l|(x,OII"+''
n 2
In the terminology of Exercise 6.42 the function R" —>■ R with
r{^) 1
X |->
71-2 (;(;2 _|_ 1) 2
issaidtohetheprobability density of the Cauchy distribution. NotethatP(x, —t) =
P(x, t). Assume that the properties of the Fourier transformation are also valid for
the function on R" given by x i-> e"'"^'", where f > 0, as can be shown by
approximation arguments.

Exercises for Chapter 6: Integration 661
(iii) Conclude that f^„ P(x, t)dx = 1, for all t ^0. More generally, prove
/ \,dx = —rvw (f e R", t > 0),
and show that one therefore has the following, known as Laplace's integrals:
(compare with Exercise 2.85):
cosfx
— dx =
r cos f A
Jr., ^^i
r X sin f.
Jr., x^ + t
ne-'^
1 t
^.-'?
(f > 0, f > 0),
(f > 0, r > 0).
Use the identity J^ ^^ dx = ^ from, for instance. Example 2.10.14 to
obtain the validity of the latter formula for t = 0.
(iv) Verify that we have the following semigroup property, for all ^i and t2 > 0:
P(x, ti +t2)= / P{x - y, fi) P(y, t2)dy.
Jr"
(v) Let 5 > 0 be arbitrary. Check that f. j.gRnr rrj-rr^^i lUII"' d^ < oo> ^iid use
this to prove
lim / P(x,t)dx = 0.
'-*°J{xsR"\\\x\\>S]
Define R±+' = {(x, t) e R"+' \ ±t > 0} (note that in the remainder of this
exercise the meaning of R'^^ differs from the usual one).
(vi) Prove by Example 7.8.4 that the functions
1(^,011
(x, t) b^ log II(x, Oil (« = 1); {x, t) b^ „,__ ^,„„_, (« > 1),
are harmonic on R"+'. Verify that, leaving scalars aside, P on R^"^' is the
partial derivative with respect to t of the preceding function, and conclude
that P is a harmonic function on R±^ .
Now let h e C(R") be bounded, and define Poisson's integral {Ph : Rj.+' ->■ R of
h by
({Ph)(x, t) = (h * P(; t))(x) = f h(x - y)P(y, t)dy.
JR"
(vii) Prove that 3^h is a well-defined harmonic function on R^+'.

662 Exercises for Chapter 6: Integration
(viii) Assume (x,t) e R±^' and prove by means of part (iii) that
\({Ph)(x,t)-h(x)\ <[ \h{x-y)-h{x)\P{y,t)dy
+2sup\h(x)\l P(y,t)dy.
A-sR" ■'l|y||>^
Using part (v) show that, uniformly for x in compact sets in R",
lim(^h)(x,t) = h(x).
±ti,o
Background. Evidently, in the terminology of Example 7.9.7, the function/ = ^h
is a solution of the following Dirichlet problem on ^ = R^^ or R^^ :
A/ = 0 with /|9^ = h.
(ix) On the basis of the results from Exercise 7.21.(ii), show
(m(x,t)=- ^^ f ^T^zShdy ((x,0 €Rr').
hyperarea„(S")yR« IK}', 1)11"+'
(x) Prove that there exists a harmonic function u on R^+' with
limM(x, t) — limM(x, t) = h(x) (x e R").
Hint: Write h = \h - i-\h).
Background. Evidently, a bounded continuous function on R" can be represented
by means of the jump along R" x {0} made by a suitably chosen harmonic function
on R±+'. This point of view is of importance in the theory of hyperfunctions.
Exercise 6.100 (Mellin transformation - sequel to Exercise 6.86 - needed for
Exercise 6.101). Let -5*(R+) be the linear space of functions / : R_|_ —>■ C with
the property that the function /* : R —>■ C given by f*(x) = f{e^) is contained in
-5(R). For / e -5*(R_|_) we define the function Mf : R —>■ C, the Mellin transform
of/, by
r dx
(Mf)(^) = / x-'^f(x) — (f € R).
We now prove the following, known as Mellin'sformula, valid for all / e -5*(R+)
andx e R+:
W f(x) = ^j^x'HMf)(^)d^.

Exercises for Chapter 6: Integration 663
(i) Check that /^(f) = (Mf)(^), for all / € S*(R+) and f € R. Conclude
that the Mellin transformation M belongs to Lin (-5*(R_|_), -5(R)), and also
that (•) is obtained. Hence, for g e -5(R) and x e R_|_,
{M-'g){x) = ^ f x'^g(^)dt
271 J^
Define Hermitian inner products on -5*(R+) and -5(R), respectively, hy ( f, g) =
/r, fM'sM f and (/, g) = J; /^ f(x)'^dx.
(ii) Prove by the Parseval-Plancherel identity from Exercise 6.86, for all / e
Jr.,. X 2n ^R
Conclude that M. € Lin(-5*(R+), -5(R)) is unitary, that is, M preserves the
Hermitian inner products.
(iii) Introduce the differential operator 9^. = jx~, and let f e R. Verify that
the function u : x h> x'^ is the unique solution of the eigenvalue problem
(djcu)(x) = f u(x), for X e R+, and u(\) = 1.
(iv) Introduce the inner product (/, g) = f^ f(x)g(x) ~ and prove that 9^. is a
self-adjoint linear operator with respect to it; in other words, verify that, for
all / e -5*(R_|_) and all bounded C' functions g,
f (9./)W^—= f f(x)(d.g)(x) —.
Jr.,. X Jr+ X
(v) Using parts (iii) and (iv), show that, for all / e -5*(R+) and f € R,
M(d.f)(^) = f (Mf)(^), M(-f log-)(^) = (^j^(Mf))(^).
Now assume / and g e -5*(R+). Define the convolution fOg: R_,_ —>- C of / and
(fOg)M= f f(-)g(y)—.
Jr.,. ^y'' y
(vi) Demonstrate that cW(/ng) = (Mf) (J^ig).
Background. The formula (M o 9. o M~^)f(^) = f /(f) shows that the
differential operator jx-^ acting on -5*(R+) can be diagonalized by conjugation with the
Mellin transformation M, and that the action of the conjugated operator in S (R)
is that of multiplication by the coordinate f. In number theory one usually calls

664 Exercises for Chapter 6: Integration
/r ^'/C^) ^' that is (Mf)(is), the MeHin transform of / evaluated at s. In that
disciphne one studies functions f(x) = X!hsN ^n^~" ^'^^ computes
4^ (-■/«-= E5.
which is called a Dirichlet series. In this case the inversion formula takes a different
form.
Exercise 6.101 (Harmonic function on sector - sequel to Exercises 3.8 and
6.100). Assume 0 < kq < jt, let V C R^ be the half-strip {(r, a) e R^ | r e
R+, \oi\ < oio], and let (p : V ->- f/ := ^(V) be the substitution of polar
coordinates as in Example 3.1.1. Then U is an open sector in R^. We look for
a solution / e C^{U) n C{U) for the following partial differential equation with
boundary condition:
A/ = 0 onf/, /l9f/=/',
where h{x\,X2) = h(xi, —X2), forx e dU and (h o >!')(•, uq) e -5*(R+), in the
notation of Exercise 6.100.
(i) Conclude by Exercise 3.8.(v) that / o ^P on V must satisfy
We now try to find / with the property that / o >!' (•, a) e -5*(R+), for all la | < kq;
we then write g(-, a) e -5(R) for the Mellin transform M(f o >!')(•, a).
(ii) Use Exercise 6.100.(v) to verify that, with |a!| < ao fixed for the moment,
W (-f' + J^)s(^^ «) = 0 (f € R).
(iii) Now interpret (•) as a differential equation with respect to the variable a,
and infer the existence of functions a and Z> : R —>■ C such that g(^, a) =
fl(f) cosh(f a) + Z>(f) sinh(f a), for all |a!| < a^. Now choose in particular
a = ztttQ and derive
cosh (^ a)
g(f, a) = M(h o «I/)(f, ao)
cosh(fa!o)

Exercises for Chapter 6: Integration 665
(iv) Prove by Mellin's formula from Exercise 6.100
1 f .. coshCfa)
/o«P(r,«) = — / r-^ ;; \M(hoM^)(^,ao)d^ ((r, «) € V).
In Jr cosh(f tto)
Conclude that, for x e f/,
f(x) = —- / ||x|rcosh(2f arctan( —-)) . ,, , d^.
In J^ \ xi + 1 / cosh(f tto)
Exercise 6.102 (Hankel transformation - sequel to Exercises 6.66 and 8.20 -
needed for Exercise 7.30). We employ the notation from Exercise 6.66. Let S (R+)
bethelinearsubspaceof-5(R)consistingoftheevenfunctionsin-5(R). For A. > —|
and / e -5(R+) we define the function M\f : R+ —>■ R, the Hankel transform of
/ of order A., by
This integral is well-defined, see Exercise 8.20. We now prove the following, known
as Hankel's formula:
Hy = I,
that is, for all f € S (R+) and x e R+ we have
Under the substitution g{x) = x^f(x) we obtain the following variant of Hankel's
formula:
8M= f f 8(y)My^)ydyMx^)^d^.
Jr.,. Jr.,.
See Exercise 7.30 for the relation between the Hankel and Fourier transformations.
(i) Introduce the differential operator with variable coefficients
1 d
A.x =
X2^+1 dx
i^-iy
Aa-?)
Let f e R+. Use Exercise 6.66.(iv) to verify that the function u : x ^-^ -^
is a solution of the eigenvalue problem
(A.v,am)(^) = ^^u(x) (x e R+).

666 Exercises for Chapter 6: Integration
(ii) Prove that Ax,x is a self-adjoint linear operator with respect to the inner
product (/, g) = f^ /(x)g(x)x^^+' dx; that is, prove for all / e -5(R_|_),
and all bounded C^ functions g,
f (A,,>,f)(x)g(x)x^^+'dx= f f(x)(A,,>,g)(x)x^'+'dx.
Jr.,. ^r+
(iii) Verify, by parts (i) and (ii), for all / e -5(R+) and f e R+,
^,(A,,/)(f) = f'(^x/)(f), ^A(-'/)(f) = (Aj,,(^,/))(f).
Conclude that J{x : ■S(R+) -^ -5(R+).
(iv) Verify J^xk = k ifk(x) = e-^^\
Hint: There are two possibilities: apply the series expansion of Jx from the
Exercise 6.66.(ix); alternatively, check that both k and J{xk (use part (iii))
satisfy the differential equation
(A,,,/)(x) = (2A + 2 - x^)f(x) (X € R+).
Prove by Exercise 6.66.(vi) that (J^xfYi^) = -f (^A+i/)(f), and conclude
by part (viii) of the same exercise that
k(0) = lim(J{^k)(^) = 1, k'(0) = lim(^x^)'(f) = 0.
Assume / e -5(R_(_) and Xq e R_|_, and write h(x) = f(x) — f(xQ)e2^ok(x). On
account of part (iv), to prove the formula (J^xf)(^o) = fi-Xo) it suffices to show
(J^lh)(xo) = 0.
(v) Check that h(xQ) = h(—XQ) = 0, and write h(x) = (x^ — x^)g(x), with
g € -5(R+) suitably chosen. Use part (iii) to prove
(J^xh)(^) = (A^,x-4)(^>^§)(^)-
Conclude by parts (ii) and (i) that
= f 'n^ ,.^.^^A(Xog)^2^+l
= \ iHxgm{A,,x-xl)(^-^y^^U^=^.
Jr., ^ (-^0?) ^
Jr., (xo^r
Background. The formula (J^x o A., a o M^ )fi^) = f ^/(f) shows that the
differential operator —^2i+r^(j'^^^'''' ^) acting on -5(R+) is diagonalized by conjugation
with the Hankel transformation Mx, and that the action of the conjugated operator
in -5(R+) is that of multiplication by f ^.

Exercises for Chapter 6: Integration 667
Exercise 6.103 (Weierstrass' Approximation Theorem - sequel to Exercise 6.92
- needed for Exercise 8.36). This theorem asserts that a differentiable function on
a compact set can be uniformly approximated by a polynomial function, and that a
finite number of derivatives of the function can then also be uniformly approximated
by the corresponding derivatives of that polynomial function. In a more exact
formulation, let ^ C R" be open and let ^ € No; then for every function/ € C'^(^),
for every compact set K CL Q., and for every e > 0, there exists a polynomial
function p : R" —> R such that, for every multi-index a = (ai, ..., a,,) € NjJ with
\a\ < k, one has
sup{ |D"/(x) - D"pix)\ \x€K} <e.
For a proof in the case of a continuous function on an interval in R, see Exercise 1.55.
We now give an outline of the proof. It is left to the reader to check and add
its details. Let x ^ C^(R") with supp(x) C ^ and x = 1 on a neighborhood
of K. Then fx € C^(R"), and thus we may henceforth assume / € C^(R"). In
particular, the integrations below are over the compact set supp(/). Now employ
the notation from Exercise 6.92 with ^ = 1, and use the arguments applied in that
exercise, with / replaced by D" f. One finds
M„(x, t) := (D"/) * grix) = i4nt)-"^^ f D'^/OOe""^'"^"'^''' dy.
Jr"
Now, firom the Continuity Theorem 2.10.2, it follows that m„ : AT x [0, 1 ] ^^ R
is a continuous function on a compact set. But then Ua is uniformly continuous on
that set, on account of Theorem 1.8.15. Consequently, for every e > 0 there exists
a number t > 0 such that, for every multi-index a € Ng with |a!| < ^, and for all
X € K,
\D"fix) - M„(X, 01 = Waix, 0) - Uaix, 01 < -.
One has
^ i\i-4t)J '
But, for X and }' in compact sets in R", this series can be uniformly approximated by
its partial sums pjv (^ > 3'). where the summation runs from 0 to A^ € N. By choosing
A^ sufficiently large, we can find a polynomial function p := pi^ -.R" ^> R with
pix) := i4nt)-"'^ T .,/, ,, f fiyKixr - }',)' + • • • + (x„ - y„)y' dy,
such that, for every multi-index a € NJj with |a!| < ^ and for all x € K,
\uAx, 0 - D"pix)\ < i4nt)-"'^ f D"fiy) T "''■~/''/ dy < ^.
Jr" ^j y!(40^ 2

668 Exercises for Chapter 6: Integration
Exercise 6.104 (Sequel to Exercise 6.50 - needed for Exercises 6.105 and 6.106).
In S (R") the convolution operation * (see Example 6.11.5) defines a multiplication
* by
x) = I fix
(*) (/. g)^ f*g with / * g(x) = fix- y)g(y) dy.
Jr-
(i) Show this multiplication to be commutative and associative.
For 5 > 0, define the function 0j : R —> R by
^s-i f x*"\ X > 0;
(Psix) = -=^, where -^r' = I
r(5) [ 0, X <o.
(ii) Verify that, for s, t > 0, the convolution 0^ * 0, is also well-defined by (•),
and that it is given by
T{s)T{t) Jo
Conclude by Exercise 6.50.(iv) that 0j * 0, = (j)s+,, for s, t > 0.
Exercise 6.105 (Fractional integration and differentiation - sequel to
Exercises 2.75 and 6.104 - needed for Exercises 6.106,6.107,6.108 and 8.35). Define
C^(R) = {/ € C°°(R) I supp(/) c R+},
and let D € End (C^(R)) be the operator of differentiation. Define
D-'€End(C^(R)) by D-\f {x) = j f {t) dt.
Jo
(i) Prove that D"' is in fact the inverse of D.
Define D-' = (D"^)", for w € N.
(ii) From Exercise 2.75.(i) deduce that D^" f = f * (j),„ for w € N and / €
C^(R), with 0„ as in Exercise 6.104.
(iii) Prove by Taylor's formula from Lemma 2.8.2 or Exercise 2.75.(iii) that D^"
is the inverse of D", that is, D-" = (D")"'.
For 5 > 0 we define the operator D"' € End (C^(R)) of fractional integration
from 0 of order s by (compare with part (ii))
D-'f = f * 0,; and so D-\fix) = -}- /" /(>-)(x - y)'-' dy.
"(5) Jo

Exercises for Chapter 6: Integration 669
(iv) By Exercise 6.104 prove that D-' o D'' = D'' o D'" = D-^''+'\ for all s,
t > 0.
For f > 0 we define D' € End (C^(R)) of fractional differentiation of order t as
follows. We have the unique decomposition t =n — s with w € N and 0 < 5 < 1.
Now define
D' = D" o D-'; hence D'f = (/ * (f),)^"\
(v) Prove D° = I, and demonstrate by integration by parts that D' = D'" o D^'',
if t = m — r with m € N and 0 < r < m.
(vi) Examine the validity of the following assertions. For t > 0 one has D' =
D^^ o D"; therefore D' f = /^"^ * 0^, and for all 5, f € R one has the group
property D' o D' = D' o D' = D'+'.
Background. In distribution theory the arguments above are generalized in a far-
reaching manner and the one-parameter group (0')5gR of linear operators, together
with its generalizations, becomes an aid in solving differential equations.
Exercise 6.106 (Hypergeometric function and fractional integration - sequel
to Exercises 6.69, 6.104 and 6.105). Let the notation be as in these exercises and
set ^a,b{x) = (1 — x)^''(j)bix). Now substitute zt = y in the integral for the
hypergeometric function to obtain on ] 0, 1 [
F(a, b;c: ■) = r = 7 •
Exercise 6.107 (Bessel function and fractional integration and differentiation
- sequel to Exercises 6.66 and 6.105).
(i) Verify that, for A. > — j and x € R4.,
2
1
AW = ,,,.,,. ,.(|) j (l-ty-icosxtdt
r(i)r(A + i)
2 1 /"t 91_i
y ) 2 cosydy.
^)^ Jo
1-2 _ ^,2^^-^
■Im-'/'i I 1
r(i)r(A + i)(2
(ii) Conclude that, in the terminology of Exercise 6.105, this formula may be
recognized as a fractional integration from 0 of order A. + j:
OS V
7^
1 1 /COS v^\

670 Exercises for Chapter 6: Integration
(iii) Prove
(*) (2v^)^ A(v^) = D-^(^M^-)) (x).
Let fj, > 0. From Exercise 6.105.(vi) conclude that
(2v^)^+'^A+^(v^) = D-''(i2^)'-MV^))ix).
Demonstrate that in integral form this becomes
2'^(v^)^+'^A+^(v^) = -^ f\x - yr-HVy)'MVy)dy.
i \f^) Jo
Now substitute x = z^,y = z^ sin^ a and subsequently z = x. This gives the
following, known as Sowme's/ormw/fl, valid for A. > — j, M > Oandx € R+i
2''"'r(//)A+^(x) =x'' / A(xsina!)sin^+'a!cos2''"'a!£?a!.
Note that we can use the left-hand side of formula (•) to define the Bessel function
A, for A. < —J. Indeed, for these values of A. the expression on the right-hand
side in (•) can be interpreted as the fractional derivative of order —A. of the function
X i-> Jqi-Jx).
(iv) Let A. > 0. Prove, by means of Exercise 6.105.(vi), that both J^ and J_x
satisfy Bessel's differential equation from Exercise 6.66.(vii).
(v) Prove, by means of Exercise 6.66.(vi), that
(-l)"(2v^)-"J„(v^) = D"(jo(v^))(x) (n € N, X € R+).
Conclude that
J_„ = (-1)"J„ («€N).
Thus, in this case the second solution J_„ is linearly dependent on the first
solution J„.
Exercise 6.108 (Abel's integral equation - sequel to Exercises 0.6, 2.75 and
6.58). Let fl > 0 and let g € C' ([0, a ]) be given with g(0) = 0. We want to
solve the following, known as Abel's integral equation, for an unknown continuous
function / : [0, a ] —> R:
g(x)= r ^^dt (x€[0,a]).
Jo s/x - t
We prove in four steps that a solution / is given by
W f(y) = - r 4^"^^ iy€[0,a]).
^ Jo \/y -X

Exercises for Chapter 6: Integration 671
(i) Multiply both parts of the integral equation by J^ to conclude that, for
y€[0,a'\,
r^^,, r [' m
^0 -Jy -X Jo Jo ^(y - x)ix - t)
n n dx
= / fit) / =dt.
Jo Ji \'(x-t)(y-x)
(ii) The value ofthe inner integral follows from Exercise 0.6.(v). For}' € [0, a],
conclude that
whence
f -4^dx=7T f f{t)dt-
Jo Vy--^ ^0
^fiy) = -J- / y dx.
dy Jo -Jy -X
Straightforward application of Exercise 2.74 is not possible, because it leads to
expressions ofthe form oo — oo.
(iii) Substitute x = yt to prove that
d f'yygiyt),^ r'^8iyt) + t^/y^iyt)
^fiy) =T-/ n ^^= / n= ^'
dy Jo ^l-t Jo Vl-f
'^' gix) + Ixg'ix)
Jo
■dx.
lyjy - X
(iv) Finally, use integration by parts to prove
1 r six) ^ [y g'jx) ^ fy xg'jx) ^
— / ax = I dx — / — dx
y Jo 2^3' - X Jo ^y -X Jo y^y - x
Now conclude that (•) holds.
LetO < s < 1 and consider the following generalization of Abel's integral equation:
gix)= r-^^dt (x€[o,«]).
Jo ix - ty
(v) Now multiply by ^_^ ,_,, and conclude by Exercise 6.58.(iv) that the inner
integral becomes
/■'
Jo
1
^-'-\\-uy-^du = T{\ -s)T{s) = —
si\\{s7T)
Conclude that
sin(5;7r) /"
^ Jo
sin(5;7r) fy g'ix)
iy-xy-^
fiy) = ^-^ , ^ ' dx (yeLO,^]).

672 Exercises for Chapter 6: Integration
It turns out a posteriori that the problem above can be elegantly formulated as
follows. For 0 < 5 < 1, define
^s-i f x^"\ X > 0;
(t>s{x) = -^, where -^r' = I
^(■s) [ 0, X <0.
Then the convolution equation for / (see Example 6.11.5; and our present treatment
is formal, that is, it is assumed that here, too, convolution is well-defined) g = f*(j)s
has the solution f = g' *^i-s-
(vi) Examine the relationship between this solution technique and that of
Exercise 2.75.
Exercise 6.109 (Convolution and spline - needed for Exercise 7.39). The
definition
/ * six) = fix- y)giy) dy
of convolution * from Example 6.11.5 turns out to be meaningful for a much larger
class of functions / and g than those from -^(R"). The convolution operation
"improves smoothness properties" and can be used to construct C* functions with
compact support, for ^ € N.
Let X be the characteristic function of the interval [ — 2. 2 ] ^ ^ ^'^^ ^^^ ^" ~
X *... * X be the w-fold convolution product, for « € N, according to the definition
above.
(i) Prove
/ x(x)dx = l, I xx(x)dx = 0, I x^x(x)dx = —,
X„ix) = Xni-x) (xeR), x„ix)dx = l (neN).
Jr
(ii) Prove that xi is the continuous function on R given by
Xiix) =
0, X < -\
x +1, -1 < X < 0
-X + 1, 0 < X < 1
0, 1 < X.
Hint;
= ^2W =/„^R|,_is,s.,+i,X(>')rf3'.

Exercises for Chapter 6: Integration 673
(iii) Prove that xs is the C' function on R given by xs (x) is equal to
0,
(A-+ 1)2
2! '
2! +(,-^+2-'"'~2!'
2! ^ 2'' ^^ 2!'
-|<
-|<
1^
X
X
X
X
X.
<
<
<
<
3
2
1
2
1
2
3
2
1
3!
2
3
2
3
1
3!
0,
(2 +
-X2
-X2
(2-
x)3
-ix3
+ ^X3
x)3
l-x^ =-2
0,
(iv) Prove that Xa is the C^ function on R given by X4(x) is equal to
0, X < -2
^, -2< X < -1:
-3^ + ^ + l(x + l) + l, -1< X < 0:
3^-2# + ^, 0< X < 1:
_ fci)i+(£r>)!_±(^_l) + l, 1< X < 2:
2 < X.
We now prove by mathematical induction on w € N that x„ is a piecewise polynomial
(^n-2 function on R whose support is the interval [ — f, f ]• The intervals
/«,o:=]-oo, -^], I„^k-=[-'^ + k-l,-'^ + k] il<k<n),
P w r
are the maximal intervals / such that Xn | r is a polynomial function. Let
Pn,k-=Xn\j iO<k<n + iy,
hi, k
then p„j., for 1 < ^ < w, is a polynomial function of degree « — 1, while p„ q =
Pn,n+\ = 0. Now write
^ (x-i-'i+k-l))'
(*) Pn,k(x)= 2l, P"''''' n •
0</<H- 1
We shall want to prove, for 1 < ^ < w and 0 < / < « — 1,
(-) p..'- = ^^3|-yy E (-iv(;)(^-i-y)"-'-'-.

674 Exercises for Chapter 6: Integration
(v) Verify that, for < ^ < w + 1 and x € In+i, i,
p„+i,k(x) = p„^k-\{y) dy + j p„,k(y)dy.
Jx-\ J-j+k-1
Conclude by means of (•) that
a . IV + 2^^Pn,k,i-\-P.,k-\.i-l) 7^ •
0S'S"-1 l<f<"
(vi) Derive the following recursion relations between the coefficients Pn+i, k, i '■
Pn+i,k,o= Yl TrrrtT (i<^<« + !);
P„+i,k,i = Pn,k,i-i-Pn,k-i,!-i (i <k <n + l, l<i<n).
Then go on to show that (••) is satisfied, using (.",) + (") = ("t ).
(vii) Prove
0<j<[.v+|] ^■''^
(viii) The function Xn is not (« — 1) times differentiable on [ — |, | ]; more precisely.
2' 2
prove, for 1 <k<n + l,
lim T^W- lim i^(x) = (-\)
xi-^+k-l d" 'x .vt-|+H d" 'x
"G-O-
(ix) We want to give an independent proof that the following does indeed hold,
for 0 <i < n — I:
For this we consider
note that 0 = f<-"-^-'\0) = (n - I - iV-xPCj), for 0 < / < « - 1.

Exercises for Chapter 6: Integration
675
Background. In numerical mathematics, C"^^ functions on R which are piecewise
polynomial of degree n are known as splines of degree n, after the flexible metal
strips used to draw smooth curves.
Exercise 6.110. Define
fk :R+ ^ R
by fk(x) =
1 „
-, 0 < X < ^ ;
k
0, k^ < X.
Show limi_>oo fki^) = 0, for all x € R4-, and this is even true uniformly on R4.,
while we have limi_>oo /r fki^) dx = 00.
Background. Note that majorizing functions like g(x) = 1 for 0 < x < 1, and
g(x) = -j^, for 1 < X are not absolutely Riemann integrable over R4.. Hence
Arzela's Dominated Convergence Theorem 6.12.3 does not apply.

Exercises for Chapter 7: Integration over Manifolds 677
Exercises for Chapter 7
Exercise 7.1 (Formula of Binet-Cauchy and Pythagoras'Theorem), held < n,
and suppose A € Mat(« x rf, R) and B € MaA{d x w, R). Denote by a,- € R'',
for 1 < z < (i, the row vectors of A, and by bj € R'', for I < j < d, the column
vectors of B. Write AT for the rf-tuple (^i,..., k^i) from the set {1,2,..., «}; and
further Xj for the set of all such rf-tuples; write A'^ and Bk in Mat(rf, R) for the
matrix with a|c^, ...,ak^ as its row vectors and with Z>i,,..., Z>jt^ as its column vectors,
respectively. Finally, set
a^ =detA^, bK =det Bk-
(i) Prove the following/ormM/a of Binet—Cauchy:
det(BA)= Y^ bKa'^.
Here i^ is the subset of Xd consisting of the strictly ascending rf-tuples, that
is, of the K = (ki,..., kj) satisfying 1 < ^i < • • • < ^^ < « (compare with
Definition 8.6.3).
Hint: If C = BA € Mat(rf, R), then c,-; = J2i<k<n ^ikCkj, for 1 < i, j < d.
For a € Sj, the permutation group on d elements, write sgn(cr) for the sign
of cr. Then, by a well-known formula for the determinant,
dctiBA) = J2 sgn(o^) n E
bik;ak,a(i)
(TSS^ l<i<d l<kj<n
= j2 Y[ ^'*'- Ji ^g"^"') n ^^.'-^o)
l<k[,...,kj<n l<i<d (tsSj ISjSd
= E ^' n ^-.-
Ks.Kj \<i<d
Note that a'^ ^0 only if K consists of mutually distinct numbers, therefore
the summation can be performed only over such K. Since fl('^(*i)--'^(*<')) =
sgn(cr) fl^*'""'*''^, for cr € S^, we obtain
dctiBA) = ^ ^'^ ^ sgn(or) Yl bia(k) = E ^^ ^''-
K^lj trsSj IS'Sd Kslj
(ii) Let V and w eR"; and let A € Mat(« x 2, R) and B € Mat(2 x n, R) have
V and w as column and row vectors, respectively. From part (i) deduce the
equality in Exercise 5.26.(i).

678 Exercises for Chapter 7: Integration over Manifolds
(iii) Suppose B = A'. Using part (i) conclude the following, known as
Pythagoras' Theorem, which generalizes the identity ||fl||^ = X!i</<n '^j' valid for a
vector a € R":
detiA'A) = J2 (det A^f (A € Mat(w x d, R)).
Note that the column vectors of A'^, for K € Ij, are the projections of the
column vectors of A onto the (i-dimensional linear subspace of R" spanned
by the standard basis vectors ei,,...,^^^. In other words, in the terminology
of Section 7.3, the square of the rf-volume of a rf-dimensional parallelepiped
in R" is equal to the sum of the squares of the rf-volumes of its projections
onto all the rf-dimensional linear subspaces of R" given by the vanishing of
some of the coordinate functions.
Exercise 7.2. Calculate the length of that part of the graph of the function log which
lies between (1, 0) and (x, logx), forx > 0.
Hint: Substitute t = tan a to obtain
I ^^' dt = Vl + f2 + logr - log(l + Vl + f2) (t>0).
Exercise 7.3 (Sequel to Exercise 4.8). Consider the part V = (j)Q0,27r [) of the
helixfromExercise4.8for which 0(f) = (cost, sint, t). Prove
I
2, 2nV2i3 + 4n^)
IMl'dix =
V
Exercise 7.4 (Semicubic parabola). This is defined as the zero-set in R^ ofg(x) =
Xj — ^2. Calculate the length of the part of this curve lying between the points
(t, -t^!^) and {t, t^'^), for t > 0.
Exercise 7.5 (Lemniscate, r(^) and elliptic integrals of first kind - sequel to
Exercises 3.44, 4.34, 6.50 and 6.69 - needed for Exercise 7.6). Consider the
lemniscate from Example 6.6.4, which in polar coordinates {r, a) for R^ is given
by r^ = cos 2a. Let 0 < (j) < j and let x = x((j)) = ^cos20 > 0. Then the arc
of the lemniscate determined by x is understood to mean that part of the lemniscate
which lies between the origin in R^ and (r, a) = ix((j)), 0).
(i) Prove that the length of the arc of the lemniscate determined by x = x(0) is
given by
/ dtv = / dt
J A Vcos 2a! Jo ^/\ — t'^
Hint: Substitute cos 2a! = t-^.

Exercises for Chapter 7: Integration over Manifolds 679
Exercise 3.44 or 4.34.(iii) now implies the following. If x and y are chosen positive
and sufficiently close to 0, then the sum of the lengths of the arcs determined by
X and y equals the length of the arc determined hy a{x,y), in the notation of the
latter exercise. It is a remarkable fact that a{x, y) can be constructed from x and y
by means of ruler and compasses.
Let Y be the length of the arc determined by 1, that is
— =/ da = / dt =—r=r{-) ,
2 Jo Vcos2a! Jo Vl -1* 4^2^ ^4/
see Exercise 6.50.(vi). Then the length of the lemniscate equals lur. Note that
^ = fo r-— dt and that the length of the unit circle equals 2ii.
(ii) Use the substitution cos 2a! = cos^ 0 to show that
UT
= V2 r , ^ d(l> = V2k(-^),
J^ yi-isin2 0 ^V2^
where
/•! 1
■.d<p (0<k<l).
Kik):= r
^0
Vl -Jfe^sin^^
K(k) is said to be Legendre'sform of the complete elliptic integral of the first kind
with modulus k (compare with Example 7.4.1). Combination of parts (i) and (ii)
now gives
r(-y = 4^k(-^Y ^(-^ = 3.625609908221 ••• .
In this calculation the following result is used,
(iii) Verify that, in the notation of Exercise 6.69,
2\
In particular.
_^( y^/(2n-l)(2n-3)---3-l A^\
2^ ^^ 2n(2n-2)...4.2 ;;
= -{1 + -jfe^ H r H k^ + •••).
2V 4 64 256 /
v4/ V2 2 2>
Moreover, deduce that K satisfies a hypergeometric differential equation.

680 Exercises for Chapter 7: Integration over Manifolds
Exercise 7.6 (Elliptic integrals and Catalan's constant - sequel to Exercises 6.39
and 7.5). We employ the notation from Exercise 7.5. Prove by changing the order
of integration that
/" K{k)dk= (' -^d(p.
Jo Jo sm (f)
Now use the substitution <p = 2 arctanx to show, see Exercise 6.39.(iv),
.1 /.I
Jo Jo ^
Conclude that
G = - A + y^ 1 n2n-l)i2n-3)---3-iy\
4\ ^2n+l\ 2n{2n-2)---A-2 ))'
neN
Exercise 7.7. Define 0 : ] 0, 1 [ ^^ R^ by 0(0 = {cos2nt'^, sm2nt'^, 2nt'^).
(i) Calculate the Euclidean length of the curve V = im(0).
(ii) Prove that i/f : ] 0, 2:7rV2 [ ^^ R^ with i/f (0 = (cos -^, sin -^, -^), is the
parametrization for V with the arc length as parameter.
Exercise 7.8. Let K be the intersection of the solid cylinders {x € R^ | x^ +x| <
1} and {x € R^ | x^ + x| < 1} (compare with Example 6.5.2). Prove that
area(9A:) = 16.
Exercise 7.9. Let V = {x € R^ I ||x|| = 1}. Prove
An
I
xfd2X = Y (2 = 1,2,3).
Exercise 7.10. Let V C R^ be the triangle with vertices (1, 0, 0), (0, 1,0) and
(0,0,1). Prove
/ xi dix = -\3.
Jv 6
Exercise 7.11. Let V C RHe the graph of / : ] 0, 1 [ x ] -1, 1 [ ^^ R given by
/(x) = x^ + X2. Prove
/ Xi d2X = Vo VZ.
Jv 3

Exercises for Chapter 7: Integration over Manifolds 681
Exercise 7.12. Let a > 0. Consider that part of the cyhnder {x € R^ | x^ + x| =
a^ } that Hes inside the cyhnder [x ^Vt? \ x^ + x^ = 2ax2} and inside the octant
{x € R^ I Xi > 0, X2 > 0, X3 > 0}. Prove that its area equals 2a^.
Exercise 7.13 (Girard's formula). A subset D c S^ is said to be a spherical
diangle with angle a if D is bounded by two half great circles whose tangent
vectors at a point of intersection include an angle a. These great circles are said to
be the sides of D.
(i) Prove that area(D) = 2a.
A subset A C S^ is said to be a spherical triangle if A is the intersection of three
spherical diangles that pairwise have a part of a side in common (compare with
Exercise 5.27). Assume that the spherical diangles have angles ai, a.^ and 0:3,
respectively.
(ii) Show that area(A) = X^k/o ^i ~ ^•
Exercise 7.14. Assume m > 0 and h > 0. Let V C R^ be that part of the conical
surface
{x € R^ I x| = m^(xf + X2)}
that lies between the planes {x € R^ | X3 = 0} and {x € R^ \ x^ = h}.
(i) Show that the Euclidean area of V in R^ equals n^^/m^ + 1.
Next, consider the surface V C R^ that is formed when one straight line segment
lying on V is removed from the conical surface V. Then unroll this surface V into
a plane without stretching, shrinking or tearing it; thus is obtained a circular sector
inR2.
(ii) Calculate the Euclidean area of this circular sector in R^.
Let IV be a plane in R^ with the following two properties: W is parallel to a tangent
plane (at (0, l,m), for example) to the conical surface V, and W goes through
(0,0, h), the center of the top circle of V.
(iii) Prove that the length of the conic V n IV equals
.1
m J-I
+ (m2+ l)t^dt.
Exercise 7.15 (Viviani's solid). This is the set L C R^ formed as the intersection
of the unit ball in R^ and a solid cylinder, more precisely L = {x€R^|||x||<
^2 _l_ y.2
"1 "I"-^2
1, x^ + x| < xi}. Then 9L = Vi U V2, with
Vi = {x€R3 I ||x|| = l,x2+x|<xi},
V2 = {x€R3| ||x|| < l,x2 + x| = x,}.

682
Exercises for Chapter 7: Integration over Manifolds
Illustration for Exercise 7.15: Viviani's solid
(i) Verify that Vi n V2 = im(0), where 0 : ] —n, ;7r [ —> R^ is defined by
(j){a) = (cos^a, cos a sin a, sin a) (—n <a < n).
(ii) Prove that the restriction of 0 to ] —tt, 0 [ and to ] 0, ;7r [, respectively, are C°
embeddings.
(iii) Show that the length of Vi n V2 equals the following complete elliptic integral
of the second kind:
^^/v
1 — J sin^ a da.
(iv) Demonstrate that the orthogonal projection of Vi onto the plane {x € R^
X3 = 0} equals the set D x {0} C R^ with
TT
TT
= {r(cosa!, sina) € R | —— <«<—-, 0<r< cosa }.
Verify L = {iy, X3) eR'lyeD, \xs\ < /T^JyT}.
(v) Calculate vol3(L) = Jtt - ^.
Hint: /g- sin ada = ^.
(vi) Prove that the area of Vi equals 2tt —4 (see Example 7.4.10).

Exercises for Chapter 7: Integration over Manifolds
683
(vii) Demonstrate that the area of V2 equals 4. The area of the boundary of'Wviani's
soHd thus equals that of the half unit sphere.
Exercise 7.16. Let the notation be that of Example 7.4.8. Prove, for all 1 < y < 3
and X € R-^ \ i4.
(j>Aj{x) = -^ [ ^,, ^ ..d2y =
4n Ja R\\x-y\\
1
1 R'
X j J
3\\xV '
\\x\\ < R;
lUII > R.
Verify that (J)aj can be continuously continued over A.
Hint: Show
if J'
yWdiy =
^ ^Ri\\xf + 3R^),
3 \\x\\
\\x\\ < R;
lUII > R.
Then note that ^ ||x — yll = f^rrfr according to Example 2.4.8, and conclude that
2
47TRjA\\y-x\\
-3^;,
||x|| < R;
^(r^-l), \M\>R-
x\\ \3\\x\\^ /
XiR/ R^
Background. In the theory of dielectrics from electromagnetism one encounters the
potential of a charged sphere where the charge density is proportional to ^ = sin 0,
varying from +1 at the north pole to —1 at the south pole.
Exercise 7.17 (Unrolling a cone - needed for Exercise 7.18). Let / = ] 0, / [
and let y : / —> R^ be a C' mapping. Assume that y{s) and y'{s) are linearly
independent, for all 5 € /. Next define
()!):D:=/x]0, 1[^^R^ by (j){s,t) = ty^s),
and consider the cone C = im(0) in R^ determined by y.
(i) Show that 0 : D —> R^ is a C ^ immersion.
(ii) From now on assume 0 to be an C' embedding, which implies that C is a C'
manifold. Then prove that area(C) = j// Wyis) x y'(s)\\ds.
Introduce the angle function a : / —> Rbya!(5) = /J u (Jl\\2 do. By shrinking
/ if necessary, we may arrange that«(/) <7t. Furthermore, define
T:D^ V:=R+x]-;r,;r[ by Tisj) = {t\\yis)\\, ais)).

684 Exercises for Chapter 7: Integration over Manifolds
(iii) Prove that the angle function is monotonically increasing and denote by ^ its
inverse. Deduce that T is a C' diffeomorphism onto its image {(r, a) € V \
0 < a < «(/), 0 < r < r{a) }, where we write r{a) = \\y o ;6(a!)||. Define
4' : V —> t/ with ^(r, a) = r(cosa, sina) as in Example 3.1.1 and show
that
V := <if o r o (j)-^ : C ^ u(C) C U
with
v{ty(s)) = t\\y(s)\\ (cosa!(5), sina!(5))
is a bijection that maps a ruling on C to a line segment in R^ of the same
length that issues from the origin.
(iv) On the strength of Example 6.6.4 verify
area(C) = area(u(C)) = - / r{af da.
In other words, v is an unrolling of the cone C C R^ onto v (C) C R^ without
distortion of area.
Exercise 7.18 (Area of tangent cluster is area of tangent sweep - sequel to
Exercise 7.17). Let J = ] 0, / [, and let / € C(J) be given. Suppose that y : J ^^
R^ is a C^ parametrization by arc length of the curve 'vca.{y) and define, for c € R,
0c : D := {(5,0 e R^ I 5 € J, 0 < f < f{s)} -^ R^
by
^c{s,t)=cy{s) + ty'{s).
We refer to the conical surface C := im(0o) C R^ as the towgewf ctoter determined
by y and /, and to the surface S := im(0i) C R^ as the corresponding tangent
sweep. Note that the difference between C and S corresponds precisely to that
between tangent vectors and geometric tangent vectors. Denote by k: (5) = ||y"(5)||
the curvature of im()/) at y{s) as in Definition 5.8.1 and assume that k: (5) > 0, for
all 5 € J.
(i) Show that 0c : D —> R^ is a C^ immersion.
(ii) From now on suppose that 0o and 0i are embeddings. Prove (compare with
Exercise 7.17.(ii))
area(C) = area(S) = - / f{sfK{s)ds.
= \jjisf.i
(iii) Verify that the angle function a from Exercise 7.17 equals a {s) = f^ ic(a)da
in this case. And using part (iv) from that exercise show area (S) = area(C) =
area (u(C)), where v is the unrolling of C into R^.

Exercises for Chapter 7: Integration over Manifolds 685
Exercise 7.19 (Pseudosphere and non-Euclidean geometry - sequel to
Exercise 5.51). Consider the tractrix and the pseudosphere from Exercise 5.51.
(i) Prove that the length of the part of the tractrix lying between (x, ±/(x)) and
(1,0) equals log {\), for all x € ] 0, 1 ].
(ii) Conclude that the tractrix does not have finite length.
(iii) Calculate the area of the pseudosphere. Why does your answer not surprise?
Let V+ be the upper half of the pseudosphere.
(iv) Verify that we obtain a parametrization of an open part of V+ by means of
0 : ] —n, ;7r [ X [ 1, oo [ —> R^ satisfying
~ /cosM sinM r- 'Jv'^ ~ 1\
(j){u,v) = \ , , log(i; +V^^ - 1) )•
\ V V ^ /
Hint: Setx = ^ in Exercise 5.51.(i) and rotate.
(v) Assume M and n : / —> Rare two C'functions ofrange such that y : / —> V+
is a well-defined C' curve if y(f) = (j){u(t), v(t)). Verify that the length of
y is given by (assuming convergence of the integral)
f -^Vu'(tr + v'(t)^dt.
JI vit)
Background. The properties (iv) and (v) imply that this open part of V+ can in
fact be regarded as subset of the upper half-plane
^ = {x €R^|x2>0}~{z€C|Imz>0}.
Here J{ is endowed with the metric, that is, the concept of distance, which assigns
to a C' curve y : I ^ M the length
The set M with this metric is a standard model of a non-Euclidean geometry (more
precisely, a two-dimensional one, of constant curvature —1).
(vi) Verify that, with respect to this metric, the arcs in M of concentric circles
centered on the Xi-axis that lie within a (Euclidean) fixed angle from the
center are all of the same length.
This makes it plausible that in this geometry these arcs play the role of parallel line
segments. We shall now prove this. Define J : ^ —> C by J(z) = 1/z.
(vii) Prove that J : M ^ M, and that J is an involution, that is, J^ = I. Verify
that J is an isometry or a metric-preserving mapping of J{ into itself, that is,
L(y) = L(Jy), for every C' curve y : I ^ M.

686 Exercises for Chapter 7: Integration over Manifolds
A straight line in R^ is the set of fixed points of the reflection in that straight
line; consequently, straight lines in R^ are one-dimensional sets of fixed points of
nontrivial involutive isometries of R^. We now define straight lines in J{ to be
those one-dimensional submanifolds that occur as sets of fixed points of nontrivial
involutive isometries of J{.
(viii) Deduce from part (vii) that the segment in J{ of the unit circle in R^ about 0
is a straight line in J{. Verify that the mappings M ^ M, with ta{z) = z + a
for fl € R, and dr{z) = rz for r > 0, are isometries of J{. Conclude that the
segment in J{ of the circle in R^ about (a, 0) of radius r is a straight line in
J{ (consider the involutive isometry tad,Jd^^t^^).
Background. The semicircles in J{ associated with circles in R^ centered on the
xi-axis are therefore straight lines in this geometry on J{. It is now obvious that
Euclid's parallel postulate is violated in J{; indeed, given a point in J{ and a straight
line in J{ which does not contain that point, there exist many straight lines in J{ that
contain the given point but do not intersect the given straight line (and are therefore
"parallel" to it).
Exercise 7.20 (Identity by spherical coordinates). Suppose / € C'(R") has
compact support.
(i) Show (see Exercise 7.21 forhyperarea„_i(S""'))
■" ' hyperarea„_i(S''-0/r" Ilyll"
Hint: We have/(x) =—/g f'{x— roi)dr = J^ (gj:ad f(x — rM),M) dr,
for every m € S"^^. Next integrate this equality over m € S"^^ and change
from spherical to rectangular coordinates in R" (see Example 7.4.12).
(ii) Deduce
'•'' "-hyperarea„_,(S"-OyR" llyll"-'
Exercise 7.21 (Spherical coordinates in R" and hyperarea of (n — l)-sphere -
sequel to Exercises 3.18 and 6.50 - needed for Exercises 7.22, 7.23, 7.24, 7.25,
7.26, 7.28, 7.29, 7.30 and 7.53). Consider the (n - l)-dimensional unit sphere
S"-' = {x € R" I Ikll = 1} and the w-dimensional unit ball B" = [x € R" \
lkll<l}.
(i) Prove that Formula (7.26) is also valid for / : R" ^^ R with f(x) = e-"^"'.

Exercises for Chapter 7: Integration over Manifolds
687
(ii) Prove
hyperarea„_,(S" )
2:7r2
{n € N).
Hint: Calculate J^„ e H*"" dx by the use of Cartesian coordinates and
Example 6.10.8, and then also by part (i). Subsequently apply Exercise 6.50.(i).
(iii) Verify that (ii) is consistent with Example 7.4.11.
(iv) Proven vol,, (5") = hyperarea„_i(S""'), forw € N.
Hint: A ball is a union of concentric spheres.
(v) Calculate vol,,(5"), and verify that the answer is consistent with that found
in Exercise 6.50.(viii).
(vi) Prove
d
— vol,, {B"{r)) = — ""'' r" = ^^.
dr "^ ^ ^^ drT{}^ + \) r(f)
-r"-'=hyperarea„_i(S"-'(r))
Here B" (r) and S" ^ (r) are the ball and the sphere, respectively, in R" about
the origin and of radius r.
(vii) Verify that Exercise 3.18.(iv) implies that the mapping
0: ] -;r,;r[ X ] - |, | ['"'^ S""'\ {x € R" | xi < 0, X2 = 0}
is a C°° embedding, if
/ a \
<t>
\ ^"-2 /
/ COS a cos 6i cos ^2 * * • cos 0„_3 COS 0„_2 \
sin a cos 6i cos ^2 * * • cos 0„_3 cos 0„_2
sin 6i cos ^2 * * • cos 0„_3 cos 0„_2
sm6'„_3 cos6'„_2
sin0„_2 /
(viii) Use the hint from part (ii) and Exercise 3.18.(iii) to conclude that
M^ia, 6*1, ..., e„^2) = cos6*1 cos^92-■• cos" ^0„_2
Further, verify that by means of Exercise 6.50.(iii) one obtains
hyperarea„_, (S'-') = 2;r F (-) f]
11-2 T-.// + l_^
2/ ^JrAy
j=l

688 Exercises for Chapter 7: Integration over Manifolds
The result from part (viii) can be generalized as follows. Set S'^' = {}' € R'| |
\\y\\ = 1}. By analogy with Exercise 6.50 we then define the generalized Beta
function B : R^J. ^^ R by
B(pi, ..., p„) :=
r(pi + --- + pn)'
(ix) Prove that we now have the following generalization of the formulae from
Exercise 6.50.(iii) and Exercise 6.64.(iii) (see also Example 7.4.10), for p €
r;:
'Pi +1 p„ + l-
2n-ypt"+'^i<j<«Pjy
Consider this formula in the particular case that pi = • • ■ = p„ = 0, and
verify that the result is consistent with that from part (ii).
Hint: Imitate the technique from part (ii), or use part (viii).
(x) Suppose that pj € 2No, for 1 < j < n. Deduce that the average of the
monomial function 3' i-> yi' = Ok/oi)'/^ overS""' satisfies, in the notation
of Exercise 6.50.(vii) and with (—1)!! = 1,
/
hyperarea„_,(S"-')yy.-.-' "" ''' no</<!£!(«+2/)
yi'dn-iy = _-'- ^—-—- € Q.
Prove this equality also by application of the technique from part (ii) to the
function x i-> x''e""^" on R".
(xi) Using spherical coordinates and part (iv), prove the following identity, which
also is a direct consequence of part (x) and Exercise 6.65.(vi)
„(5")A" hyperarea„_i(S''-') A"-'
_n + \p\
vol
Exercise 7.22 (Pizzetti's formula - sequel to Exercises 2.52, 6.66 and 7.21).
Suppose / € C°°(R") is equal to its MacLaurin series. Verify for r € R+ and
yeS"-^'
f(ry) =Y^r'Y^ ^D"fiO),
(teNo \p\=k P-

Exercises for Chapter 7: Integration over Manifolds 689
where p € Ng. Using Exercise 7.21.(x) and the Multinomial Theorem from
Exercise 2.52.(ii) prove Pizzetti's formula for the spherical mean of / over the sphere
of center 0 and radius r
1 r
f(ry)d„_iy
a„-i(S''-0 A«-
hyperarea,,
y^ r^'' y^ D^i'fjO)
~ ^ 2k nn</^t(« + 2/) ^ p!
= y" ^ aV(o)
AV(0) /r^^k
/n\fr-sJ^^V^'^- ,
Here A denotes the Laplace operator and J" _i the Bessel function of order | — 1 as
in Exercise 6.66. Furthermore, the last equality is obtained by a formal substitution
in the power series for the Bessel function from part (ix) of the latter exercise. See
Exercise 7.54 for a different proof.
Exercise 7.23 (Sequel to Exercises 6.50 and 7.21). Prove (compare with
Exercise 6.99.(iii))
-^dx = - hyperarea,, (S").
/.
R" (1 + ||x|p)-i-
Hint: Substitute spherical coordinates first, and subsequently r = tan^S.
More generally, prove by means of Exercises 6.50.(iv) and 7.21.(ii)
L
1 , a rip-".)
n.
, „ ■dx=ni —^^ — (p > -).
'r„(1 + ||x|P)p Tip) ^P l'
Exercise 7.24 (Sequel to Exercises 6.50 and 7.21). Using Exercise 6.50.(iv) and
Legendre's duplication formula from Exercise 6.50.(vi), show
I /"(f2 + (l + ||x||Y)-("+'^rff^x = ^ (n€N).

690 Exercises for Chapter 7: Integration over Manifolds
Exercise 7.25 (Sequel to Exercise 7.21). Let B" c R" be the unit ball and let
^n-i ^ ^n i^g j-jjg jjjjjj. spjiej-g jn physics the moment of inertia Ii of B" about the
Xj-axis is defined by
/,■ = / V" xjdx (1 < i < n).
(i) Show that /,■ is in fact independent of i.
(ii) Using Exercise 7.21.(iv) prove, for I < i <n,
^' = -7^hyperarea„_i(S"-^) = ^vol„(5").
n{n + 2) w + 2
Hint: One has T^ T^ x? = (« — 1) T^ x^.
1 </<n l<j<n, j^i 1 <i<n
Exercise 7.26 (Generalized Beta function and standard (« — 1 )-simplex - sequel
to Exercises 6.64 and 7.21 - needed for Exercise 7.52). Prove, for p € R'|,
= B(pi, ...,p„) =
r(i:is;5«Pj)'
Hint: Substitute y„_i = (1 — X!i</<n-2 ^j) ^ ^"^ ^^^ innermost integral, and apply
mathematical induction over w € N. Alternatively, apply Dirichlet's formula from
Exercise 6.64 to the two outermost integrals, and proceed by induction in this case
also. A third possibility is to apply Exercise 3.19.
The standard (n — \)-simplex S""^ in R" is defined by
S"-' = {y € R" I J2 yj = l, yj > 0 for 1 < ;• < « }.
Prove, for p € R^,
/ P7 yj'^^^i-i}'= \/«B(pi,...,p„).
In particular, therefore,
hyperarea„_i(S" ) =
«-K Vn
(n - 1)!

Exercises for Chapter 7: Integration over Manifolds 691
Exercise 7.27 (Simplex coordinates in R" - sequel to Exercise 3.19 - needed for
Exercises 7.28 and 7.38). The standard (n - I)-simplex S""Mn R" is defined by
S"-^ = {x € R" I Yl ^J = ^' ^J ^ 0 for 1 < y < « }.
i<y<n
Let (j) : D —> S""\ with D C R""' open, be a C^ parametrization of an open part
of S""^ with neghgible complement (see part (vi) for explicit formulae).
(i) Verify that the mapping 4' : R+ x D —> R'| given by 4'(c, y) = c(j)(y) is sl
C' diffeomorphism on an open dense subset in R^. To verify the injectivity
of 4', use {(piy), (1,..., 1)) = 1. Show that
|detDvI/(c,30l =c"-'|det(D0(>-) | <t>iy))\.
(ii) Define e = -^(1,..., 1) € R". Demonstrate
(*) {<t>iy),e) = -^, {Dj<t>iy),e)=0 (y€D,l<j<n).
Conclude that e is a unit vector in R" that is orthogonal to S""' at every point
of S""^ Verify that the vectors e and Dj<p(y), for 1 < j < n, together form
a basis for R"; and prove by means of (•) that there exist numbers cj € R,
for 1 < y < «, with
\ln
(iii) Now prove by means of parts (i) and (ii)
1
detDvI/(c,30l = ^c"-'ftj4>-),
'n
where oi is the Euclidean {n — l)-dimensional density on S" '. Conclude,
for/€ C^(R:|.),that
\ f{x)dx =-^f c"-' f f{cy)d„_,ydc
= -^f f c"-'f{cy)dcd„_,y.
(iv) Verify that the formula from part (iii) also applies to the function f{x) =
e"^i<;<n *V^ forx € R'\.. Conclude that the formula for hyperarea„_j(S""^)
from Exercise 7.26 follows immediately.

692 Exercises for Chapter 7: Integration over Manifolds
(v) Verify that S^ C R^ is congruent with a regular triangle in R^ having edges
of length Vz. Give a geometrical proof of
11 1
areafS^) = V6 • V2 = -VS.
2 2 2
Likewise, verify that S^ C R'' is congruent with a regular tetrahedron in R^
having edges of length vz. Give a geometrical proof of
1111
vol(S') = ---Vl2--V3 = -.
3 3 2 3
Finally, we prove the existence of an embedding 0 as above,
(vi) Verify that Exercise 3.19.(i) implies that the mapping
0: ]0,1["-^^{>'€R:[| ^ >'; = 1}
\SjSn
is a C°° embedding, if we define 0 {y) as
{yiyiy^ •••yn-i, (l-yOyzya •••>'«-i, ••• , (1->'„-2)>'h-i, (l-y,,-!)).
Prove, by part (ii) and Exercise 3.19.(iii),
o^^fiy) = ^y2yl--- 3Cf iy ^ ] o, i {"-').
Exercise 7.28 (Feynman's formulae - sequel to Exercises 6.50, 7.21 and 7.27).
sfine Sr' ■'
(i) Prove
Define S" ' as in Exercise 7.21. Let a € R+.
Hint: One has
/ e^"'""' dxj = — (1 < /■ < n).
use Formula (7.26) and properties of the Gamma function from Exercise 6.50.
(ii) Now show, for p € R^J.,
f ni<j<„>';' _ B(pi, p2, ..■,Pn)
where B stands for the generalized Beta function from Exercise 7.21.
Hint: One has
f .-y-v;^;^-'rfx, = ^ il<j<n).

Exercises for Chapter 7: Integration over Manifolds 693
Next let S""^ be as in Exercise 7.27.
(iii) Verify, for p € R'|,
ni<;<« y'j' , r- B(pi, P2,...,Pn)
Js'-' (a, y)^'<J<"Pj
Hint: See Exercise 7.27.(iii).
Background. In quantum electrodynamics the Feynman formulae above are
important tools in making calculations concerning Feynman diagrams.
Exercise 7.29 (Sequel to Exercise 7.21 - needed for Exercise 730). Let S" C
R"+^ be the unit sphere, let / in C(R) be continuous and let x € R""^'. One then
has the following, known as Poisson's formula:
f fi{x, y)) d„y = hyperarea„_,(S"-^) j fi\\x\\t) (1 - t^)i-' dt.
Hint: The expression on the left-hand side is independent of the direction of
X € R""^^; therefore choose x = (0,..., 0, ||Ji:||). Now use Exercise 7.21.(viii).
Exercise 7.30 (Fourier transform of radial function and Hankel's formula -
sequel to Exercises 0.8, 6.51, 6.66, 6.99, 6.102, 7.21 and 7.29 - needed for
Exercises 7.31 and 8.20). Assume that / € -^(R") has the property that there
exists a function /q € -^(R) with f{x) = /o(||x||), for all x € R".
(i) Make use of spherical coordinates x = ry, where r € R4. and y € S""\ to
prove
m= f Mr)r"-' f e-'^^y-^Un-iydr (f € R").
(ii) Using Exercise 7.29, show that
f{^)=l /o(r)r"-'hyperarea„_2(S"-^) /" e-""*!" (1 - f^)'^ rff rfr.
(iii) Use Exercises 7.21.(ii) and 6.66 to prove that
Mf^-'m = i2n)i f r'ifoir)Ja_iiM\\r)dr (f € R").

694 Exercises for Chapter 7: Integration over Manifolds
(iv) Conclude that / € -^(R") also is a radial function,
(v) Demonstrate, by means of Example 6.11.4, that part (iii) implies
p"e-iP'= rie^i^Jn._i(r)dr (p € R+).
Prove that one obtains in particular, using Exercise 6.66.(x) and (v), for p € R,
17T 12 / 1 "^
'—e-iP = I e^^'" cos(pr) dr.
—pe^2p = j re^^'' sm(pr)dr.
See also Exercise 6.83.
(vi) Prove by the Fourier Inversion Theorem, for x € R",
\M\i-'fix)=f pJ|_,(||x||p) f rifoir)J,_,ipr)drdp.
JR.,. JR.,.
Use Exercise 6.66.(x) to prove that, in the case n = \, this formula takes the
following form:
f(x)=- cos(xf) / cosier) f(r)drd^ (x € R).
^ JR+ JR+
Now also prove the latter directly from the Fourier Inversion Theorem.
In what follows we write x instead of ||x|| and A. instead of | — 1; and, finally
g(x) = \\x\\i-'fo(\\x\\).
(vii) Now verify that, using (vi), one obtains Hankel's formula, known from
Exercise 6.102,
8ix)= \ / gir)J^irp)rdr J^ixp)pdp (x > 0).
Background. Here we have proved Hankel's formula for A. € { —^, 0, |, 1,...}.
Actually, the formula is valid for all A. > — j and suitably chosen functions g on
R4-, as was shown in Exercise 6.102.
(viii) Combining part (iii) and Exercise 6.99.(ii), prove the following, known as
Gegenbauer's formula:
2tr(^) fpl-'
V^ (f2 + p2)^
In particular (compare with Exercise 0.8),
JRj
e-"-r-2j:i_^{pr)dr = "J ../ \.^ (f > 0, p > 0).
/ e "'cos prdr = — -, / re "'Jo(pr)dr = r-
Jr.,. t^ + p^ U, it^ + p^)2

Exercises for Chapter 7: Integration over Manifolds
695
(ix) Using Exercise 6.66.(vi), show
r ds\ A J c4^ V 2/
Deduce, for ^ € N with k—k> — j.
_y T ^ r / ^ds ^ ^ 5 ~2~
Apply this identity with A. = | — 1 and ^ = A. + j, or ^ = A., if « is odd or
even, respectively. Using furthermore Exercise 6.66.(x), prove
1
2\S^/ 2 \hd- ^
(VI)^-' '
(—y (-r) MVsr), weven.
Now deduce from part (iii), that /(f) equals, for f € R",
i-\)'^2"n'^(— ) ' / /o(r)cos(v^r)^r, «odd;
^ds ^iijip/ A^
even.
Conversely, knowing the values of the integrals in part (viii), we can obtain
the identity in Exercise 6.99.(ii) by means of the formula above.
Exercise 7.31 (Sequel to Exercises 6.62, 6.86 and 7.30). For 0 < 5 < «, define
2^11 ■ ir*
/, : t/:= R" \ {0} ^ R by f,=
Then we have the following claim:
r(^)
/, = (2;r)^/-„_,.
(i) Verify that /, is a well-defined function, use Exercise 7.30.(iv) to show that
fs is radial, and prove it to be of degree of homogeneity s —n. Deduce there
exists a constant c{n, 5) € C such that /s(f) = c{n, s)\\^Y^'\ for ^ € U.
Next use the Parseval-Plancherel identity from Exercise 6.86.(iv) to evaluate
/^„/,(x)e-2ll-»ll-^xand prove the claim.

696 Exercises for Chapter 7: Integration over Manifolds
(ii) Suppose n = \ and recall the functional equation for the zeta function from
Exercise 6.62 or 6.89. Show that g^ : f/ —> R defined by
gs = -TT satisfies g, = y/2ji gi_, (0 < s < 1).
Exercise 7.32. Assume, for i = 1,2, that Vi are compact C* submanifolds in R"'
of dimension dj, where rf,- < «,-.
(i) Prove that V := Vi x V2 is a compact C* submanifold in R"' +"2 of dimension
d := di+ d2.
Let / : V —> R be a continuous function.
(ii) Prove that xi i-> f f(xi, X2) dii^X2, for xi € Vi, defines a continuous
function on Vi, and that
/ f(x)ddX= I I f(Xi,X2)dj^X2dd^Xi.
Jv Jv, Jvi
In particular, verify that volj(V) = vol^, (Vi) vol^, (V2).
Define the n-dimensional torus T" in R^" by T" = {x € R^" | ^l^t-i ~^ ^2k ~
1 il<k<n)}.
(iii) Verify that T" is a compact C°° submanifold in R^" of dimension n, and
calculate the Euclidean w-dimensional volume of T".
Exercise 7.33. (See Exercise 8.27.) Let V be a C' submanifold in R^ of dimension
d. Let p : V —> R be a continuous function. In electrostatics one defines the field
determined by the charge density p on V as the mapping £ : R^ \ V —> R^ with
EiM = ^ f „''' ~ ^^3 Piy) ddy (!</<«).
An Jv \\x-yr
(i) Let V be the plane xi = 0 in R^ and let p = 1 on V. Prove E{x) =
:^(sgnxi, 0, 0), for x ^ V, and conclude that the field is perpendicular to V
and of constant magnitude j.
>3
(ii) Let V be the xs-axis in R and let p = 1 on V. Prove
1
lTl{x\Jrxi)
^(•^) = ^_/..2 2T (-^1' -^2' ^) (-^ ^ ^^^
and conclude that, with cylindrical symmetry about V, the field is
perpendicular to V, and of magnitude equal to the reciprocal of 2ii times the distance
fromx to V.
Hint: Use a substitution of the form u = tan v.

Exercises for Chapter 7: Integration over Manifolds 697
Now assume that V is compact, and define (compare with Example 7.4.8 and
Exercise 8.28.(iv)) Xh& potential of V with density p as the function 0 : R^ \ V —> R
with
(t>{x) = — I -day.
An Jv \\x - y\\
(iii) Use the Differentiation Theorem 2.10.4 to prove E = — grad 0.
Exercise 7.34. Let / € Cc(R") be a C* function, for^ € N, and define /, = /*i/f,,
for f > 0, similarly as in Formula (7.55). For each a € Ng with \a\ ■<k, prove that
the functions D" f converge uniformly in R" to D" f, as t | 0.
Exercise 7.35 (Hyperarea as derivative of volume - sequel to Exercise 5.11).
Assume V C R" to be a C^ hypersurface. According to Exercise 5.11.(iii) there
exist for every x € V a number ^ > 0, an open set D C R""\ a C^ embedding
0 : D —> R" with im(0) C V, and an open neighborhood U(x) of x in R", such
that
%:]-S,S[xD^ U(x) with %(s, y) = 4,(y) + s vi4>(y)),
is a C' diffeomorphism (see Sequel to Example 5.3.11); here v((j)(y)) is as in
Formula (7.37).
(i) Assume that V is compact. Then prove that there exists a number t > 0 such
that
"^ ■.]-t,t[ X V ^ V, with '^(s,x) =x+ sv(x),
is a C^ diffeomorphism onto an open neighborhood V, of V in R". Prove
Vr = [z €R" I there exists x € V with ||x - z|| < t],
that is, the open tubular neighborhood V, of V of radius t equals the open
neighborhood of V consisting of the points whose distance to V is less than
t.
Let /■ : V —> R be continuous. Then define a continuation f : V, —> R of / to V,
by ■
fix + sv(x)) = f(x) (x€V, \s\<t).
(ii) Prove that / is well-defined on Vr and continuous.

698 Exercises for Chapter 7: Integration over Manifolds
(iii) Using that | det D^^(0, y)\ = co^iy) show
^tA / /W^-^= / fi-x)d„-ix,
^"■'^\t=aJv, Jv
whence
1 d I
tTT: vol„(Vr) =hyperarea„_i(V).
Background. The results above explain the formula from Exercise 7.21.(vi).
Exercise 7.36 (Integration over tubular neighborhood of level hypersurface
- needed for Exercises 7.37, 7.38 and 7.39). Let Uq C R" be an open subset,
and let g € C*(f/o), for ^ € N. Assume that x° € f/o and g(x°) = c° (that is,
x° € N(c°) = {x € Uq \ g(x) = c°}), and assume g to be a submersion at x°.
Finally, let / € C(Uq). We now prove that there exist a number y > 0 and an open
neighborhood U of x° in Uq such that
/f(x)dx= I I 7z—d„_ix dc.
.iec/o||g(A-)-c°i<}'}nc/ Jc°-Y yA^wnc/ II gradg(x)||
Heuristic argument: Let x € N(c) be fixed for the moment, and assume that
grad g(x) points in the direction of the x„-axis. Then T-^Nic) is spanned by the set
of standard basis vectors ej € R", with 1 < y < «. If g(x + dx) = c + dc, then
c + dc = gix) + {gradgix), dx}-\ = c+ \\ gradgix)\\ dx„ -\ ,
and therefore
dc
dc = \\ grad g(x)\\dx„, that is, dx„ =
I grad g{x)
Indicating a point in N{c) by x, we obtain from the identity dx = dxi-- ■ dx„ =
(dxi ■ ■ ■ dx„_i) dXn the equality
dx = d„-ixdx„ = —— dn-\x dc.
Ilgradg(5)||
The following observation is an additional argument to show that the formula above
must contain the reciprocal of || grad g(x) ||. If || grad g(x)\\ is large, the volume of
a rectangular domain between the level hypersurfaces A^ (c) and N(c + dc) is small.
But this volume can be calculated by multiplication ofd„_iX, the hyperarea (which
has a normal value) of the basis, and .. ^^f^ f^u, the height (which is small).
(i) Prove by the Submersion Theorem 4.5.2.(iv) that we have an open
neighborhood U of x° in f/o and a C* diffeomorphism 4' : V —> f/ of open sets in R"
such that
(*) (go4')(xi,...,x„_i, c) = c.

Exercises for Chapter 7: Integration over Manifolds 699
With the notation
•^ = (.^1,. ■., Ji^n-i) € R ,
V((x°)'; v) = {x'€ R"-' I \\x' - ix°y\\ < 7?},
and the Remark at the end of the proof of the Submersion Theorem one has,
for suitably chosen numbers t] and y > 0,
V = {y=(x\c ) € R"-^ X R I x' € V((x°)'; ??), \c - c°\ < y },
U = {x = (x', Xn) € R"-' X R I x' € V((x°)'; r)), \g{x) - c°| < y }.
(ii) Assume \c - c°\ < y. Verify that V((x°)'; rf) B x' \-^ *c(-^') := *(-^', c) is
a parametrization of N(c) Pi U; that is,
N(c) nU = imC^/^).
(iii) Write x = ^(y), for y € V. Conclude by the chain rule that (•) implies
Dg(x)D^(y) = (0,..., 0,1). Use this to prove that there exist numbers
Cj € R such that, for 1 < y < n,
gradg(x)_LD^vI/(y);
^"*(^> = llgradgWIP ^'''^(^> + ,IJ„'^^■^^■*(^>-
(iv) Conclude that the vectors gradg(x) and (Di^' x- ■ -x D„_i^)(y) are linearly
dependent in R", and that therefore
IdetD^/Cj)!
1
|gradg(x)|p
1
lgradg(x)
1
llgradg(x)
(v) Derive from parts (ii) - (iv)
|det(DiVl/(>-) .-. D„-i^(y) gradg(x))|
-|((DiVl/ X ■■• X A,-i*)(>'), gradg(x))|
■||(DiVl/x-..xD„_iVl/)(y)||.
I det Dvl/(x', c)| = ^''j'''] {(x', c) € V).
||gradg(4'(x',c))||

700
Exercises for Chapter 7: Integration over Manifolds
Now prove by the Change of Variables Theorem 6.6.1 and Corollary 6.4.3
f fix)dx = f if o\if)ix\c)\detD<ifix',c)\dx' dc
Ju Jv
L-y M.vOy;,)'' ''||gradg(vl/,(x'))||
c°+y r fix)
Jc°-y J^
Y JNic]
llgradg(x)
dn-{x dc.
The inner integral is said to be the (« — l)-dimensional integral of / over the level
I —I
hypersurface N(c)nU with respect to |^|, the GeVfand—Leray density associated
with g on N{c). One therefore has
/ f(x)dx= / / f(T)
Ju Jc°-y JN(c)nU
dx
'dg
(x) dx dc.
The result above is a local one. In order to arrive at global assertions we now
assume the following: N(c°) is compact and g is a submersion at every point of
Nic°).
(vi) Prove that there exists a y > 0 such that, for every c € R with |c — c°| < y,
the set N(c) is compact and g is a submersion at every point of N(c).
(vii) Use a partition of unity to prove that, for every c € R with |c — c°| < y,
dx
/ ,
Jn(c)
f&)
dg
(x) dx.
the (n — l)-dimensional integral of / over N(c) with respect to the density
7^ on N(c), is well-defined and is a continuous function of c.
(viii) Prove
//.c +y /.
f(x)dx= / f(x)
.xsUo\\g{x)-cO\<y} JcO-y Jn{c)
dx
'dg
(x) dx dc.
(ix) Conclude that, for all c € R with |c — c°| < y,
54,0 2S
lim — / f(x) dx = I f(x) ^ (x) dx.
-O J{xeUo I |g(A-)-cOl<5 } " Jn(c) '
dx
'd'g

Exercises for Chapter 7: Integration over Manifolds 701
Exercise 7.37 (Sequel to Exercises 6.64 and 7.36). Using Exercise 7.36, we now
give anotherproof of Exercise 6.64.(iv). That is, let A^ = {x € R^ | X1+X2 < 1},
assume pi, p2 € R4., and let /z : [0, 1 ] —> R be a continuous function. We then
have the following, known as Dirichlet's formula:
[ X^'-'x^'-'hix, +X2)dx = ^^P'^^^P^^ f xP^+>"--'h{x)dx.
Ja^ r(pi + P2) Jo
From this point onward the notation is that of Exercise 7.36. Verify the correctness
of the following assertions. Defining g : R^ —> R by g(x) = xi + X2 one has, for
all X € R2, that || gradg(x)|| = Vl, and A^ = {x € R^ | 0 < gix) < 1}. For
0 < c < 1 the following holds:
N(c) n A^ = {x € R^ I Xi +X2 = c} = im(0c),
where 0c : ]0, c[ —> R^ with (pdy) = iy,c — y). Consequently, co^^y) =
||^(j)||=V2, and therefore
/ II '^^^~m/'^= / f(y'C-y)dy.
JN{c)nA'- II grad g(x) \\ Jq
It therefore follows, with f(x) = xf" •^2^" ^(-^i + -^2),
/ xf'"^xf"^/2(xi+X2)^x =11 y'"-^ (c - y)!"--^ h(c) dy dc
Ja^ Jo Jo
= I h{c) cP'+'"--^ dc I t'"-^{\-t)>"--''dt.
Jo Jo
Exercise 7.38 (Sequel to Exercises 7.27 and 7.36). Using Exercise 7.36, we now
give a different proof of the formula from Exercise 7.27.(iii). That is, for every
/ e C.(R'I),
/" /(x)^x =-^l c"-' I ficy)d„_iydc
JrI v« ^r+ ^e«-'
= -^i f c"-\f(cy)dcdn-iy.
Define g : R^ —> R by g(x) = xi + h x„, then
||gradg(x)||=V;;: (x€R'|), R'| = \J Nic).
0<c<oo
For c > 0 we have
Nic) = {x € R:^ I ^ x^- = c} = {ex € R'l I X € ;V(1)} = im(0,),

702 Exercises for Chapter 7: Integration over Manifolds
with
0c : ] 0, 1 ["-' -^ R'l given by My) = cMy)-
Note that e:=4=(l,...,l)isa unit vector in R" which at every point of N(c) is
perpendicular to N(c), for all c > 0. Therefore we obtain, for y € ] 0, 1 ["~\
co^Ay) = d^iiDMy) I e) = c"-' det(D0i(>-) | e) = c"-'co^,iy),
where m is the Euclidean (n — 1)-dimensional density on N(c). As a result
f f{x)dx =11 ficMy)/' "^p^^^dydc
JK'^ Jr.,. J]o,i["-^ V"
= ^ f c"-' f f{cy)d,_,ydc.
Exercise 7.39 (Intersection of hypercube - sequel to Exercises 6.109 and 7.36).
Let g : R" —> R be defined by g(x) = ^i^j^„Xj, and let N(c) be the level
hypersurface in R" for g determined by c € R. The intersection of the hypercube
[ — J, J ]" with N(c) is a bounded subset S""' (c) of the hypercube, perpendicular
to the diagonal in the direction (1,..., 1)
3"-'(c) = [-i,i]"f];v(c).
As in Exercise 6.109 we write x for the characteristic function of the interval
[ — J, J ] C R, and /„ : R —> R for the «-fold convolution of/ with itself. Finally
we define the function x (8> • ■ • (8> x : R" —> R by
X 0 ■ ■ • 0 X(x) = x(xi) ■ • ■ x(x„) (x € R").
(i) Calculate the (n — l)-dimensional integral of x (g> • ■ ■ (g> x over N(c) with
respect to the density | j^ | on N(c), and conclude that the following formula
holds for the (n — 1)-dimensional area of S"~' (c):
hyperarea„_i (3"-\c)) = ^Xn{c) (c € R).
We recall the result from Exercise 7.27.(iv) that hyperarea„_j(S"~^) = "^" ,.
(ii) Now demonstrate, for c € R,
hyperarea„_i(S""^(c))
= hyperarea„_,(S"-^) ^ (_iy(p(c + ^ - ;)"-'.
0<J<[c+|]

Exercises for Chapter 7: Integration over Manifolds 703
In particular, for c € R satisfying c + | € No,
hyperarea„_i {3"-\c)) = hyperarea„_, (S"-^) /!(« - 1, c + ^),
where the Euler numbers A(n, k) known from combinatorics, not to be
confused with the Euler numbers E„ from Exercise 6.29, are given by
Ain,k)= ^(-iyY" + ^V/:-;■)«.
OSjSk \ J /
(iii) Verify that 3^(0) C R^ is congruent with a regular hexagon (e^ = six) in R^
with edges of length j V2, and that S^(0) is the disjoint union of six regular
triangles, all of them congruent with jS^- Likewise, verify that 3^(0) C R"*
is congruent with a regular octahedron (oxtw = eight) in R^ with edges of
length Vl.
Exercise 7.40. Let S = {x € R^ I ||x|| = 1} and let / : R^ ^^ RHe defined by
f(x) = (2xi, x|, x|). Prove that
/,
8:?r
{f,v)iy)d2y = —-.
s J
Exercise 7.41. Let S = {x € RM Ikll = 1} and let g : R^ ^^ R be defined
by gi-x) = x^ + X2 + xs. Prove by Gauss' Divergence Theorem (compare with
Exercise 7.9)
/ 8iy)d2y = —.
Exercise 7.42. Consider the cylinder ^ = {x € R^ | x^+x| < 1, — 1 < X3 < 1
and / : R^ —> R^ with f(x) = (xix|, x^X2, X2). Prove
/ {f,v)iy)d2y = ^.
Jdn
Exercise 7.43. The mapping 4' : R^ —> R^ is given by
^(y) = ((3 + B siny2) cosyi, (3 + ys sin3)2) sinyi, ys cosy2).
Furthermore,
C = {y €R^ \0 <yi <7T, 0 <y2<27T,0<y3 <l},
D = {y €R^ \0 <yi < tt, 0 < y2 <27t, ys = I}.
Finally, define the vector field / : R^ —> R^ by f(x) = (xs — xi, 2x2 + 2, X1X2).

704 Exercises for Chapter 7: Integration over Manifolds
(i) Calculate the volume of 4'(C).
(ii) Describe the boundary of 4'(C).
(iii) Calculate f^,Q){f, v}(y)d2y, where v is the outer normal to ^(D) with
respect to 4'(C).
Exercise 7.44. Let ^ C R^ be the bounded open subset bounded by
the unit sphere {x € R^ | ||x|| = 1};
the cylindrical surface {x € R^ | x^+| = 1};
the two horizontal planes {x € R^ | X3 = /z,-} (1 < z < 2),
where —1 < hi < /?2 < 1- Thus the boundary 9^ is the union of a shell Si on
the sphere, a shell S2 on the cylindrical surface, and two plane regions. We want to
prove in two different ways that the areas of the shells Si and S2 are equal. (This
result does not generalize to dimensions n ^3.)
(i) Show this by applying Gauss' Divergence Theorem to the vector field / :
R3 -^ R3 with
/(•^) = (2/2' 2 1^ 2' Oj-
(ii) Now give the proof by calculating the two areas.
Exercise 7.45. Let Q. C R" be as in Theorem 7.6.1.
(i) Prove in two different ways
«vol„(^)= / {y, v(y))dn-\y.
Verify that Formula (8.26) is a special case of the formula above.
Hint: Assume 0 € ^ and let f/be an open neighborhood of y in R". Consider
the solid cone with apex at 0 and base 9^ n f/.
(ii) Prove n vol„(5") = hyperarea„_j(S""') (compare with Exercise 7.21.(iv)).
Exercise 7.46 (Sequel to Exercise 2.32). Let / : R" —> R be positively
homogeneous of degree rf € R, that is /(fx) = t^ fix), for all t € R4- and x € R". Using
Exercise 2.32.(ii) prove
/ Afix)dx =d fiy)d„^iy.
Deduce (compare with Exercise 7.21.(x))
yj d„_iy = Yol„iB") il<j<n).
/
Js"-i

Exercises for Chapter 7: Integration over Manifolds 705
Exercise 7.47 (Generalization of Example 7.9.4 - needed for Exercise 8.24).
Let V be a bounded C' hypersurface in R" that occurs as an image under one
parametrization. Assume that 0 ^ V, and that every half-line originating from 0
has at most one point in common with V, but is not tangent to V at that point.
Let / : X i-> n. , ,x : R" \ {0} —> R" be the Newton vector field from
Example 7.9.4. Prove
X
solid angle subtended by V from 0
(/, v)(y)d„_iy = ^ 1 .
V total solid angle from 0
Hint: There exists an e > 0 such that ||x|| < e implies x ^ V. Consider
^ = { Ax I X € V and -—- < A < 1},
Ik II
and note that this determines an orientation of V, while fg^if, v) (y) d„^iy = 0.
Exercise 7.48. Let g € C^(R") and assume
^ = {x € R" I gix) <0}^&; 9^ = {x € R" I g(x) = 0};
Q is bounded; || gradg(x)|| = 1 (x € dQ).
(i) Prove the following: 9^ is a compact C' submanifold in R" of dimension
w — 1; ^ is Jordan measurable and vol„(^) > 0; ^ lies at one side of 9^.
(ii) Show that hyperarea„_, (9^) = f^ Ag(x) dx.
Exercise 7.49. Let B(R) and S(R) be the open ball and the sphere, respectively,
in R" about 0 and of radius R, and let / : B(R) —> R be a differentiable function
such that / and Djf, for I < j < n, can be extended to continuous functions on
BiR). Prove
f iDf(x)x + nf(x))dx = R f f(y)dn-iy.
Jb(R) Js(,r)
Exercise 7.50. Let Q C R" be as in Theorem 7.6.1, let / € C^(R") and let
grad / = 0 on 9^. Prove that
fiAf)Hx)dx=f J2 (DiDjfix)fdx.
•'^ •'^ ISiJSn
Exercise 7.51. Let Q C R" be as in Theorem 7.6.1 and let k>0.

706 Exercises for Chapter 7: Integration over Manifolds
(i) Prove
(n + k + l) f x„\\xfdx= f y„\\yf{y, viy))d„_iy.
Let a > 0 and assume next that Q = Q(a) is given hy Q(a) = {x € R" | ||x|| <
a, 0 < x„ }. Then note that 9^ = V(a)U V'(a) is a disjoint union if
Via) = {y€R"\ \\y\\=a, 0<y„},
V'ia) = {y€R"\ \\y\\<a, 0 = j„ }.
(ii) Show
r , fl*+2 /•
/ x„\\x\\''dx = — - ■ / v„{y)d„^iy.
Ja{a) n+k+ i J Via)
2(a) n -\-k+ 1 Jv{a)
(iii) Let B"-^ be the unit ball in R"-^ Verify
ii+k+\
k J '^ „„i rnn-ls
I Xn\\xrdx= " vol„_i(g"-0.
Hint: See Illustration for 7.4.111: Hyperarea.
Exercise 7.52 (Sequel to Exercises 6.65 and 7.26). Let the notation be as in those
exercises, but assume that p,- > 1, for 1 < z < w. Verify that S""^ C 9 A", and that
the function y i-> Wi^j^„y''j' vanishes at the points of 9 A" \S""^ Nowproceed
to prove, by Theorem 7.7.3, that in this case the formulae from Exercise 6.65.(iv)
and Exercise 7.26.(ii) are equivalent.
Exercise 7.53 (Spherical mean, Mean Value Theorem for a harmonic function,
Darboux's equation, and Liouville's Theorem - sequel to Exercise 7.21 - needed
for Exercises 7.54, 7.67 and 8.33). Let S{x\ r) and B{x; r) be the sphere and the
ball, respectively, in R" about x € R" and of radius r > 0. Let Q. C R" be
as in Theorem 7.6.1. For every / € C{Q.), x & Q. and every r > 0 for which
S{x\ r) C ^, we define the spherical mean mf(x, r) of / over the sphere in R"
about X € R" and of radius r > 0, by
'"/(•^' '■) = r Tz;—^ / f(y) dn-iy-
hyperarea„_, (S(x; r)) Js^,.,^
(i) Verify that
'"/(•^''■)= i^;;ZTT / ^f(x + ry)d„_iy.
In this connection, see Exercise 7.21.(ii) for |S""^| := hyperarea„_,(S""').

Exercises for Chapter 7: Integration over Manifolds 707
(ii) For all / € C^(^), x € ^ and r > 0 for which Six; r) C Q., prove that
dnif 1 /•
or \S" 'I 7^,-1
(iii) Let B" be the unit ball in R", and g : B" ^ R" the vector field x' i->
grad f{x + rx'). Check that
divg(x') = r Af(x + rx') (x' € B").
(iv) Then conclude that
9«V/ N ^
9r ^'''''^~ |S''-H
We say that / € C(Q) possesses the mean value property on ^ if, for all x € ^
and all r > 0 for which Six; r) C ^,
fix) = nifix, r).
The Mean Value Theorem for harmonic functions then asserts that / satisfies the
mean value property on Q if and only if / is harmonic on Q (that is. A/ = 0 on
Q).
(v) Using part (iv), prove the Mean Value Theorem for harmonic functions (see
Exercise 7.69.(iv) for another proof).
We now want to prove that the function mf : Qx R+ D-> R satisfies the following,
known as Darboux's equation:
d^m f n — \ dm f
(*) ^-T"(-^''') + -^ix,r) = Aj,mfix,r).
dr^ r dr
Here A,- is the Laplace operator with respect to the variable x € Q.
(vi) Verify
r" f fix + rx')dx' = \S"-^\ f p"-^ mfix, p)dp.
Jb" Jo
(vii) Now show by means of part (iv)
dmf
and
(•^' '■^ = '^ ^^- ( / '°" ' "^f^^' P)dp],
l(,"-i ^(;,, ,)) = A,(r"-' mfix, r)).
Then show that the differential equation (•) holds.

708 Exercises for Chapter 7: Integration over Manifolds
To conclude, we prove Liouville's Theorem which asserts that a bounded harmonic
function on R" is constant. (See Exercise 7.70.(viii) for a different proof.)
(viii) Assume / is bounded and harmonic on R". Prove by part (vi), for every
X € R" and r € R4-,
Use this to demonstrate, for every x € R" and r € R+, with ||/|| the supre-
mumof / on R",
l/W-/(0)|
= ^^ ,, / f{x')dx'-i f{x")dx'
\f\\( f dx'+f dx')
V||A-'-;t|l<r,||.v'||>r y||A-'||<)-, ||;t'-A-||>r /
r"\S"-^
f"-+\\x\\
dx'=^\\f\\ I
\U'\\<r+\]
n C n C+wn
^^T;;znll/ll / ^^' = -11/11/ p"-'dp
r \o I ^r-ii.v||<||A-'|l<)-+ii.v|i r J,—\\x\\
= ^WfWdr + \M\)" -ir- llxlD") = &(^), r -^ 00.
Verify that this proves Liouville's Theorem.
Exercise 7.54 (Pizzetti's formula - sequel to Exercise 7.53). Suppose / €
C°°(R") is equal to its MacLaurin series. We will give another proof of Pizzetti's
formula from Exercise 7.22
1 /■ r^
hyperarea,_,(S"-.) L f^''^'-" = £ 2*1 n„_<,<.(» + 2,) ^'^C''
To this end, note that mf(r), the spherical mean of / over the sphere of center 0
and radius |r|, is an odd function of r € R. Deduce from the MacLaurin series for
/ the existence of c^ € R such that mf(r) = X!ieN ^k'"'^- Substituting this series
in Darboux's equation from Exercise 7.53 prove
m^fir) = ^ 2(p + \){2p + n)Cp+ir^P.
peNo
Next show by mathematical induction over ^ € N
peNo ^' 0<l<k
and deduce Pizzetti's formula by taking r = 0.

Exercises for Chapter 7: Integration over Manifolds
709
Illustration for Exercise 7.55: Isoperimetric inequality
Exercise 7.55 (Isoperimetric inequality - sequel to Exercise 539). As was shown
in Example 7.9.1,
w vol„(^) = hyperarea„_,(9^),
if ^ = B". We now want to prove that, in general, one has for a set ^ C R" as in
Theorem 7.6.1 the following, known as the isoperimetric inequality:
/vol„(^)\"-^ ^ /hyperarea„_,(9^)\"
Vvol„(5")/ ~ Vhyperarea„_i(S"-')/ '
That is, among all permissible sets Q for which hyperarea„_| (9^) has a prescribed
value, the ball (of suitable radius) has the largest volume. Assume for the moment
vol„(^) =vol„(5").
(i) Prove that
ti^vol„iB"r\{y€R" \yi <t})
is a monotonically increasing continuous function on R.
(ii) Use the Intermediate Value Theorem 1.9.5 and mathematical induction over
j = 1, 2,..., « to find numbers
fjix) = fjixi, ...,Xj)€R (1 <;■<«)
such that
fix):={Mx),...,Mx))€B";
vol„+,_,(5" n {J € R" I Ji = /i(^),..., yj-i = fj-xix), yj < fj(x)})
= vol„4.i_^(^ n {j € R" I yi = -^1, ■ • ■, yj-i = Xj-i, yj < Xj }).
This can be worded as follows. The hyperplane 7/„_i(x) through x orthogonal
to the xi-axis divides Q into two parts, with volumes v^^ and v^p. Let H„^i(x)
be the hyperplane through (fiix), 0,..., 0) orthogonal to the xi-axis, such that
7/„_i(x) divides B" into two parts, with volumes n,^/^ and v^^\ respectively. Now

710 Exercises for Chapter 7: Integration over Manifolds
let Hn^2{x) be the intersection of ^ n 7/„_i(x) with the hyperplane through x
orthogonal to thex2-axis. Then H„_2{x) divides Q. n H„_i(x) into two parts, with
(w—l)-dimensional volumes v^^_, and v^_^, respectively. Let H„_2(x) be the («—2)-
dimensional affine submanifold through (/i(Ji:), f2ix), 0,..., 0) orthogonal to the
xi-axis and the X2-axis, such that Hn-2{x) divides 5" n 7/„_i(x) into two parts,
with (« — l)-dimensional volumes v\_^ and v\^_^, respectively. We continue in this
way, until we have found lines Hi{x) and H\{x), and finally points Ha{x) = {x}
and Hq(x) = {fix)}. This then constitutes the definition of fix).
We assume that the mapping / : ^ —> 5" thus defined is differentiable
(problems may occur here, for example, if 9^ contains "plane" regions). Moreover,
almost by definition, / is volume-preserving, that is, one has | det Dfix)\ = 1.
(iii) Verify that, with Xj (x) = Dj fj (x) > 0,
^ Xiix) • ■ ■ ■ • \
0 X2ix) '■■ ■
Dfix) =
*
V 0 ■■■ 0 X„ix) /
(iv) Prove by means of Exercise 5.39.(ii)
1/"
l<j<n ISjSn
«=«( n ^^w) "^ I] xjix)=diYfix).
(v) Conclude that
wvol„(^)</ {f,v}iy)d„_iy < rf„_ij = hyperarea„_,(9^).
(vi) Prove the isoperimetric inequality in the general case.
Exercise 7.56 (Normal derivative in polar coordinates - needed for
Exercise 7.57). Define 4' : R^ ^^ R^ by 4'(r, a) = /-(cosa, sina). Let r > 0
and dr > 0, and assume that a and a + da satisfy a and a+da € ] —tt, tt [. Let
^ C R^ be given by
Q = {4'(r', a') \ r < r' < r + dr, a < a' < a+da}.
(i) Verify that 9^ is the union of the following four sets:
{ 4'(r, a') \oi < a' < oi + doi], {*(/*', a.) \ r < r' < r + dr],
{ 4'(r +dr,oi')\oi<oi'<oi+ da.}, {*(r', a + da) \ r < r' < r +dr}.

Exercises for Chapter 7: Integration over Manifolds 111
(ii) Show that
v{^{r,a')) = (—cos a', —sin a'),
v(^(r',a)) = ( sin a, —cos a),
v(^(r +dr,a')) = ( cos a', sin a'),
v(^(r',a+da)) = (—sin(a! + rfa), cos(a +da)).
(iii) Prove by analogy with Exercise 3.8.(iv)
(Dif \ / cos a —sin a \ I 9^
Dif / \ sina cos a / I 1 9(/o 4')
r da
J
(iv) Verify
df
^{^(r, a')) = -cos a' D, fi^lfir, a')) - sin a' D2f{^ir, a'))
av
= ^ ir,a),
dr
dv r da
^{<l>ir+dr,a')) = ^^•^°^\r + dr,a'),
dv dr
9/ , 19(/o4') ,
/-(vl'(r',a + rfa)) = -^ Hr\a + da).
dv r da
Exercise 7.57 (Laplacian in polar coordinates - sequel to Exercise 7.56). (See
Exercise 3.8.(v).) Letx € R" and assume that the sets ^ below satisfy the conditions
of Theorem 7.6.1, and x € Q. Let the function / be as in Example 7.9.6.
(i) Prove
Afix) = lim ——— / Afix)dx = lim ——— / J-iy)d„
^^' s2|wvol„(^)L ■' am^oln{^)Jaadv^^'"
Define 4' : R^ ^^ R^ and ^ C R^ as in Exercise 7.56.
-ly-

712 Exercises for Chapter 7: Integration over Manifolds
(ii) Prove by means of Exercise 7.56.(iv) that, for small values of dr > 0 and
da > 0, modulo terms of higher order in dr and da,
/ T-(y)d„-iy = (r,a)rda (r,a)dr
Jdn ov or r da
d(fo^) id(fo^)
-\ ^ (r + dr, a) (r +dr)da -\ (r, a + da) dr
dr r da
\dr \ or / r oa-^ )
(iii) Prove the following formula for the Laplacian in polar coordinates:
S2
1 // 9 \2 32 \
Exercise 7.58 (Divergence in spherical coordinates). Let
vl/ : V = R+ X ] -;r, ;r [ X ] -|, I [ ^ R3
be the substitution of variables x = 4'(r, a, 0) = r (cos a cos 0, sin a cos 0, sin 6).
Let e,-, ea and eg be the orthonormal vectors as defined in Exercise 3.8.(iv), and let
/ : R^ —> R^ be a vector field. Then on V we define the component functions /,-,
fa and fg of / in spherical coordinates by
(/ o \if)(r, a, 9) = fr{r, a, 9) e,. + /„(r, a, 9) e„ + fg{r, a, 9) eg.
Consider numbers r > 0 and dr > Q;a and da > 0, with a and a+da € ] —n, n [;
and 6> and rf6> > 0 with 6> and 6> + rf6> € ] -f, f [. Let ^ C RHe defined by
Q. = {\if(r', a',9') \r <r' <r+dr, a <a' <a+da, 9 <9' <9+d9 }.
(i) Verify that 9^ is the union of six smooth surfaces, given by the condition
that precisely one of the r',a' or 9' is constant. Prove that the areas of these
respective surfaces, for small values of dr > 0, da > 0 and d9 > 0, modulo
terms of higher order in dr, da and d9, are given by (see the Illustration for
Example 7.4.11)
(r+ drf cos 9 da d9 {r' = r+dr), r^ cos 9 da d9 (r'= r),
r dr d9 (a'= a + da), r dr d9 (a'= a),
r cos{9 + d9)dr da {9' = 9+d9), cos9drda (9'= 9).

Exercises for Chapter 7: Integration over Manifolds 713
(ii) Now apply Gauss' Divergence Theorem to Q, and conclude that, for small
values of dr > 0, da > 0 and dO > 0, modulo terms of higher order in dr,
da and dO,
(div /)(4' ir,a,e))r^ cos 6> dr da d9
= (frir+dr, a, 9){r +drf - fr{r,a,9)r'^) cos,9da d9
+(/„(r, a + da, 9) — fa{r,a,9))rdrd9
+{fg (r, a, 9 + d9) cos(0 + d9) - fg (r, a, 9) cos 9) rdr da
9('■'/-■). .. , 9/«
= (cos6> — (r, a, 6>)+ /•-—(/•, a, 6>)
V dr da
dr da
9(cos0 fa) \
+r ^ •' '{r, a, 9)\ dr da d9.
(iii) Conclude, by taking a limit, that one has the following formula for the
divergence in spherical coordinates:
M- r^ ,T, ^ 9(rV,) , 1 9/ 1 d(cos9fg)
(div /) o w = — H 1 .
r^ dr rcos9 da rcos9 d9
(iv) Calculating in two different ways, demonstrate that for the identity mapping
/ : R^ ^^ R3 one has div / = 3.
(v) Define on V the component functions /'', /" and /^ of / in spherical
coordinates by
■' ■' dr ■' da ■' d9
Verify that /'' = /,, f = j^f„ and f = }.fg, and conclude that
1 /dif'A&iD^) d{f"A&iD^) d{fA&iD^)\
(div f)o^ = (^^ L^^il L^^il i).
■' detD^/V dr da d9 I
See Exercise 7.60 for a generalization of this result.
Exercise 7.59 (Outer normal on the boundary of an image - needed for
Exercise 7.60). Let 4' : V —> f/ be a C' diffeomorphism of open subsets of R".
Assume a and Z> € V are such that the rectangle B = B(a,b) C V, where
Bia, b) = {y€R"\aj< yj < bj (1 < y < n)}.
Define 9/,±5, the "smooth parts of the (n — l)-dimensional faces of B" by
9/,±5 = {y€R"\aj<yj < bj (j ^ i), yi = "' } (1 < / < n).
a;
Let Q = ^(B) C U. We now give a formula for the outer normal v(x) on 9^, for
X in the "smooth part" of 9^.

714 Exercises for Chapter 7: Integration over Manifolds
(i) Verify that S := dB \ [J^ ^ di^±B is an (n — l)-dimensional negligible set in
R". Prove that 4'S C ^ is an (« — l)-dimensional neghgible set in R", and
that dQ\<ifS = 4/(95 \S) = U/,± *(9,-,±5)-
Let I < i <n, and parametrize the (n — l)-dimensional manifold 4'(9/,±5) in R"
by the embedding, defined on an open subset of R""\
*/,± : (yi,.•., yi-i,yi+i,.•.,yn) ^ ^iyi, • • •,yi-i, ' ,yi+i, ■■■, y,,)-
(ii) Verify that Q lies at one side of 4'(9,- ±5) everywhere. Next, assume w > 3.
Prove that on 9/_± B one has the following equality of vector fields in R":
= ±sgn(detD4')(-l)''"'Di4' x ■■• x A_i* x A+i* x •■■ x D„4'.
To do so, use properties of the cross product. Formula (7.25), and
(±A4', ±sgn(detD4')(-l)'~'Di4' x •■■ x A-i* x A+i* x ••■ x A.*)
= sgn(detD4')(-iy-'det(D/4' Di^/• ■ ■ A^i^/ A+i* ■ ■ ■ A,*)
= \detD^\ > 0.
(iii) Show in a sketch, with B a rectangle in R^ and 4' = /, that the formulae
from part (ii) do indeed yield the outer normals.
Exercise 7.60 (Divergence in arbitrary coordinates - sequel to Exercises 3.14
and 7.59). Let the notation be as in Exercise 3.14. In the exercise we derived the
following formula for (div /) o 4' : V —> R, in terms of the component functions
with respect to the moving frame determined by 4':
(*) (div /) o vi/ = J^ div(Vi^vi/*/) = -^j2 ^'(^f'^y
Now we obtain (•) by means of integration, see Exercise 8.40 for another proof. To
verify (•), we choose y € V and dyi > 0, for I < i < n, such that the rectangle
B = B(y, y + dy) C V, in the notation of Exercise 7.59. We also employ the
other notations and results from that exercise.
(i) Prove by successive application of the Change of Variables Theorem 6.6.1 and
Gauss' Divergence Theorem 7.8.5 that, for small values of the dyt, modulo
terms of higher order in the dyi,
(div /) o vl/(y) ^{y) dyi... dy„ ^ I (div /) o ^{y')\ det D^{y')\ dy'
JB
= I Aivf{x')dx'= I {fv)(u')d„-\u'.
Jn ' Jdn

Exercises for Chapter 7: Integration over Manifolds 715
(ii) Conclude from Exercise 7.59.(i) that
f if, W("')d„_iu' = Tf if' vKu')d„_,u'.
(iii) Let I < i < n and let 4',-± be the parametrization of 4'(9,-,±5) from
Exercise 7.59; verify that the definition of integration over a hypersurface implies
f if, V)(«') rf„_iM' = f {/ o vl/,. ±, V o vl/,. ±)(/) co^^_^(/) dy'.
J^{di,.±B) JBi,.±B
(iv) Conclude by Exercise 7.59.(ii) that the following equality of functions holds
on di^±B:
{ f o vl/,. ±, V o vl/,. ±) co^^^ = ±/(0 I det Dvl/|.
(v) Now prove, for small values of dyi, and modulo terms of higher order in dyi,
Tf {fo vl/,-±, V o vl/,. ± ) iy') OJ^^ ^ iy') dy'
± Jdi,±B
= tI ±/«(y)|detDvi/(y)i^y
-dyi ■■■dy„.
dyi
Here we write y + dyt for (yi,..., >',_i, j,- + rfj,-, Jz+i, ...,>-„)€ 9/,+5.
(vi) Finally, prove (•) by taking the limit for dyi | 0, for 1 < i <n.
Exercise 7.61 (Laplacian in arbitrary coordinates - sequel to Exercises 3.8 and
3.16). In Exercise 3.16 we obtained the following formula for A on f/ in the new
coordinates in V, valid for every / € C^{U):
(A/) o vl/ = J^ div i^G-' grad(/ o^)) = ^ V D,{^g'^ DjXf o vf/).
We now prove this identity by means of integration,
(i) Verify that, for every h € CdU),
f h{x)dx= I {ho^){y)^{y)dy.
Ju Jv

716 Exercises for Chapter 7: Integration over Manifolds
(ii) Prove by means of formula (•) from Exercise 3.8.(ii) that, for / and h €
C'(f/),
{(grad /) o vl/, (grad h) o vf/ )(y) = D(/o vl/)(y) o Giy)-' ograd(/7, o vl/)(y).
Show that writing out into matrix coefficients one finds the identity of
functions on V
{(grad /) o 4/, (grad h)o<l>)= ^ g'JDj (/ o <l>) Dj (h o <l>).
(iii) Prove that for every x € U there exists an open set Q C R" that satisfies the
conditions from Theorem 7.6.1, and additionally x € Q C ^ C U.
(iv) Let / € C^iU) and h € C^i^) be chosen arbitrarily. Apply Green's first
identity from Example 7.9.6 to Q, and conclude by parts (i) - (iii) that
f (A/) o vl/(y) (h o vl/Xy) ^(y) dy= f (A/)(x) hix) dx
Jv Ju
= - I {grad /, grad h ) (x) dx
Ju
= - {(grad /) o vl/, (grad h)o^) (y) ^(y) dy
Jv
= -Jlf{jl^S'' Di(f o vI/)V>-) Dj(h o ^)(y)dy
= / E 0;(Vi^^''A(/ovi/))(j)(/2ovi/)(3,)rfy.
In the last equality Corollary 7.6.2 has been used.
(v) Finally prove the desired identity, making use of the continuity of the
integrands in (iv) and of the freedom in the choice of the function h.
Exercise 7.62 (Hamilton's equation - sequel to Exercise 2.32). In this exercise
we write x € R^" as
X = (q,p) = (qi,--., q,,, Pi,-.-, p„) € R" x R".
Let Q. C R^" be an open subset and H : ^ —> R a C' submersion. Here H is
said to be a Hamiltonian, while the (2« — 1)-dimensional C^ submanifold N(c) =
{x € Q I H(x) = c} is said to be the energy surface for H with energy c € R.

Exercises for Chapter 7: Integration over Manifolds 111
Furthermore, hh : ^ —> R^' is said to be the Hamiltonian vector field determined
by 7/if
\dpi dp„ dqi dq„ )
(i) Prove that (grad 7/(x), iih(^) ) = 0, for all x € ^, and conclude that
vh{x) € T,NiH(x)) (X € ^).
That is, at every point of Q the Hamiltonian vector field vh is tangent to the
energy surface for H containing that point.
Let X : J —> ^ with x : t h^ x(t) be a C' curve with x(0) = x° € Q. Assume
that t H^ x(t) isan integral curve of the following (ordinary) differential equation,
known as Hamilton's equation:
x'(t) = vaixit)) it € J),
that is,
dH , dH
(ii) Now show that the image of the curve x lies in the energy surface N(H (x°))
by proving
^MM^ = (grad H(xit)), x'it)) = 0 (t € J),
dt
(iii) Let w = 1 and 7/(x) = ^(q^ + p^) for x € R^ \ {0}. Verify that an
integral curve of Hamilton's equation which satisfies x(0) = (1, 0) is uniquely
determined, and is given by
x(t) = (cost, — sint) (t eR).
Now let w = 1, and assume that the function H is positively homogeneous of degree
m. In the notation of Example 6.6.8 we write Pj C R^ for the area swept out during
the interval of time J.
(iv) (Generalization of Exercise 0.5). Prove area(P7) = ^mH(x°) length(J).
Hint: Use Euler's identity from Exercise 2.32.(ii).

718 Exercises for Chapter 7: Integration over Manifolds
Exercise 7.63 (Archimedes' law). At a point x in a homogeneous and
incompressible liquid at rest a hydrostatic pressure p(x) € [0, oo[ exists, caused by the weight
of the "column" of liquid above x. In more precise formulation, the rate of increase
of this pressure is highest in the direction of gravity; we therefore have the equality
of vectors in R^
Vp(x) = p{x)g,
where p{x) is the mass density of the liquid at x, and g the acceleration due to
gravity.
Assume a body is submerged into the liquid and that, after it has come to rest,
it occupies the set ^ C R^. According to Pascal's law, at the point y € 9^ a force
of magnitude p{y) and direction —v{y) is exerted on the body, in particular, the
magnitude of this force is independent of the direction of the surface 9^ in j.
(i) Prove that the total force experienced by the body is described by the equality
of vectors
/ p{y)v{y)d2y = -g / p{x)dx.
Jdn Jn
(ii) Conclude that one has Archimedes' law: the force experienced by a body at
rest in a liquid is the opposite of the gravitational force acting on that mass
of the liquid that could fill the space taken up by the body.
Exercise 7.64 (Self-adjointness of Laplacian). Assume that the set Q c R"
satisfies the conditions of Theorem 7.6.1. Write C^(Q), with 0 < ^ < oo, for the
linear space of the C* functions / : ^ —> R satisfying supp(/) C ^.
(i) Verify that the integral inner product
{f^8)= f fix)six)dx (/, g € C°(^))
Jn
is well-defined on the linear space C°(Q).
(ii) Show that functions / and g € C^(^) satisfy the conditions from
Theorem 7.6.1. Demonstrate that the Laplace operator A € Lin (C^(Q), C°(Q))
is a self-adjoint operator with respect to this inner product on C°(Q), that is,
for / and g € C^iQ),
{Afg)=[ Afix)gix)dx = I fix)Agix)dx = if, Ag).
Jn Jn
(iii) Prove that, for every / € C^(^),
f fix) Afix) dx = - f \\ grad /(x)f dx.
Jn Jn
Assume that there exist a nontrivial / € C^(^) and a number A. € C with
—A/ = A./. Then conclude that A. = /x^ with /x > 0.

Exercises for Chapter 7: Integration over Manifolds 719
Exercise 7.65 (DiricWet's principle). Let Q be as in Theorem 7.6.1, and denote
by F the linear space consisting of the functions / € C(Q) such that all Djf, for
i < j < n, exist and satisfy the conditions of Theorem 7.6.1. Given k € C(dQ),
write Fj^ for the subset of all / € F with /Igo = ^; iii particular, consider Fq.
(i) Using Green's first identity show that f € F is harmonic on Q if and only if
we have
(*) f {grad fix), grad g(x) )dx = 0 (g € Fq).
Jn
Suppose ^ € C(9^) is fixed.
(ii) Consider f € Ft. Prove that / is harmonic on Q (in other words, / solves the
Dirichlet problem A/ = 0 and satisfies the prescribed boundary condition)
if and only if
(**) f II grad fix) fdx<f\\ grad gix) f dx (g € F,).
Jn Jn
Hint: Suppose the inequality holds. Then g € Fq implies f + t g € F^, for
all f € R. It follows that the function
D : R ^ R given by D(f) = /" || grad(/ + t g)ix) f dx
attains its minimum at 0. Then verify by differentiation that (•) is satisfied
and conclude on the strength of part (i) that / is harmonic. Conversely,
suppose / is harmonic on Q.. If g € Fj. is arbitrary, then h = g — f € Fq.
From the equality
f Wgmd gix) fdx
Jn
= /"(||grad/(x)||2+||grad/2(x)||2 + 2{grad/(x), grad/7,(x) ))rfx
Jn
in conjunction with part (i) we obtain that (••) holds.
Exercise 7.66 (Orthogonality of spherical harmonic functions of different
degree - sequel to Exercise 2.32). Let / € No and let Jii be the linear space over R
of the harmonic polynomials on R" that are homogeneous of degree /. Consider
pi € Jii and p„, € Ji,„ with /, m € Nq. Let S"~^ = {x € R" \ \\x\\ = I}. Apply
Green's second identity to show

720 Exercises for Chapter 7: Integration over Manifolds
Deduce from Euler's identity in Exercise 2.32.(ii)
^"'" (y) = {grad p„,(y), y)=m p„,(y) (y € S"-').
9v
Conclude that
/ Piiy)P,niy) dn-iy = 0 (/, m eNo, l^ m).
Js-^
Background. See Exercises 3.17 and 5.61 for more properties of the spaces J{i.
Exercise 7.67 (Nevrton's potential and Poisson's equation - sequel to
Exercise 7.53 - needed for Exercises 7.68, 7.69, 7.70, 8.28 and 8.35). Define, for
every x € R", Newton's potential p = p^ : R" \ {x} —> R of x by
1
-—logllx-x'll, «=2;
n^2.
Pix') = Pxix') =
i2-n)\S
H-n
Inthisconnection, seeExercise7.21.(ii)for |S" '| := hyperarea„_j(S" '). Inwhat
follows all differentiations of p are with respect to the variable x' € R" \ {x}.
(i) For all X € R", verify that p^ is a C°° function on R" \ {x}. Verify that, as
a consequence of Example 7.8.4, px is harmonic on R" \ {x}, that is, for all
X € R",
Ap^ = 0 on R" \ {x}.
Let Six; r) be the sphere in R" about x and of radius r, with outer normal,
(ii) Use Example 7.9.4 to prove that, for all x € R",
lim / p(x,y)dn-iy = 0, lim / —(x, >')4,_i>'= 1.
'■J-O Js{x;r) ''J-O JS{x;r) O^
(iii) Verify, for all x € R", that p^. and Djp^. are locally Riemann integrable on
all of R", but that this is not true of Djp-^, for 1 < j <n.
Hint: Use spherical coordinates with respect to x for x'.
Let Q. C R" be as in Theorem 7.6.1. Let / € C^{Q.), and assume / and Df can
be extended to continuous mappings on Q..
(iv) Using Green's second identity, and applying parts (i) and (ii), prove that, with
Pj. being Newton's potential of x,
fix) = f (Px A/)(x') dx' + f (/ ^ - Px ^)iy) d„-iy (X € Q).
In other words, the value of / at the point x can be expressed in terms of the
values of A/ on Q., and those of / and ^ on 9^.
Hint: See Example 7.9.4.

Exercises for Chapter 7: Integration over Manifolds 721
(v) Assume that, in addition, / is harmonic on Q; then prove
Conclude by part (i) that / € C°°(^), that is, a harmonic function is infinitely
differentiable.
Next, let / € C^(R") and define A^evi'tow 'spotential (j) of / by (see Example 6.11.5)
<p = p(0, ■)* f : R" -^ R, that is, <p(x) = / p(x, x') f {x') dx'.
Jr"
(vi) Prove by part (iii) that 0 is well-defined on all of R", that 0 € C^(R"), and
that (j) on R" satisfies the following, known as Poisson's equation:
(*) A0 = /.
Hint: Show that A0 may be calculated by differentiation under the integral
sign, that is
A0(x)= / p(0,x')Afix-x')dx' = / ipj,Af)ix')dx'.
Jr" Jr"
Then choose^ C R" as in Theorem 7.6.1 such that x € ^andsupp(/) C ^;
then /|gj2 = 0 and g^ | =0. Conclude by part (iv) that (•) is true.
(vii) Use Liou ville' s Theorem from Exercise 7.53. (viii) or from Exercise 7.70. (viii)
to show that 0 is the unique C^ solution of (•) that satisfies the boundary
condition at infinity
lim 0(x) = 0.
Exercise 7.68 (Another computation of Newton's potential of a ball - sequel to
Exercises 3.10 and 7.67). Let A C R^ denote the closed ball about the origin, of
radius R > 0. Without integration we shall compute Newton's potential (J)a from
Example 6.6.7,
<t>Aix) = --L f ^ dy i\\x\\>R).
An J A \\x - y\\
(i) Deduce from Exercise 7.67.(vi) that 0^ is harmonic on R" \ A.
(ii) Using Exercise 3.10.(i) or (iii) establish the existence of a and Z> € R with
4,^(x)=-^+b i\\x\\>R).
Verify limyj^ii^oo(PaM = 0, and deduce b = 0.

722 Exercises for Chapter 7: Integration over Manifolds
(iii) Prove lim||j:||_>oo \u.'_,\\ = 1 uniformly for y € A, and show
,. 1 f \M\ ,
a= hm --— / ;:dy =
in J A \\x - ■■"
voliA)
Wxf-^'oo An J A \\x - jII "■' 4n '
Background. There are still other ways to arrive at this result. Verify that the
definition of 0a (^) makes sense for x € A, and use Exercise 3.10.(iii) to prove
that there exist p and ^ € R with (pAix) = |||x||^ ~ ifiii "I" ^' ^°^ ^ "^ H-^H "^ ■'^■
Show 0a(O) = —^R^ using an elementary integration over A by means of spherical
coordinates. Deduce
0^W = -^(3/?'-IUf) i\M\<R).
o
Finally use the continuity of ^^(x) for ||x|| = R.
(iv) Prove similarly the result from Example 7.4.8.
Exercise 7.69 (Green's function, Poisson's kernel and Mean Value Theorem for
harmonic function - sequel to Exercise 7.67 - needed for Exercise 7.70). We
employ the notation from Exercise 7.67. Let Q C R" be as in Theorem 7.6.1. Let
p = p^ be Newton's potential of x € ^, and make the assumption that for every
X & Q. there exists a function q = q^-c € C^(^) such that q^ and Dq^ can be
continuously continued to 9^, while
Aq,=0 on Q, ^^Ign = "^^Ian-
According to Example 7.9.7, this uniquely determines q-^. The diagonal A^ of Q
is the set {(x,x) \ x € Q}. Green's function Gfor Q is defined by
G = Gn:i^x^)\AQ^ R, G(x,x') = Gn.xix') = pAx') + q^W).
Poisson 's kernel P for Q is defined by
P = Pn : Q X dQ ^ R with P(x, y) = —^(y).
dv
(i) Verify that G.v has the properties which p^ was shown to possess in parts (i)
and (ii) of Exercise 7.67, and that Gx\^^ = 0. Conclude that, for every / as
in Exercise 7.67.(iv),
fix) = f G(x, x') A/(x') dx' + f P(x, y) f(y) d„^iy {x € Q).

Exercises for Chapter 7: Integration over Manifolds 723
(ii) Let B(x; r) be the closed ball in R" about x and of radius r > 0. Assume x,
x' € Q, and let Q' be obtained from Q by leaving out B(x; r) and B(x'; r),
for r sufficiently small. Apply the identity from part (i), with Q replaced by
Q', and fhyz i-> Gix, z) and z i-> Gix', z), respectively. Thus demonstrate
the following symmetry of Green's function:
G(x, x') = G(x', x) ((x, x') € (^ X ^) \ A^).
(iii) Verify that in this case the Dirichlet problem A/ = g on ^ and f = h on
dQ from Example 7.9.7 can be solved by means of part (i)
f(x)= f G(x,x')g(x')dx'+ f P(x,y)h(y)d„
_iy (x € Q).
(iv) Prove the Mean Value Theorem/or harmonic functions (see Exercise 7.53 for
its formulation).
Hint: Letx € ^andr > 0, and apply (i) with ^ = {x' € R" | ||x'-x|| < r}.
In this case. Green's function G for Q differs by a constant from Newton's
potential of x. Therefore, with nif the spherical mean of /,
x) = mf(x, r) + I Gix, x') Af(x')dx'.
Jn
/(
Exercise 7.70 (Green's function and Poisson's kernel for half-space and ball,
Poisson's integral, Schwarz' Theorem and Liouville's Theorem - sequel to
Exercises 1.3 and 7.69). Consider the special case of Exercise 7.69 where we
defined, violating our standard convention,
Q^ = R^+' := {(x, t) € R"+' I X € R", ±f > 0}.
Then 9^± = {(x, 0) € R""^^}, and the outer normal at every point of 9^± is given
by T(0, 1).
(i) Verify that, for ((x, t), (x', t')) € (^± x ^±) \ A^±, Green's function G±
has the value G±((x, t), (x', t')) given by
^{log\\ix-x',t-t')\\-log\\ix-x',t+t')\\) (n = l);
(1
n)\S"\\Ux-x',t-t')r-' ||(x-x', f + fOII"-'/ ^''^^^'

724
Exercises for Chapter 7: Integration over Manifolds
(ii) NowprovebydifferentiationthatonehasforPoisson'skernelP±,for(x, t) €
^± and (y, 0) € dQ±,
P^iix, t), (y, 0)) =
±f
|S"|||(x-j, OII"+''
Conclude (compare with Exercises 6.99.(iii) and 7.23) that
L
1
R„ II (X -J, Oil
" + 1
rf>' =
1 ;r
|f| r(^)
(xeR", t^O).
Let r > 0. Then consider the special case of Exercise 7.69 where
^ = B"ir):={x€R" \ \\x\\ <r},
and thus 9^ = S"-'(r) := {j € R" | llyll = /• }•
(iii) Let x € B"(r) \ {0}. Using the symmetry identity from Exercise 1.3 prove,
for all J € S"-'(r),
ll^->'ll =
r \\x\\
X y
lUII r ■"
r
Verify x := -f-^x ^ B"(r). (Actually, x is the inverse of x with respect to
the sphere S"^^ir), that is, x € R" is the vector in the direction of x for which
||x||||x||=r2.)
(iv) For (x, x') € iB"(r) x B"(r)) \ AB"(r), if x y^ 0, show that G(x, x') equals
r \\x\\ ,
X X
lUII r
)
^{\og\\x-x'\\-\og
J ( \ \ ^
n)\S"-'\\\\x - x'W"-^ ||^;,_M;,.||"-V
(2
and further, if x = 0,
(« = 2);
in ^ 2);
G(0, x') =
1
lit
, \w\\
log
r
1
(2
-i (^ L^
(n = 2);
(n ^ 2).
Now prove by differentiation
^(^,>') =
\S"-'\r Wx-yW"
(x € B"(r), y € S"-\r)).
Assume that n = 2 and let y € S^ir). Prove that the level curves of the
function x h^ P(x, y) : B^{r) —> R are circles tangent to S^{r) at the point
y-

Exercises for Chapter 7: Integration over Manifolds 725
Given h € C(S""^(r)), define Poisson's integral 3^h : B"{r) -^Rofh by
|S"-V A-wlU-jll"
(v) Prove that ^h € C°°(5"(r)). Check that a solution / of the Dirichlet
problem A/ = 0 on B"(r) with /|™-ifri = ^' is given by / = ^h, known
as Poisson 's integral formula.
(vi) Conclude that
/
1 27r~ r
dn-iy = z—- , „ „, (x € 5"(r)).
(vii) Prove Schwarz' Theorem, which asserts
lim (^/2)(x) = hiy) (y € S"-\r)).
Hint: Let y° € S""^(r) be fixed. Prove by part (vi), for x € 5"(r),
'(r)
{(Ph){x) - hiy ) = / —-^-ly.
\S" 'k A«-'(r) lU-yll"
Let e > 0 be chosen arbitrarily. By virtue of the continuity ofh, among other
arguments, there exist a partition of S"^^ir) into subsets Si and S2, and a
number c > 0 such that
ye Si =^ \hiy)-hiy°)\<^, y€S2 =^ ||>'-/||>2c.
Hence, for all x € B"(r),
r^-\\xf f \h{y)-hif)\, €
\ / «n-lV < -.
15"-'k A, \\x-y\\" 2
Assume next that x € B" (r) satisfies ||x — y°\\ < c. Then, for y € S2,
\\y-n > \\y -/ll - 11/ -^11 >2c-c = c.
Let ||/?,|| be the supremum ofh on S"-'(r); then \hiy) - hiy°)\ < 2\\h\\, for
all y € S"-'(r). Ergo, there exists 0 < S < c such that for all x € B"(r)
with ||x -y°\\ < S, and for all y € S2,
r^-\\xf r"-' ^ „ „ 2r"-' e
< ('•- Ikll)—— <
r ||x-j||" c" 4||/2||
This gives, for all x € B"(r) with ||x - /|| < ^,
; / «n-lV < -.
15"-'I r 4 llx-yll" '-^ 2

726 Exercises for Chapter 7: Integration over Manifolds
(viii) Prove Liouville's Theorem from Exercise 7.53.(viii).
Hint: Let x € R". Apply Poisson's integral formula from part (v), with
r > ||x|| arbitrary, and with / itself as the function h on the boundary.
Calculate the partial derivatives Djf, for 1 < y < w, by differentiation
under the integral sign. One obtains, with ||/|| the supremum of / on R",
\DJ{x)\<2\\f\\ ""^l' 11,+"11/11',,''"'" /" -. ^^-ly.
Subsequently, use ||x — yH > r — ||x||, for y € S" {r), to estimate the
integral, and show
\Djf{x)\=0{^), r^oo.
Exercise 7.71 (Uniqueness of solution of heat equation). Let Q. c R" be as in
Theorem 7.6.1 and let T > 0. Let C C R"+' be the open cylinder ^ x ] 0, T [. Let
/ : C —> R be a continuous function for which the following continuous derivatives
exist on C:
9/ . 9V . . 9/
dxj ~ ~ ' dXidXj — ' — ' Qj •
Assume that / satisfies on C the heat equation from Example 7.9.5, with ^ = 1.
(i) Prove, for every f € ] 0, T [,
+ I ||grad,/(x,Of ^x- I (f^)iy,t)d„^iy.
Here the gradient is calculated with respect to the variable x € R".
Now assume /(x, 0) = 0, forx € ^, and f{y, f) = 0, for y € 9^ and f € ] 0, T [.
(ii) Prove, for all 0 < f < T,
0>-r- I fi^,t)^dx; and conclude that / f{x,t)^dx=Q.
of Jn Jn
Show fix, t) = 0, forx € ^ and f € [0, r ].
We derive some additional results, under the assumption that / satisfies the heat
equation and the Neumann boundary condition
^(j,0 = o (j€9^, f€ ]0,r[).
dv

Exercises for Chapter 7: Integration over Manifolds 727
(iii) Prove, foiO <t <T,
I fix, t)dx= I fix, 0)dx, I fix, tfdx < j fix, Qfdx.
Jn Jn Jn Jn
Exercise 7.72 (Needed for Exercise 7.73). Let x° € R^ be chosen arbitrarily, and
let / : R3 \ {x°} ^^ R be defined by
/(x) = log||x-x°||.
(i) Using Example 2.4.8 prove, for x € R^ \ {x°} (see also Exercise 3.8),
grad/(x)= ^oii2(^-^°)' ^/W
X -x"|r ' \\x -x°P'
(ii) Prove that A/ is absolutely Riemann integrable over R^ \ {x°}.
Now let g € C' (R^) be a function with compact support. Part (ii) then leads to
2
— dx < oo.
In fact, we have the following inequality: for every x° € R^ and every g € C^ (R^)
with compact support one has
W f II ^^''^112^^^^/' llgradg(x)||2rfx.
We now prove (•).
(iii) Use Green's first identity to prove
f gi^f , ^ f gM , 0 . , s\.
/ ir^ dx =—2 I —— (x — X , grad ?(x))(2x.
Then derive from this
and conclude that (•) holds.

728 Exercises for Chapter 7: Integration over Manifolds
Exercise 7.73 (Stability of an atom - sequel to Exercise 7.72). According to
quantum physics an atom consists of an atomic nucleus with a positive electric
charge, and negatively charged electrons moving around that nucleus. Assume that
the magnitude of the electric charge of the nucleus is a multiple Z € N of the
electronic charge. An electron moving around the nucleus then finds itself in an
electric field, caused by the atomic nucleus, which is described (to within a constant)
by Newton's potential (see Exercise 7.67)
^W = -A (x€R^\{0}).
It has been found that an electron moves in certain orbits only. Furthermore, an
electron can spontaneously jump from such an orbit to another orbit lying closer to
the atomic nucleus, under simultaneous emission of an amount of energy. According
to Schrodinger these orbits are parametrized by numbers A. € R that are eigenvalues
of the Schrodinger operator with Newton's potential; that is, those A. for which there
exists a function g € C^(R^) with/j^, g(x)^rfx = 1 satisfying the following, known
as Schrodinger's equation:
(*) - l-Agix) -Z^= Xgix) ix€R'\ {0}).
2 \\x\\
Then the amount of energy emitted upon the transition from an orbit with parameter
A. to one with parameter /x is proportional to A. — /x. If the collection of eigenvalues
A. has a lower bound, this opens up the possibility for the electrons not to continually
emit energy, but instead to remain in a ground state of minimal energy; this is then
referred to as the stability of the atom. We now prove this. To avoid convergence
problems we restrict ourselves to g € C^(R^) with compact support.
(i) Multiply both sides of (•) by gix), to deduce
l f \\ giadgix) fdx-Z f ^dx=^ f gixfdx.
(ii) Using (•) from Exercise 7.72, prove that
Iff- ^L'-^-L^-^'-T
< -^ /" ||gradg(x)||2 + 2Z /" g{xfdx.
(iii) Conclude that a number A. € R for which there exists a nontrivial g € C^(R^)
with compact support satisfying (•) has the property A. > —2Z^.

Exercises for Chapter 8: Oriented Integration 729
Exercises for Chapter 8
Exercise 8.1. Let / C R be a closed interval and y : / —> R" a C ^ mapping.
(i) Prove that there exists a C' mapping S : [k_,k^] —> R" with im(y) =
im(S) and also S'(k^) = S'(k^) = 0. (This mapping S is not an immersion
everywhere, and a fortiori not an embedding.)
Hint: Define the C' mapping ^ : J := [0, 1 ] ^^ R by i/r(0 = t\2 - tf.
Then i/f(0) = 0 and T/f(l) = 1, and thus J C im(i/f) according to the
Intermediate Value Theorem 1.9.5. Moreover f(t) =4t(l - t)(2 - t) > 0
for 0 < t < \, and therefore i/f is monotonically strictly increasing on J.
Hence i/f : J —> J is a C ^ mapping that preserves the order of the endpoints
for which ^'(0) = ^'(1) = 0.
(ii) Using part (i) deduce that the image under a piecewise C ^ mapping can be
written as the image under a C' mapping.
Exercise 8.2. Let U C R" be an open rectangle with 0 € f/, and let / : f/ ^^ R"
be a C ^ vector field with Af = 0. Then g : f/ —> R is a scalar potential for / if
gix)= ^ / fiiXi,X2, ...,X!^i,t,0...,0)dt.
Prove this by direct calculation, and also by using that g(x) = f {/(s), dis},
where yx : [ 0, w ] —> R" is the curve from 0 to x successively following the
directions of the standard basis vectors e;, for 1 < i <n,
YxiO = (xi,X2, ...,x,_i,(f-/ + l)x/,0...,0) (1 < / <«, 0 <f-/ + l < 1).
Exercise 8.3. Define f : R'^ ^ R'^ hy f(x) = (xixj, xi + X2), and let ^ c R^
be the open subset in the first quadrant bounded by the curves {x € R^ | xi = X2 }
and {x € R^ | x^ = X2}. Prove by integration over Q, and also by using Green's
Integral Theorem
1
curl f(x)dx = —.
/
Exercise 8.4 (Astroid - sequel to Exercise 5.19). This is the curve A defined as
the zero-set in R^ of g(x) = x, + x^ — 1.
(i) Verify that the length of A equals 6.
(ii) Prove that the area of the bounded set in R^ bounded by A equals \n.

730 Exercises for Chapter 8: Oriented Integration
Exercise 8.5 (Quadrature of parabola). Define 0 : R —> R^ by 0(f) = (f, t^);
then P = im(0) is a parabola. Consider arbitrary t_^ and f_ € R with t_^ > t_, and
define
t+ - f_ f, +1_
(i) Let f € R with ?+>?>?_. Demonstrate that the area of the triangle in R^
with vertices (j)it_^), (j)(t) and 0(f_) equals
tl - tl f+ - t^
Calculate the unique t such that the corresponding triangle A(t_^, tJ) has
maximal area, and show that this area equals 6^.
(ii) Find the value of t for which the direction of the tangent space to P at 0 {t)
is the same as that of the straight line l{t+, f_) through 0 (^4.) and (pit-).
(iii) Prove in three different ways, including the use of successive integration and
of Green's Integral Theorem, that area (Sit+, f_)) = IS\ if Sit+, f_) is the
sector of the parabola P with base /(?+, f_), that is, the bounded part of R^
which is bounded by the parabola P and the straight line /(?+, f_).
Hint: Check that
S(f+, f_) = {x € R^ I f_ < xi < t+, x^ <X2< (to - Sf + Itoixi - t-)}.
Further, use |(f^ - tl) = JS^ + 2t^S.
The quadrature of the parabola according to Archimedes follows from parts (i) and
(iii): for all t+ and f_ € R with t+ > f_ one has
rcr. , ^^ 4area(A(r+, r_))
area(S(r+, f_)) = .
The area of a sector of a parabola therefore equals four thirds the area of the inscribed
triangle having the same base as the sector of the parabola, and whose apex is the
point at which the tangent line to the parabola runs parallel to that base.
We give another, direct proof of this result. Define 4' : R^ —> R^ by
*(>-)= ]=(f)ito) + S[ y.
\ti + 2toSy,+S^y2 ) "^^ ' \ 2to S )'
(iv) Demonstrate that 4' is a C°° diffeomorphism. Verify 4' o 0(f) = (j)(tQ+S t),
for all f € R. Now conclude that 4' maps the parabola P into itself; and that
the triangle A(f4., f_) from part (i) is the image under 4' of the triangle with
vertices (1, 1), (0, 0) and (-1, 1).

Exercises for Chapter 8: Oriented Integration 731
(v) Prove that det D'^{y) = S^, for every y € R^. Then conclude by part (iv)
that the quadrature of the parabola has been reduced to a special case.
Exercise 8.6 (Steiner's hypocycloid and Kakeya's needle problem - sequel to
Exercise 5.35). Let Z> > 0; let 0 : R ^^ R^ and Steiner's hypocycloid H cR^he
defined by, respectively,
/ 2 cos a + cos 2a! \ u • /J.^
(p(a)=b{ ^ . . ^ and H = im((p).
^^ ' V 2 sin a - sin 2a! / ^^'
(i) Prove that the length of H equals 16Z>, that is, 16 times the radius of the
incircleof 7/.
(ii) Prove that area of the bounded set in R^ bounded by H equals Inb^, that is,
twice the area of the incircle of H.
Background. Consider the special case where Z> = |. Exercise 5.35.(vii) then
implies that a needle of length 1 can be continuously rotated in the interior of H
by an angle In; moreover, during the rotation the needle is always tangent to H.
The area of that interior of H equals ^, while that of a circle of diameter 1 equals
^. Thus arises Kakeya's needle problem: what is the minimal area of a subset of
R^ within which a needle of length 1 (lying in R^) can be continuously rotated.
For a long time it was thought that |- would be the answer. It has been shown by
Besicovitch, however, that there are sets in R^ of arbitrarily small area > 0 that
have the desired property. ^
Exercises.?. LetC C R^ be the intersection of the cylinder {x € R^ | x^+x^ = 1}
and the plane [x € R^ | Xi +X2 +Xj, = 1}, and let C be oriented by the requirement
that the tangent vector at the point (1, 0, 0) have a positive X3-component. Let
/ : R3 ^^ RHe defined by f{x) = (-xj, x^, -xj).
(i) Prove
f{fis),dis} = 3 f
Jc J{x
fdx=^-^.
eR^\\\x\\<l}
(ii) Also calculate f^ifis), dis ) by means of a parametrization of C.
Hint: cos'* a + sin a = j(3 + cos 4a!).
See Section 3.5 in Krantz, S. G.: A Panorama of Harmonic Analysis. Mathematical Association
of America, Washington 1999.

732 Exercises for Chapter 8: Oriented Integration
Exercise 8.8. Let the surface S C R^ be the union of the cyhnder {x € R^ |
x^ + ^2 = 1, 0 < X3 < 1} and the hemisphere [x € R^ \ x^ + x^ + {xj,— 1)^ =
1, 1 < X3 }. Let S be oriented by the requirement that the normal at (0, 0, 2) point
away from the origin. Define / : R-^ —> R-^ by f{x) = (xi + XiX^ + X2x|, X2 +
xiX2x|, x^Xj). Prove
L
{curl fix), d2X ) = 0.
Exercise 8.9. Let 0 : R^ —> R^ be the mapping given by
(j)(r,a) = (r cos 4a!, r sin 4a!, cos a).
Further, let
Q = {(r,a) € R^ | a < tt, r < sina }, S = (piQ),
Y\ = {(sin a. cos 4a!, sin a. sin 4a!, cos a) € R^ | 0 < a < tt },
Y2 = {x € R^ I xi = X2 = 0, — 1 < X3 < 1}.
Let Y2 be oriented by the requirement that the tangent vector at (0, 0, 0) have a
positive X3-component.
(i) Show that 9 S = yi U )/2-
Assume one is given the function g : R^ —> R and the vector field /z : R^ —> R^
defined by
g(x) = 2(X3 +X,X2)v/3+||x||2,
h{x) = ((4 + xf)x2+xiX3, (4 + x|)xi+X2X3, 4 + XI+X1X2X3).
(ii) Demonstrate that grad g restricted to the unit sphere S^ C R^ equals the
restriction of h to S^, and conclude that
/ {gradg(5), d\s)= I {h(s), dis).
Jyi Jyi
(iii) Calculate y (gradg(5), dis).
Hint: Use the fact that integration by parts with respect to X3 commutes with
setting Xi and X2 equal to 0, so that the third component of grad g can easily
be integrated.
(iv) Prove that curl grad g = 0 and conclude f {h(s), dis } = —8.
Hint: It may be taken for granted that Stokes' Integral Theorem also holds
for integration over S.

Exercises for Chapter 8: Oriented Integration 733
Exercise 8.10. Let g : U := R^ \{ (0, 0, X3) | X3 € R} ^^ R^ be the vector field
satisfying
JC3
gix) = -Xsifixi,X2), 0) = ^ ^fe, -xi, 0);
here / is the gradient vector field of the argument function: R^ \( ] —00, 0] x {0}) —>
]—TT, 71 [ (see Example 8.2.4). Let i/f € [—^, ^] be fixed and suppose y is the
parametrization of the parallel on the unit sphere S^ C R^ determined by i/f, given
by
y : a \-^ (cos a cos i/f, sinacosi/f, sini/f) € S^ (—tt < a < tt).
(i) Prove f {gis), dis) = —iTTsini/f by explicit computation as well as by
application of Formula (8.18).
(ii) Demonstrate
1
cmlg(x) = — j(x,,X2,0) and {curlg(x), x) = 1 (x € U).
Let S be the cap of the sphere S^ determined by i/f as in Examples 7.4.6 and 8.5.1.
(iii) Verify by a direct computation using part (ii) and Example 7.4.6
/.
(curlg(x), d2x) = 27i{l — sini/f).
(iv) The results from parts (i) and (iii) seem to contradict Stokes' Integral
Theorem. Prove this is not the case.
Hint: The vector field g is not defined in (0, 0, 1) € S^, and a limit argument
based on the result from Example 8.2.4 gives the missing line integral having
the value 27r.
Background. The line integral in part (i) gives the angle of daily rotation of Fou-
cault'spendulum from Exercise 5.57. See Exercise 8.45 for the same computation
in terms of differential forms.
Exercise 8.11 (Cauchy's Integral Theorem by differentiation under integral
sign). As in Cauchy's original proof, verify Cauchy's Integral Theorem 8.3.12 by
means of differentiation under the integral sign.
Hint: Combine Formula (8.27), Formula (8.11) and Lemma 8.3.10.(iii).
Also perform this computation working over C. That is, consider
j f(Z2)dZ2 = j(f o r)(zu Z2)D2r(zi, Zl) dzi,
and differentiate under the integral sign. Next integrate by parts and finally use
Diif o r) D2T - D2(f o r) D,r = (/' o r) DiF DzF - (/' o r) DzF d.f = o.

734 Exercises for Chapter 8: Oriented Integration
Exercise 8.12 (Equivalence of holomorphic and complex-analytic - needed for
Exercises 8.13, 8.16, 8.17, 8.21 and 8.22). Assume that ^ C C and the
holomorphic function / : ^ —> CmeettheconditionsofCauchy'slntegralTheorems.3.11.
For every z € ^ there exists a number /?° > 0 such that the circle S{z\ R) about
z of radius 0 < /? < /?° is contained in Q.. Let Q.' = Q'(R) C C be the open set
bounded by Siz', R) and 9^.
(i) Apply Cauchy's Integral Theorem to Q' in order to conclude that, for 0 <
R < R°,
I aw = I aw;
Jdnw-z Js(z-R) w -z
h(,z;R) W -Z
here S{z\ R) has positive orientation.
Now consider the parametrization a : ] —tt, tt [ —> S{z\ R) with a : t \-^ w{t) =
z + Re".
(ii) Verify that, for all 0 < /? < R°,
/ -^^-^^ dw = i fiz + Re") dt.
Js{z:K) W-Z J-^'
(iii) Using the Continuity Theorem 2.10.2, prove the following, which is known
as Cauchy's integral formula:
2^2 J3!^. "' "
W
Let z° €Qhe fixed for the moment. Then, for every z €Q and w € dQ,
1 _ 1 I ^-^° ^ _ y- (z-z°)' ^ (^-^°\" ^
w — z w — z° w — z° w — z „, (u; — z°)*+' \w — z°) w — z
It follows that, for every z €Q,
(z - z°)' :— / , ' dw + Rn iz).
0<k<ii
with
2^2 Jdn\w -z^J w -z
(iv) Note that 9^ is compact; use this fact to prove the following. There exist
numbers p>0, 0<p<l and ^ > 0 such that, for every z € C with
\z — z°\ < p and all w € dQ,
zeQ;
z-z°
w — z°
p; \w -z\>

Exercises for Chapter 8: Oriented Integration 735
(v) Prove that, for every z € C with \z — z°\ < p, we have lim„^oo Rniz) = 0;
then conclude that
fiz)=j:iz-zy':^f -^^-
That is, the holomorphic function / is complex-analytic, which means that / can
be written as a complex power series near z° € Q.. The differentiability of a power
series on its disk of convergence gives the reverse of this assertion.
(vi) Use «-foId differentiation of the power series in part (v) to conclude that
In I Jsn (w - 2°)"+'
Exercise 8.13 (Fundamental Theorem of Algebra - sequel to Exercise 8.12).
Let p : C —> C be a complex polynomial function of degree « € N, that is
Piz) = J2osksn ^''^'^ '^^^^ cj. € C and c„ ^ 0. Suppose that p(z) i^ 0, for all
z € C. Prove that p : C —> C is a nowhere vanishing complex polynomial function
if
piz) •■= Yl ^"-''^^ = ^"^(~)' ^^ deduce p{z) = z"p{-)-
0<k<n
Let Q = {z € C \ \z\ <!}• Using Cauchy's integral formula from
Exercise 8.12.(iii), the substitution z = ^ and Cauchy's Integral Theorem 8.3.11, prove
2ni f I /■ 1 f w"-^
0 # = / dz = / r- dz = - dw = 0.
P(0) Js^zpiz) J3nz"+'pO JsnPiw)
From this contradiction obtain the Fundamental Theorem of Algebra, which states
that p must have a zero in C (see Example 8.11.5 and Exercise 3.48 for other proofs).
Exercise 8.14 (Equivalence of holomorphic and orientation-preserving confor-
mal). Let f/ C C be open; we identify U with an open set in R^ via U B z =
xi + ix2 <^ X. Let / : f/ —> C be a complex-valued function, to be identified
with the vector field / : f/ —> R^. Prove that the following four assertions are
equivalent.
(i) / : f/ —> C is a holomorphic function at z with f'iz) 7^ 0.
(ii) / : f/ —> R^ satisfies the Cauchy-Riemann equation at x and Df(x) ^ 0.
(iii) detD/(x) > 0 and there exists R = Rifx) € SO(2, R) with Dfix) =
^detDfix)R = \fiz)\R.

736 Exercises for Chapter 8: Oriented Integration
(iv) det Df(x) > 0 and there exists c = c(f, x) € R with ( Df(x)u, Df(x)v) =
c {u, v), for all u, v € R^. In other words, / is an orientation-preserving
conformal mapping at x (see Exercise 5.29).
Hint: First prove by means of Df(xyDf(x) = c I that c = det Df(x), and show
that (iv) =^ (ii).
Exercise 8.15. Let U C Che open and let / : f/ —> C be holomorphic. We
identify / with a C' vector field f :U ^R'^.
(i) Prove that for every C' curve y : I ^ U
lcngth(fiy)) = j \f'(yit))\\y'it)\dt.
(ii) Let L € ${U), and assume that /|^ is an injection with f'{xi + 2x2) ^ 0, if
X € L. Demonstrate that
= i \f'(xi+i^
area((/(L))= / \fixi+ix2)fdx.
Exercise 8.16 (Winding number and Residue Theorem - sequel to Exercise
8.12). Let / : R2 \ {0} ^^ R^ be given by /(x) = j^Jx, with J as in For-
11-^' II
mula (8.20). Let ^ C R^ be as in Green's Integral Theorem, and a €Q.
(i) Prove
-^ f {fis-a),d,s) = l.
2^ Jd<^.
Hint: There are various conceivable methods:
(a) use Example 7.9.4;
(b) prove / = grad arg with arg : R^ \ (] —00, 0] x {0}) —> R the argument
function;
(c) use Kronecker's integral from Example 8.11.9.
Lett/ C C be open and fl € f/, and let y be a closed C'curve with im()/) C U\{a}.
(ii) Prove that there exists a number w = ■w(y,a) € Z, the winding number of
y about a, with
1 r 1
—^ / dz = w(y,a).
2ni Jy z -a
Hint: Reduce to part (i), in particular Kronecker's integral, or use the
following argument. We may assume )/:/ = [0, 1]—>f/\ {a}, and we
introduce
gix)= r ^^dt (x€/).

Exercises for Chapter 8: Oriented Integration 131
Then /'—(/— a)g' = 0 on /, and hence ((y — fl)e~*)' = 0. As a result,
forx € /,
^gW^K^^U^^ in particular £«(') = !,
and so g(l) = Iniw.
(iii) Verify that Cauchy's integral formula from Exercise 8.12.(iii) under the
present conditions takes the following form. For / : f/ —> C holomorphic,
w
iy,a)fia)= —^ / rfz.
Ini Jy z— a
(iv) Now assume y in f/ \ {a} to be homotopic with the mapping yi : t \-^ a +
re^", where r > 0 has been chosen sufficiently small to ensure im()/i) C U.
Prove that «;(/, a) = i(;(yi, a) = 1.
Let / : f/ \ {fl} —> C be holomorphic. The residue ReSj=Q f{z) of f at a is
the unique number r € C such that z i-> fiz) — tz^ has an antiderivative on a
sufficiently small neighborhood of ainU \{a}.
(v) By means of part (ii), prove
f{z)dz = w(y,a)Resf(z).
— I
Ini Jy
(vi) Assume im()/) C U \{a\,..., a,,,} and / : U \{a\,..., a,„} —> C to be
holomorphic. Prove the following Residue Theorem:
^ f fiz)dz= T wiy,a,)Rcsfiz).
2ni Jy , .
Background. In complex analysis methods are developed for the efficient
calculation of residues.
Exercise 8.17 (Generalization of Cauchy's integral formula - sequel to
Exercise 8.12 - needed for Exercise 8.18). Let i = v^—T. Identify x = (xi,..., X2„) €
R^'withz = (zi,..., Zn) ^ C", whereZj = X2j-\+iX2j;y^hile'zj = X2j-i — iX2j,
for 1 < y < n. We then have the real 2w-dimensional vector space T^R^' ~ T,C",
with the partial differentiations Dj, for 1 < y < 2«, as basis vectors (see
Exercise 5.75). Let (TxR^")c be the complexification of r^R^".
(i) Show that the following vectors form a basis over C for (TxR^")c, with
9 1 9 1
(*) ^ = :^(D2j-i - i D2j), —- = -(02,-_, + i D2j).
aZj I aZj I

738 Exercises for Chapter 8: Oriented Integration
As usual, we write T*R^" for the dual vector space over R of TxR^". Then the dxj,
for 1 < y < 2n, are basis vectors for T/R^" over R. Let (r;R2")c be the dual of
(r.R^'Oc.
(ii) Prove that the basis over C for (r*R^")c, dual to that in (•), is given by
dzj = dx2j-\ + i dx2j, dzj = dx2j-\ — i dx2j (1 < y < n).
(iii) Prove
dx,AdX2A...A rfx,„_, A rfx,„ = Q"dz. Adz, A... A dz„ A dz„.
Now let / : C" ^^ C be a C^ function, and consider £?/(x) € (T*R'^")c.
(iv) Show that
l<j<n
that is.
^/ = E (f^^^^ + f ^^^) =■■ f/^^+1^^ =■■ ^f+^f^
d = d + d.
The function / is said to be holomorphic or complex-differentiable if it satisfies the
Cauchy—Riemann equation
9/ = 0, that is, i D2j-if = D2jf (!<;■< n).
(v) Let n = \, and assume / : C —> C to be holomorphic. Prove
df
df = -fdz = df.
az
Use this to show that the differential 1-form f dz is closed on C (this is
where the restriction « = 1 is important). Next, let a € C. Prove, using
^(t;^) = 0, that the following differential 1-form is closed on C \ {a}:
^ -dz.
z- a
Let fl € C and assume / : C —> C to be an arbitrary C' function,
(vi) Conclude that on C \ {a}
d(^- dz) = ^— %dzA dz.
\z —a / z — a dz

Exercises for Chapter 8: Oriented Integration 739
Let Q C C be a bounded open subset having a C' boundary 9^ and lying at one
side of 9^. Let / : ^ —> C be a C' function such that / and the total derivative
Df can be extended to continuous functions on Q.
(vii) Conclude by part (iii), and using polar coordinates (r, a) for z — a €C, that
dz Adz = —2i dxi A dx2 = —2i rdr A da.
Use this to prove that z i-> 71^ g| (z) is absolutely Riemann integrable over
(viii) Now prove, analogously to Exercise 8.12.(iii), the following generalization
of Cauchy's integral formula:
f(a) = -^ f I^dz + -^ f ^—^iz)dzAdz (aeQ).
27Tt JsnZ-a Ini J^z-adz
Exercise 8.18 (Generalization of Cauchy's Integral Theorem 8.3.12 - sequel to
Exercise 8.17). Let <I>o and <I> 1 : C —> C be two homotopic C' mappings. Suppose
/ : C —> C is a holomorphic function and let ftj = /(iz be the corresponding closed
differential 1 -form on C as in Exercise 8.17.(v). Apply the Homotopy Lemma 8.9.5
to find a C^ function g on C such that ^\a> — ^^u> = dg. Next suppose that
y : [0, 1 ] —> C is a closed C' curve, thus, in particular, y(0) = y(l). Now prove
the following generalization of Cauchy's Integral Theorem 8.3.12:
f fdz= f fdz.
Exercise 8.19 (Sequel to Exercise 6.57). In four steps we shall prove (see
Exercise 6.60.(iii) for another demonstration), for 5 € C,
cos I cos 2 (0 < Re5 < 1).
(i) Let a > 0 and apply Cauchy's Integral Theorem to the function f(z) =
e-^z'"' and the set Q which equals the (open) square with vertices 0, a,
a + ia and ia, of which the vertex 0, however, is cut off by a small quarter-
circle of radius e with 0 < e < a. Show, for 0 < Res < 1,
pa pa+ia pia
0= e-^'x'-'dx+l fiz)dz+ fiz)dz
Je J a J a-\'ia
+ r e-'yiiyy-'diiy)+ f e-'^'^'-^e'^'-^^f'd{ee''>') =: V /,-.
J" Ji l<j<5

740 Exercises for Chapter 8: Oriented Integration
(ii) Verify
I/2 + /3I < e-^+^'^'^i r(a^ + y^)^^ dy
Jo
Jo
Furthermore,
|/5|<!^eRe.^|Im.|f_
(iii) Conclude by taking limits for e | 0 and a —> 00, for 0 < Re 5 < 1, that
Vis)
= f e-'x'-^ dx = e''i f e-'>/-^ dy.
Jr^ Jr^
(iv) If we carry out the same reasoning as in parts (i) - (iii) but using the square
of vertices 0, a, a — ia and —ia, the square again being indented at 0 by a
quarter-circle of radius e, we obtain
= e-''i f e'>
Jr^
r(5)=e-"2 / e'^y'-'dy.
Jr+
Deduce the formulae in (•) by addition and subtraction, for 0 < Res < 1.
(v) Verify that the first equation in (•) is valid for —1 < Res < 1.
Exercise 8.20 (Asymptotics of Bessel function - sequel to Exercises 6.66 and
7.30). We employ the notations from Exercise 6.66. Let A. > — |. Under the
substitution v(x) = ^u(x) Bessel's equation takes the form
(*) v"(x) + (1 - -^-^jv(x) =0 (X € R+).
Neglecting terms 0(^),forx —> 00, we obtain the harmonic equation i(;" + i(; = 0,
with w(x) = a cos(x — /x), for constants a and /x € R, as the general solution.
This makes it plausible that a solution v of the equation (•) has the form
v(x) = a cos(x — fj,) + 0( — j, X —> 00.
Indeed, for the Bessel function J)^ we shall prove
(**) AW=/f-^cos(x-|A-|) + 0(-^), x^oo.

Exercises for Chapter 8: Oriented Integration 741
(i) Verify
-1
Let Z = C \ (]-oo,-1 ] U [ 1, oo [).
(ii) Check that
fj^(x) = f e""(l - fY-i dt (x € R).
g:z^e'«(l-zY-5 :Z
is a well-defined holomorphic function if we require that (1 — z^)^ ^ > Q,
forz€]-l,l[.
(iii) Let a > 0 and apply Cauchy's Integral Theorem to the function g and the set
Q C Z which equals the (open) rectangle with vertices —1, 1, 1 + ia and
— 1 + ia. Conclude, for x € R4., that
0 =f^(x) + i e'''^^+'^\y^ - 2iy)^-'^ dy
Jo
/.O
+i / e'''^-^+'^'\y^ + 2iy)^-^ dy + Ria),
J a
where linia^oo R{a) = 0. Verify that then
Mx) = I+ix) + /_(x), /±(x) = iie"^" f e-^^(>-2 ± 2iy)^-i dy.
(iv) Show
.2^.. ai .^..a i A i f e)(/+i) (0<>'<1);
[ &iy^-^) il<y <oo).
Prove that I±(x) then equals, for x —> 00,
2V X
) %^"T(A + ]r) + 0(x-^-i) + 0(e-^').
/ 2
(v) Now prove (••).
(vi) Conclude, by part (v) and Exercise 6.66.(viii), that the Hankel transform Mx f
of a function / from Exercises 6.102 and 7.30 is well-defined for / having
the property that r i-> -/r f(r) is absolutely Riemann integrable over R4..

742
Exercises for Chapter 8: Oriented Integration
xlO"
xlO"
4 li x^ ,/— cos (x - ^)
IgllMIIMftiW
xlO-
xlO"
£i
IjWi/ifwwvw
Illustration for Exercise 8.20: Asymptotics of Bessel function
Eiix) = J2(x) -J cos ^x - — j.
Esix) = MX)-^(^(l-^) cos (x
Stttx 15 .
^ , —„ ,.. — sin
128x2/ V 4 / 8x
('-f))
on [50, 250]
(vii) It is possible to formulate a stronger version of the result from part (iv). To
show this, we write
/±(x) = -(±20'+
'^^"j^b-'^-'i^-^r^-
Applying Taylor's formula for z i-> (1 + z)^~^ at 0 with the remainder
according to Lagrange (this is obtained from the integral formula for the
remainder by application of the Intermediate Value Theorem 1.9.5), we find.

Exercises for Chapter 8: Oriented Integration 743
for m € N,
0<H<m
for a f € [0, 1 ]. We now note that 11 T ^' I > 1 for the present x and y\
and so the absolute value of the remainder, for m > A. and those x and y, is
dominated by
\m + \)\lx)
The integration with respect to y over R+ can now be carried out, and we find
the following asymptotic expansion, for x —> oo:
■'^w-vf^^
(-1)2 cos(^x--A--j
(-1) 2 sin(x--A--j
where we take the upper or the lower expression behind the brace according
to whether n is even or odd, respectively. This asymptotic expansion can also
be put into another form. To do so, we introduce
<h(X,x) =x A. ,
2 4
(4^2 - 12)(4a2 - 3^)... ((4^2 - (2« - 1)2)
a(X,n) = .
«!8"
One then has
r^/ .^ „a{k,2n)
^ HeNo
5m0(A,x)2_^(-l) ^ 1, X
Exercise 8.21 (Laplace's formula for Legendre polynomial - sequel to
Exercises 0.9 and 8.12). Let z € C and choose ^ C C such that the conditions of
Cauchy's Integral Theorem 8.3.11 are met, and such that z € Q.
(i) Use Exercise 8.12.(vi) to prove the following, known as Schlafli's formula
for the Legendre polynomial P/, for / € Nq, from Exercise 0.4:
1 [ {w"^- 1)'

744 Exercises for Chapter 8: Oriented Integration
Now letx € R with \x\ > 1 and choose Q = {w € C \ \w — x\ < -y/x^ — 1}.
(ii) Check that t i-> w(t) = x + ^/x^ — 1 e" is a parametrization of 9^. Then
prove the following, known as Laplace's formula:
Pi(x) = — I (x + v/x2 - 1 cosf)' dt (x € R, |x| > 1).
The choice a = x and Z> = -y/x^ — 1 leads to a special case of the integral studied
in Exercise 0.9.
(iii) Prove by using that exercise
Pi(x) = - (x- y/x^-l COSty'-^ dt (x € R, |x| > 1).
^ Jo
Also prove, using the identity P/(x) = (—1)'P;(—x), forx €Rwith|x| > 1,
P,(x) = — / (x-y/x'^- 1 costYdt = — (x + v/x2- 1 cosO~'~'^^
n Jo n Jo
(iv) Analogously prove
Pi(x) = — (x + iy/l - x2 cosf)' ^^ (-^ e R, |x| < 1).
Conclude that the zonal spherical function Y° from Exercise 3.17 satisfies
Y^{a,e) = -\ (sine+i cos9 cos tydt (|a!| < ttt, |6>| < -).
^ Jo 2
Exercise 8.22 (Real and imaginary parts of a holomorphic function are
harmonic - sequel to Exercise 8.12 - needed for Exercise 8.23). Let / and Q be as
in Exercise 8.12, and consider /i and /2, with f = fi + 2/2, as functions on an
open subset of R^.
(i) Prove by means of the Cauchy-Riemann equation that /i and /2 are harmonic
functions. Check that the vector fields grad /i and grad /2 on Q are both
harmonic, and mutually orthogonal at every point of Q.
Conversely, let /i € C^(^) be given, with ^ C R^ open. We want to find /2 €
C-^(Q) such that / := /i + 2/2 is a complex-analytic function on ^ C C.
(ii) Prove that the Cauchy-Riemann equation for / gives the following condition
on/2:
grad f2 = J grad f.
Show that the integrability condition curl grad /2 = 0 implies the identity
div J'J grad /i = A/i = 0, that is, the function /i has to be harmonic on Q.

Exercises for Chapter 8: Oriented Integration 745
(iii) Assume /i to be harmonic on Q. and Q. to be simply connected. Check that a
scalar potential/2 : Q. —> R exists for the vector field J grad/i, and conclude
that the / thus constructed has the desired property.
Exercise 8.23 (Poisson's integral formula and Schwarz' Theorem - sequel to
Exercise 8.22). Let Q. = {z ^ C \\z\ < 1} and define, for every z € ^,
— w + z
^,:a^C by ^,{w) = —
zw + l
(i) Verify that for all z € ^ the mapping 4'. : ^ —> ^ is a C^ diffeomor-
phism with inverse 4'_.; and also that 4'. : 9^ —> 9^. Prove that we have
lim,gj2, .^e'" *j(e''^) = e'", for all a, ^ €] -n, n ].
Now let h € C(9^) be given, and define the Poisson integral r?*/?, : ^ —> R of/z by
iS'h)iz) = :^j\(%ie'^))d^.
(ii) Prove by part (i) and Arzela's Dominated Convergence Theorem 6.12.3 that
lim (^/2)a) = hie'").
zsn, z-^e'"
(iii) For all z € ^, make the (^-dependent) substitution of variables on ] —n, n ]
given by a = a(^), with e'" = ^,(e'^), and show that
T^(«) = ^
Use this to derive the following, known as Poisson's integral formula:
In !., \e" - z?
(iv) Demonstrate that
u/«_H2
ye'" -z)
and conclude by Exercise 8.22 that S'h is a harmonic function on Q..

746 Exercises for Chapter 8: Oriented Integration
Background. On ^ = {x € R^ I ||x|| < 1} a solution / of the Dirichlet problem
A/ = 0, with /|gj2 = h, is obviously given by / = ^h, where (compare with
Exercise 7.70.(v))
1 - \\xf r hiy)
Part (ii) now tells us that (Ph)(x) converges to h(y), for x € Q approaching
y € dQ; this is Schwarz' Theorem (see Exercise 7.70.(vii)).
Exercise 8.24 (Sequel to Exercise 7.47). Demonstrate that the proof in that exercise
comes down to the fact that / is solenoidal on R" \ {0}.
Exercise 8.25 (Sequel to Exercise 5.80 - needed for Exercises 8.26 and 8.29).
Assume that / and g : R^ —> R^ both are C^ vector fields.
(i) Prove, analogously to Grassmann's identity from Exercise 5.26.(ii),
Vx(Vx/) = V{V,/)-{V, V)/.
Conclude that
curl (curl/) = grad(div/) — A/;
here the Laplacian A acts by components on /. Deduce that the component
functions of a harmonic vector field on R^ are harmonic functions.
(ii) Prove from the antisymmetry of the cross product operator with respect to
the inner product, that
(V, /xg) = {Vx/, g)-(Vxg, /),
and conclude that (see Exercise 8.39.(iv) for a different proof)
div(/ X g) = (curl/, g) -{f, cmlg).
(iii) Prove, analogously to Grassmann's identity,
Vx(/xg) =(V,g)/-{V,/)g
= /{V,g) + {g,V)/-g(V,/)-(/V)g,
and conclude, using the formula from Exercise 5.80 for the commutator [•, •]
(for a different proof see Exercise 8.39.(vi)) that
curl(/ X g) = (div g)f + {g, grad)/ - (div f)g-{ /, grad) g
= (divg)/-(div/)g-[/g].
Here the differential operator {g, grad) = giDi-\ h ^3^3 acts by
components on the vector field /.

Exercises for Chapter 8: Oriented Integration 747
(iv) Let ^ C R^ be as in Theorem 7.6.1. By means of part (ii), prove
({cmlf, g)-{f, curlg))(x)dx = {f x g, v)(y)d2y
Jn Jdn
= f if,
Jdn
g X v)(y)d2y.
Write Vo'^(^), with k € Nq, for the Hnear space of the C* vector fields / : ^ ^^ R-^
satisfying supp(/) C ^.
(v) Deduce that curl : Vo'(^) —> Vq(Q) is a self-adjoint linear operator with
respect to the integral inner product on Vq(Q), that is.
Jn
{curl f,g) = (f,curlg):= {f, curlg)(x)dx (/, g € Vq'(^)).
Jn
(vi) Show (see Exercise 8.39.(viii) for another proof)
grad{/, g) = {f, grad )g+ {g, grad) / + / x curl g + gx curl /.
Exercise 8.26 (Maxwell's equations - sequel to Exercise 8.25 - needed for
Exercises 8.27, 8.28, 8.31 and 8.33). In the theory of electromagnetism three time-
dependent C ^ vector fields on R^ play a role:
the electric field E
R^xR^►R^ {x,t)h^ E{x,t)\
R^xR^^R^, {xj)h^ B{xj)\
R3xR^R3, {xj)^j{x,t)-
the magnetic field B
the current (density) j
together with a time-dependent C' function on R^:
the charge (density) p : R^ x R —> R, (x, t) h^ p(x, t).
These entities are mutually correlated by Maxwell's equations, which in addition
contain several constants with physical meaning. If these constants are equated to
1 for simplicity, the equations read, in the absence of matter,
(V,£)=p, Vx£ = -^, {V,5)=0, VxB = j + ^.
6t 6t
Here the divergence (V, •) and the curl Vx are calculated with respect to the
variable in R^.
(i) Prove that / V, y -|- ^ \ = 0, and that this leads to the continuity equation
(V,;) + |^ = 0.
at

748 Exercises for Chapter 8: Oriented Integration
Let ^ C R^ be an open set as in Theorem 7.6.1, and let S C R^ be as in Stokes'
Integral Theorem 8.4.4, that is, S is an oriented surface having the closed curve 9 S
with corresponding orientation as its boundary. Now prove the following assertions,
under the assumption that the conditions of the theorems used are met.
(ii) (Gauss' law). The flux of E across the closed surface 9^ equals the charge
inside Q, that is
/ {E{y,t), d2y)= / p{x,t)dx.
(iii) (Faraday's law). The circulation of E along the closed curve 9 S equals the
negative of the rate of change of the flux of B across the surface S, that is
jjE(s,t),dis) = -—f{B(y,t), day).
(iv) (Absence of magnetic monopoles). The flux of B across the closed surface
dQ. vanishes, that is
Jda
{B{y,t),d2y)=0.
(v) (Ampere-Maxwell law). The circulation of B along the closed curve 9 S
equals the flux of j across the surface S plus the rate of change of the flux of
E across the surface S, that is
jjB(s,t), dis)= f{j(y,t), d2y) + j(
(vi) (Law of conservation of charge). The flux of j across the closed surface 9^
equals the negative of the rate of change of the charge inside Q., that is
/ {jiy.t), d2y) = -— / p{x)dx.
Jan at Jn
The electromagnetic energy F and the Poynting vector field P are defined by,
respectively,
1 1
F = -{E,E) + -{B,B) and P = E x B.

Exercises for Chapter 8: Oriented Integration 749
(vii) (Law of conservation of energy). Prove by Exercise 8.25.(ii)
I F{xj)dx= I {P{y,t),d2y)+ \ {E{x,t), j{x,t))dx,
Jn Jdn Jn
9
"9f
that is, the flux of P across the closed surface 9^ equals the fraction due to
dissipation across 9^ of minus the rate of change of the energy inside Q.
We speak of Maxwell's equations in vacuum if p = 0 and j = 0, that is, if
(V,£) = 0, {V,5) = 0, Vx£ = -—, Vx5 = —.
St St
Assume that E and B both are C^ vector fields,
(viii) Prove by Exercise 8.25.(i) that in this case
V X (V X £•) = -AS, V X (V X 5) = -AB.
(ix) Prove
D £• = 0, 0 5=0, where D := D^ - A.,. := Df - ^ DJ.
That is, both E and B obey the wave equation for a time-dependent vector
field G on R^
nG;(x,O = 0 (1 < z < 3, (x,0 e R^ X R).
Background. This prediction, in 1864/5, on theoretical grounds of the existence of
electromagnetic waves in vacuum is one of the great triumphs of Maxwell's theory.
The existence of radio waves was experimentally verified by Hertz in 1887.
Exercise 8.27 (Sequel to Exercise 8.26). Use Exercise 8.26.(ii) to give another
proof of parts (i) and (ii) from Exercise 7.33.
Hint: Let 9^ in Exercise 8.26.(ii) be a straight circular cylinder with axis
perpendicular to the plane {x € R^ | xi = 0}, in Exercise 7.33.(i), or coinciding with the
X3-axis, in the case of Exercise 7.33.(ii).
Exercise 8.28 (Maxwell's equations: time-independent case - sequel to
Exercises 7.67 and 8.26 - needed for Exercise 8.29 and 8.51). Under the assumption
that the vector fields E and B on R^ are time-independent, one obtains Maxwell's
laws in the following form:
{V,£) = p, Vx£ = 0, {V,5)=0, VxB = j.

750 Exercises for Chapter 8: Oriented Integration
In this case E is curl-free and B is divergence-free on R^. Therefore, one may try
to find solutions E and B of the form
E = -V0, B = S/ X A,
with the function 0 : R^ —> R a scalar potential for E, and the vector field
i4 : R^ —> R^ a vector potential for B. (In physics, the minus sign for V0 is
customary.) To limit the analytical complications we assume that p € C^(R^) and
j € C2(R3, R3).
(i) Verify that 0 has to satisfy Poisson 's equation
(•) A()!) = -p,
while quite obviously V x (V0) = 0.
(ii) Demonstrate that A has to satisfy Vx(VXi4) = y, while naturally we have
{V, V x/l)=0.
(iii) Use Exercise 8.25.(i) to prove that, in addition, under the Coulomb gauge
condition
{V,A)=0,
the vector potential A has to satisfy Poisson's equation by components
(••) AA = -j.
We now inquire about solutions 0 of (•) and A of (••) that satisfy the following
boundary condition at infinity:
lim <p(x)=0, lim A(x) = 0.
||a'||—>-oo pll—>-oo
(iv) Apply Exercise 7.67.(vii) and conclude (note that, contrary to our usual
conventions for potentials, the minus sign is missing; this is because in electro-
magnetism the forces between like charges are repulsive)
'^w = 7-/' ir^^^' (^^»')-
4:t Js? \\x-x'\\
Verify that now the electric field E is described by Coulomb's law
E(x) = — f ^^^^ (x - x')dx' {x € R^).

Exercises for Chapter 8: Oriented Integration 751
(v) Prove, in similar fashion as in (iv),
Aix) = -/-/" -^-—jix')dx' (x € R^).
Verify that now the magnetic field B is described by the Biot-Savart law
B{x) = ^ f ^---j {x') X (X - x') dx' {X € R^).
(vi) Verify that the Coulomb gauge condition {V, A ) = 0 is met.
Hint: Use Corollary 7.6.2 and the continuity equation (V, y ) = 0 (see
Exercise 8.26.(i)).
Exercise 8.29 (Helmholtz-Weyl decomposition - sequel to Exercises 8.25 and
8.28 - needed for Exercise 8.30). Let A^ be the Newton vector field on R" from
Example 7.8.4, and let * be the convolution from Example 6.11.5. Demonstrate
that the results from Exercise 8.28.(iv) and (v) can be generalized as follows.
(i) A C^ vector field / on R" with Af = 0 is uniquely determined by div /, if
this function has compact support on R", via
/ = (div/)*Af.
Here the integration is carried out by components of A^.
(ii) A C^ vector field / on R" with div / = 0 is uniquely determined by Af, if
this vector field has compact support on R", via
/ = {Af) * ;V;
in more explicit notation, for 1 < z < w and x € R",
fi{x)= V I {Djf-Difj){x')Nj{x-x')dx'.
(iii) Let / be a C^ vector field on R^ with compact support. Verify there exist a
C' function g : R^ —> R and a C' vector field ft : R^ —> R^ such that we
have the following Helmholtz-Weyl decomposition:
Prove that
g and h
8i^)
I,/'v-\
/ =
= grad g
are solutions if
= -/■
1 r
\\x-
1
+ curl/2.
ix-x'))
-x'lp
j:/../\ ,,
dx',

752 Exercises for Chapter 8: Oriented Integration
Deduce / = grad g + fi with /2 divergence-free on R^.
Hint: Write / = A(/ * p), where the actions of convolution and A are
according to components of / and p(x) = —-^-a^,, for x € R^ \ {0}, and
use the identity from Exercise 8.25.(i).
Exercise 8.30 (Hodge decomposition - sequel to Exercise 8.29). We want to
determine conditions for the uniqueness of the summands /i and /2 occurring in
the decomposition / = /, + /2 = grad g + f2 from Exercise 8.29.(iii)- And we
would like to generalize this decomposition to R". Therefore we consider a set
U C R" and vector fields f\ and /z : t/ —> R" satisfying the conditions from
Theorem 7.6.1. Moreover, we assume that /i possesses a potential gonU, that /2
is divergence-free on U, and that /2 is parallel to 9t/, which means (/2, v){y) =0
for y € ^U, with v as in Theorem 7.6.1.
(i) Prove div(g/2) = {/i, /2) and use Gauss' Divergence Theorem 7.8.5 to
conclude that
f {fuf2)(x)dx=0.
Ju
(ii) Now assume /i and /2 satisfy the same conditions as /i and /2, respectively,
and /i -I- /2 = fi+ fi- Prove f^ \\ (/i - fi)(x)f dx = 0, and deduce that
/i = 7i and /2 = J2 on U.
As to the existence of /i and /2, we note that / = grad g + /2, with /2 divergence-
free on U and parallel to 9t/, implies div / = divgrad g = Ag on U as well as
{f,v) = {grad g,v) = ^ on dU. Given a C^ vector field / on U it is therefore
sufficient to determine a C^ function g onU with
(*) Ag = div/ on U, ^ = {f,v) on dU,
dv
where we also need that ^ is well-defined on dU. Indeed, f\ = gradg and
f2 = f — grad g then form a solution. A partial differential equation, together with
a boundary condition
Ag = p on U, -— = q on dU,
av
for given functions p and q, is said to be a Neumann problem on U.
(iii) Using Green's first identity, verify that the following condition is necessary
for the solvability of the Neumann problem:
/ pix)dx = / qiy)d„_iy.
Ju Jdu

Exercises for Chapter 8: Oriented Integration 753
(iv) Verify that the condition from part (iii) is satisfied in our problem (•).
We state without proof that, for sufficiently well-behaved dU, the condition from
part (iii) is also sufficient for the solution of the Neumann problem.
(v) Assume the vector field f on U satisfies the integrability conditions. Prove
that /2 is a harmonic vector field on U, and that we have the direct sum
decomposition
/ = grad g ® f2 with /2 harmonic on U and parallel to dU.
Assume « = 3. Let m = bi/ be the differential 1-form on U associated with /
according to Example 8.8.2, and similarly mi with /i, and C02 with /2.
(vi) Show Ml to be exact. Assume m is closed. Then the cohomology class of
0) in the first de Rham cohomology H^(U) has a harmonic representative,
namely C02, satisfying
[m] = [0)2] €H\U), dM2 = 0, d*M2 = 0.
Here the Hodge operator * : Q^(U) —> Q^(U) is defined by *bi = b2,
furthermore * : Q^iU) -^ Q°iU) by *(/dx) = f, and finally d* = *rf* :
Q^(U) —> Q°(U). (A more intrinsic definition is possible but is not discussed
here for lack of space.)
Background. The sum decomposition of the closed form m = mi + a>2 into an
exact form a>i and a harmonic form (i>2 is called a Hodge decomposition^ of m. It is
used to investigate under what conditions on U the de Rham cohomology H'^{U)
is a finite-dimensional vector space, for ^ € Nq.
Exercise 8.31 (Maxwell's equations in covariant form - sequel to Exercise 8.26).
We employ the notation, and make the assumptions, from Exercise 8.26. Note
that £(•, t) and B{-, t) both are C^ vector fields on R^ dependent on a parameter
f € R. In this exercise, the operators b, curl, div, grad, A and the differential form
dx = dx\ /\ dx2 A dxj, are associated with R^. The operators d and Dq := ^ are
associated with R''. Define, for (x, t) € R'',
Six,t) = b,(£(.,0)(^) e ^'(R^), Six,t) = b2iBi;t)Xx) € ^2(R'*).
(i) Taking the indices / modulo 3, verify
6= 2_, Eidxi, S= 2_. Bi dxi+\ Adxi+2-
i<;<3 i<;<3
See for more details in the case of n = 3: Cantarella, J., DeTurck, D., Gluck, H.: Vector calculus
aiid the topology of domains in 3-space. Amer. Math. Monthly 109 (2002), 409 - 442.

754 Exercises for Chapter 8: Oriented Integration
Introduce the Faraday form
r = S Adt + £ eQ^iR'^).
(ii) Demonstrate, using Formula (8.54),
dr = dSAdt+dS = \?2(cmlE + DoB)Adt + (divB)dx € ^^(R''). (1)
Define the Hodge operator * € Lin (^^(R''), ^^CR"*)) by, for 1 < z < 3,
*(rfX/_j.i A dXi_f_2) = dXj A dt, *(dXi A dt) = —dXi_f_i A dXi_^2-
(More intrinsic definitions are possible but are not discussed here for lack of space.)
Further, introduce
(iii) Verify
*r = JiAdt-3)€ Q.'^iR'^). (2)
Let
$ = -pdt + \>J eQ'iR").
Introduce the Hodge operator * € Lin (^' (R"*), ^^(R'')) by
*(dXi) = dXi+i A dXi+2 A dt, *(dt) = dx.
Then * is a bijection since it takes a basis into a basis. Therefore, define * €
Lin (^^(R''), ^' (R"*)) as the inverse of the mapping just defined.
(iv) Now prove
d(*3=^) =dJi Adt -dS) = b2(curl B - DqB) Adt - (div E) dx
(3)
= *g€ ^3(R^).
Thus, using (1) and (3), one may formulate Maxwell's equations as the following
system of equations for the Faraday form on R"*:
dr = 0 and d^r = g,
where
^* = *d* : Q.'^iR'^) -^ ^'(R").
The Hodge operators can be shown to be independent of the choice of a basis in R'',
but they do depend on the choice of an orientation. Consequently, the formulation
of Maxwell's equations given above is independent of the choice of coordinates in
R''. In physics an equation is said to be covariant if its form is independent of the
choice of coordinates used to write the equation.

Exercises for Chapter 8: Oriented Integration 755
(v) Prove that, in terms of the exterior derivative ds associated with R^, Maxwell's
equations take the form of identities between differential forms on R^ with
additional dependence on a parameter in R, as follows:
d3S> = pdx, dsS+DoS = 0, rfaS = 0, d^^M - DqS:) = \>2J.
(vi) Show that the integral theorems from Exercise 8.26.(ii) - (v) immediately
follow, by application of Stokes' Theorem 8.6.10.
Since 3^ € ^^(R'') is a closed C' differential form, it follows from Poincare's
Lemma 8.10.2 that there exists a C^ differential form %, € ^^(R'') with 5" = d%,.
(vii) Demonstrate the existence of C^ potentials 0 : R'' —> R and A : R"* —> R^
with
% = -(j)dt + \)iA€Q.\^*).
The equation 3^ = d% now leads to expressions for E and B in terms of 0
and A
E = -V(p - DqA, B = ^ X a. (4)
Use (2) and (4) to show that the equation d*3^ = $ is equivalent to
d ^ ((A+i^i+2 - A+2^/+i) dxi A dt + (A-0 + DoAi) dxj+i A dxi+2)
= *$■
(5)
Note that ^ is not completely determined by $, and that, consequently, (5) does not
completely determine 0 and A; it follows that we may impose another condition.
(viii) Try to find ^ such that the Lorentz gauge condition d*% = 0 is satisfied.
Then verify that
( V, A ) + Do0 = 0.
Demonstrate that under this assumption (5) is equivalent to the following
equations for 0 and A, for given p and j, respectively:
D 0 = p, \JA = j, where D = D^ - A (6)
is the wave operator or D'Alembertian. In general, % + df will satisfy the
Lorentz gauge condition if D / = 0.
In Exercise 8.34 we prove that solutions of (6) are given by the retarded potentials,
for(x,f) eR^'withf > 0,
^, . 1 /• p{x\t-\\x-x'\\) ^ ,
(f)(x,t) = — dx ,
^^ J\\x-x'\\<r \\X-X'\\
1 r J(x',t-\\X-X'\\) ,
Aix,t) = — -^' „ ,„ "'dx\
^^ J\\x-x'\\<t \\X-X'\\

756 Exercises for Chapter 8: Oriented Integration
Exercise 8.32 (Invariance of wave operator under Lorentz transformation and
special relativity - sequel to Exercise 2.39 - needed for Exercises 5.70 and 5.71).
Let 7„_j.i in Mat(« + 1, R) be the diagonal matrix having —1, 1, ..., 1 on the main
diagonal. The mapping
0',y)^ly,y]= y'Jn+J■■ R"+' x R"+^ ^ R
is a nondegenerate symmetric bilinear form on R""*"'. A Lorentz transformation of
R"+^ is a linear mapping L € End(R""'"^) leaving this form invariant, that is, one that
satisfies fLy, LJ] = ly,'y'\, for all }', y € R"+^ The Lorentz group lM(n + 1,R)
consists of all Lorentz transformations of R"+'.
(i) Consider (t, x) € R x R" ~ R"+'. Prove
\it,x), (T,x)l =t7- (x,x), (t, TeR, x,x € R").
Show L € Lo(« + 1, R) if and only if L'J„+iL = Jn+i- From this deduce
detL = ±1 for L € Lo(w + 1, R), furthermore that Lo(w + 1, R) indeed
satisfies the axioms of a group, and also that L € Lo(w + 1, R) if and only if
L' € Lo(« + 1, R). Let S„+i be a diagonal matrix with Sf,_^i = Jn+i- Then
L € Lo(w + 1, R) if and only if S„+iLS^^i is an orthogonal linear mapping
(with complex coefficients).
(ii) By means of part (i) and Exercise 2.39 verify that the wave operator or
D'Alembertian
n = Df- A, = Df- Y^ Dj
in R"+^ is invariant under Lorentz transformation, that is
D (/ o L) = (D/) oL (/ € C\R"+'), L € Lo(n + 1, R)).
Background. The invariance under Lorentz transformations of the wave operator,
and also of Maxwell's equations, played a role in the development of the theory of
special relativity in physics.
(iii) Assume n = 1. Then L € Lo(2, R), detL = 1 and trL > 0 if and only if
there exists a number f € R with
cosh f sinh f
/ cosnf smnf \
^ V sinhf coshf /
Verify that L^ o L^i = L^+^i and thus LZ = L_f, for all f and f' € R. The
mapping L^ is said to be the hyperbolic screw or boost in R^ with rapidity f.
Hint: If L(l, 0) = (t, x) then l(t, x), (t, x)l = t'^ - x'^ = 1, and therefore
there exists a unique number f € R with t = cosh f and x = sinh f. For
the computation of L(0,1) = (T,x) use l(t,x), (T,x)] = 0 and detL =
tx — tx = 1.

Exercises for Chapter 8: Oriented Integration 757
Next we define — 1 < t; < 1 by
tanhf = V,
then
1 1 + t; 1
f = o l°g 1 ' ^°^^ f = /. ='■ V' sinh ^ = yv.
In physics f is called the rapidity of the velocity v. Addition of rapidities
corresponds to the following relativistic law for addition of velocities:
V\ + V2
V := tanh(f 1 + ^2) = "r- (vi = tanhf,).
1 + t;it;2
(iv) Verify, if (F, x) = L_^(t, x) for t and x € R, and
/ 1 -t; \ , 7=yit-xv),
L_f = yf ), that
(v) Next we generalize to R"+^ the Lorentz transformations having the form from
part (iv). Let v € S""' = {x € R" | ||x|| = 1} and -1 < t;o < 1 be arbitrary
and write y = (I — Vq)^2, Prove that L € Lo(« + 1, R) if
/ t \ ( yit-vo{x,v)) \
Li ) = { (f€R, x€R").
\x/ \ X+ {(y - \){x,v) - yvQt)v /
Hint: Direct computation, or else proceed as follows. Write x = X[[+ x± for
the decomposition in R" of x in components parallel and perpendicular to v.
Application of part (iv) to the linear subspace in R"+^ ~ R x R" spanned by
eo and v then gives
T=y(t - vq{xh,v)), X|| = )/(x|| - t;of t;), xj_ = x_l.
Since xj = (x, t;) t; and xj_ = x — (x, t;) t; we obtain
X =x±+X[[ = x± + y(x[\ — VQt v) = X — {x,v) V + y({x, v) V — vot v).
(vi) Let t; € S""^ be arbitrary and let 0„ denote 0 € Mat(w, R). Put

758 Exercises for Chapter 8: Oriented Integration
Prove V'Jn+i + Jn+iV = 0. Note that vv' € Mat(«,R), and show by
induction
y2j
=(i„!')' '"'='' '^■"'"-
Demonstrate for f € R (see Example 2.4.10 for the definition of exp)
El / coshf sinhf v'
-,i^ vy =
j^f^^ y! V sinh f t; /„ + (-1 + cosh f) t;i
=: B^y.
Thus, for f € R and X € R",
fcoshf +(x, t;)sinhf
(t \ / fcoshf +(x, t;)smhf \
=
X / \ t sinh ^ V + {x, v) cosh ^ v + x — {x,v) v /
B^y is said to be the hyperbolic screw or boost in the direction v € S"^^ with
rapidity f. See Exercise 5.70.(xii) and (xiii) for another characterization
of a boost. Verify that {x, v) v, the component of x along v, undergoes a
hyperbolic screw in the plane spanned by e^ and v with rapidity f, and that
x—{x,v)v, the component of x perpendicular to v, remains unchanged under
the action of B^^. Now define —1 < vq < 1, the speed of the boost, by
1
tanhf = Vq, then coshf = ^ =: y, sinhf = y vq.
Verify that B^^y equals the Lorentz transformation L given in (v).
(vii) Consider (T,x) = L(t,x) as in (v). Prove that elimination of yt from the
expression for x in (v) gives
X = Jl — Vq{x, v)v +x — {x, v}v — Votv.
Deduce that for two different points with coordinates x and }', and x and y,
respectively,
||x - yil = (1 - vl) II (X - y)^^f + \\ix- y)^f.
For a stationary observer objects that move with velocity vqv € R^ contract by
a factor (1 — v^)^ along the direction of motion while there is no contraction
perpendicular to the direction of motion; this is the FitzGerald-Lorentz contraction
of space.

Exercises for Chapter 8: Oriented Integration 759
(viii) Now assume v = ei € R^. Then similarly elimination of yxi from the
expression for t in (v) gives
t = ^l - V^t - VqXi.
Deduce that for two different moments with coordinates t and u, and t and
u, respectively.
t — u = JI — Vgit — u).
For a stationary observer clocks that move with velocity hqi; € R^ run slower by a
factor (1 — iio)2; this is the dilatation of time. In particular, long journeys across
cosmic distances would be instantaneous for an observer traveling with the speed
1 of light (in our usual normalization).
Exercise 8.33 (Wave equation in three space variables - sequel to Exercises 3.22,
7.53, 8.26 - needed for Exercise 8.34). Consider the wave equation, which we
encountered in Maxwell's theory, in particular in Exercise 8.26.(ix)
(•) — Dfu(x, t) = Ajcu(x, 0 = X! ^7"(-^' 0 (c > 0),
for a C^ function m : R^ x R —> R, with (x, t) i-> u(x, t). We want to solve the
initial value problem for this equation, that is, we look for solutions u of (•) which
in addition satisfy the following initial conditions, for t = 0:
(••) u(x,0) = fix), D,u(x,0) = gix) (xeR^),
for given functions / € C^(R^) and g € C^(R^).
Form the spherical means with respect to the space variable, as in Exercise 7.53,
for the functions u, f and g, and write the resulting functions as
m„ : R^ X R X R —> R and nif, mg : R^ x R —> R, respectively.
In particular,
iu(x,r,t) =-— I u(x+ry, t)d2y.
47r y||_,,||=i
m,;
(i) Prove
(ii) Show
m„ix, 0, t) = u(x, t), m„(x, r, 0) = mf(x, r),
Dtm„(x, r, 0) = mg{x, r).
1 ,
— D,m„{x, r, t) = Aj^m^ix, r, t).

760 Exercises for Chapter 8: Oriented Integration
(iii) Prove by means of Exercise 7.53 that, for x € R^ fixed, the function (r, t) i->
m,j(x, r, t) satisfies the following partial differential equation:
1 2
— Dfm„ix, r, t) = Dlmi,{x, r,t) + - Z),.m„(x, r, t).
c^ r
Conclude that {r, t) i-> rmi,{x, r, t) satisfies the wave equation in one space
variable
1
— Df {rm^,{x, r, t)) = D^ (rm„ix, r, t)),
rm^iix, r, 0) = rnifix, r), D, {rm„ix, r, 0)) = rnigix, r).
(iv) Prove, by means of Exercise 3.22.(iii),
rmi,{x, r, t) = -{(r + ct)mf(x, r + ct) + (r — ct)mf(x, r — ct))
■[ pr+ct
+— / smg(x, s)ds.
^C J!•_
igy
-ct
The definitions of mf(x, r) and mg(x, r) show that the functions r i-> mf(x, r)
and r i-> mg(x, r) are well-defined on all of R, and are even functions.
(v) On the basis of the foregoing observation, prove that
(ct + r)mf(x, ct + r) — (ct — r)mf(x, ct — r)
m„(x,r,t) =
2r
"cr+r
1 r"+'
-\ / sm„(x,s)ds.
2cr j„_.
Hint: J"^',,^) smg(x, s) ds = 0.
(vi) In the formula in (v), take the limit for r —> 0, and prove by (i)
u(x,t) = Dp\^^^^^{pmf(x,p))-\-tmg(x,ct)
= D,{tmf(x, ct))-\-tmg(x, ct).
That is, u is given by the following, known as Kirchhoff's formula:
(***) u(x,t)
= :i~ ifi^ + cty) + r g(x + cty) + tD,f(x + cty)) d2y.
47r y 11^, 11=1

Exercises for Chapter 8: Oriented Integration 761
(vii) Conclude that formula (• • •) gives the unique solution of the initial value
problem (•) and (••).
In the following we shall assume that there exists a bounded set AT C R^ such that
supp(/) C K, supp(g) C K.
(viii) Let t > 0 and u(x, t) ^ 0. Prove that then there exists az € K such that x
lies on the sphere in R^ of center z and radius ct. Thus, in particular, there
exists, for all f > 0, an open ball 5, in R^ about the origin and off-dependent
radius such that
X ^ Bf =^ u(-, t) = 0 in a neighborhood of x.
Note that according to formula (•••) the solution u may be one order less differen-
tiable than the initial / and g. This is a "focusing effect": irregularities from various
places in the initial data are focused, thus leading to caustics, that is, (smaller) sets
of stronger irregularity. Nevertheless the solution u is "on average well-behaved",
as becomes evident from the following. Define the energy E(t) of the solution u at
time t by
Eit) = - lU-Dru) +||grad,M||M(x, t)dx,
with grady the gradient with respect to the variable x € R^.
(ix) Prove that £ is a conserved quantity, that is, t i-> E(t) is a constant function.
Hint: One has
-j^iO = (^ iD,u) (Dfu) + {grad, u, grad,(Am) ))(x, t) dx
= I DfU (— Dfu — AxUj(x,t)dx
+ f -^iy,t)DrUiy,t)d2y = 0,
by Green's first identity from Example 7.9.6, and part (viii).
Exercise 834 (Inhomogeneous wave equation - sequel to Exercises 2.74 and
8.33 - needed for Exercise 8.35). We want to find a C^ function m : R'' —> R
satisfying, for a given function g € C^(R'^), the inhomogeneous wave equation
(•) D M = g.

762 Exercises for Chapter 8: Oriented Integration
Let T € R be chosen arbitrarily, and let (x, t) i-> v(x, t\ t) be a solution of the
initial value problem
D11 = 0, v{x, t) = 0, Dqv{x, t) := —{x, t) = g{x, t) (x € R ).
dt
On account of Kirchhoff's formula from Exercise 8.33.(vi), this is satisfied by
v{x, t; t) = ——^ / g{x + (f - T)y; t) rfz}' ((•^. 0 € R^ x R).
47r y||^,||=i
Now define, assuming convergence,
(•) m(x, t) = I v(x, t; t) rfr ((x, t) € R^ x R).
Jo
Then u(x,0) = 0, and we find, by means of Exercise 2.74,
(••) Dou(x,t) = v(x,t;t)+ I Dov(x,t;T)dT= j Dov(x,t;T)dT,
Jo Jo
because v(x,t;t) = Om view of the initial condition on v. Hence, Dou(x, 0) = 0.
Furthermore, differentiation of (••) gives
Dlu(x,t) = DQv(x,t;t)+ D^v(x,t;T)dT=g(x,t)+ Dlv{x,t\T)dT.
Jo Jo
And, from (•),
Au{x,t)= / Av{x,t;T)dT= I D^v{x,t;T)dT.
Jo Jo
Upon subtracting these results we obtain
D u{x, t) = g{x, t), u(x, 0) = 0, Dou(x, 0) = 0.
Therefore
uix,t) =— (t-r) g{x + (t - T)y; t)d2y dr
^^ Jo J\\y\\=i
^^ Jo J\\y\\=\
Substitution of y = 4'(x') = -(x' — x) leads to the retarded potential from
Exercise 8.31
1 r' 1
gix',t-sgn(t)\\x-x'\
u{x,t) =— - g(x';t-T)d2x'd
4-7T Jo T y||.v'_j;|| = |T|
-/■
4:t yii,_w||<|r| l|Ji:-Ji:'ll
dx'.

Exercises for Chapter 8: Oriented Integration
763
Background. The value of the potential u at the point (x, t) € R'' with t > 0
is exclusively determined by the charge density g at points (x', t') € R'' with
t > \\x — x'W = t — t' >_ 0. These points {x', t') lie on that part of a "rearward"
conical surface in R'' which has apex {x, t) and lies in the "positive" half-space in
R''. The method used here to solve an inhomogeneous equation is a special case of
Duhamel's principle.
Exercise 8.35 (Fundamental solution - sequel to Exercises 6.49, 6.68, 6.92,
6.105, 7.67 and 8.34). Let / € C^(R") be given. Prove that the inhomogeneous
partial differential equation P(D)m = / on R" has a solution m = /*£ € C^(R"),
where * denotes convolution and E € C°°(R" \ {0}) satisfies the homogeneous
partial differential equation P(D)E = 0 on R" \ {0}, in the following cases:
variable
PiD)
E{x)
Exer.
(i)
(ii)
(iii)
(iv)
(V)
(vi)
X € R
X € R2
X € R2
x € R" (« 7^ 2)
x€R^
(x, t) € R" X R
D' {s > 0)
\{D^+iD2)
T{s)
1 1
71 X\ + iX2
^loglkll
A + iJ,^
i-'^-^
(2-«)|S"-M ||x||"-2
1 e±'>ll.v|l
4n \\x\\
1
{AnktYl"^
0,
e «' {t > 0)
(r<0)
6.105
6.49
7.67
7.67
6.68
6.92
Such a solution E is said to be a fundamental solution for the partial
differential operator P{D). The operator in (i) is fractional differentiation; in (ii) it is
the Cauchy-Riemann operator; in (iii) and (iv) the Laplace operator; in (v) the
Helmholtz operator (see the technique of the Exercise 7.67); and in (vi) the heat
operator.
Background. In the case of the wave operator D} — A^ from Exercise 8.34 the
situation is more complicated; it turns out that E never is a differentiable function
on R"+^ \ {0}. For instance, for « = 1 a solution E is given by the function with
the constant value | on the forward cone {(x, f) € R^ | |x| < t}, and the value 0
elsewhere. And E even fails to be a function for larger values of «; nevertheless
it always is a distribution, a generalization of the notion of a function. For that
reason the retarded potential from Exercise 8.34 where n = 3 can not immediately
be recognized as a convolution product.

764 Exercises for Chapter 8: Oriented Integration
Exercise 8.36 (Brouwer's Fixed-point Theorem - sequel to Exercise 6.103). The
assertion from Example 8.8.3 holds for an arbitrary continuous mapping / : t/ —>
R" instead of a C^ mapping /. We now prove this, leaving it to the reader to fill in
the details.
Suppose f(x) ^ X, for all x € B". The continuous function x i-> ||/(x) — x||
then reaches a minimum of value 4m > 0 on B". By application of Weierstrass'
Approximation Theorem from Exercise 6.103 by components, for example, / can
be approximated by means of a polynomial function p : 5" —> R" such that in the
uniform norm || ■ || on B"
||/-p|| <m.
This gives ||p|| < I +m. We therefore have p := j^^p '■ B" —> B", while
(1 \ nt
1 - -T—-)\\P\\ <m + ——(1 +m) = 2m.
1 +m/ I +m
Consequently, for all X € B",
IIpW-^11 = ll/W-x-{f-p){x)\\ > ||/(x)-x|| - 11/-pii
> Am — 2m = 2m > 0.
By Example 8.8.3, the polynomial function p does have a fixed point x € B", and
this implies a contradiction.
Exercise 8.37 (Sequel to Exercise 6.23). Use Exercise 6.23.(i) in order to show
that Brouwer's Fixed-point Theorem is false for open balls.
Exercise 8.38. Prove that Formula (8.49) can be written as
= — X! ^g"^^') ^^ ° °')('^i' ■ • •. »^/t) (?? o cr)ivk+i, ..., Vk+i).
Hint: Note that in this formula {o-(l),... ,a(k)} and {a(k + I),... ,a(k+ 1)}
are not ordered.
Exercise 839 (Vector analysis in R^). We derive the formulae in Exercise 8.25.(ii)
and (iii) using differential forms. If t; is a vector field on R^, let bi t; € ^' (R^) and
b2ii € ^^(R^), be the corresponding 1-form and 2-form, respectively, as in
Example 8.8.2. Further, denote by iy the contraction with the vector field v as in
Formula (8.57), and by L„ the Lie derivative in the direction of v as in Formula (8.55).
(i) Recall that iydx = \)2V, and deduce from Formula (5.29) that Lydx =
divvdx € ^^(R^).

Exercises for Chapter 8: Oriented Integration 765
(ii) Prove, for vector fields vi and 112 on R^,
Further, show
bi(t;i xt;2) = -ivi^ivi = iv2^2Vi, biHi Abit;2 = \>2iviy.V2) = iv,xv2dx.
(iii) Suppose 113 is a vector field on R^. Compute «„, (bit;2 A bi 113) € ^' (R^) using
(ii) and the antiderivation property of z„,, and deduce Grassmann's formula
from Exercise 5.26.(ii).
(iv) Compute dQ>iVi A bit;2) € ^^(R^) by means of (ii) and the results in
Formula (8.54), and deduce the identity in Exercise 8.25.(ii).
(v) Prove [L„,, iy^] = i[vi,v2]- To this end, note that [L„,, iy^] is an antiderivation
that takes ^-forms to (^ — 1 )-forms and that vanishes on Q° (R^). It is sufficient
therefore to establish the identity on ^^(R^).
(vi) Apply the homotopy formula to b2(curl(t;i x 112)) = d{bi(vi x 112)) =
divi^i^i- Then, using part (i), note that L„,b2iii = L^iy^dx and apply part
(v). Finally, deduce the identity in Exercise 8.25.(iii).
(vii) In the same way as above show curl(/t;) = /curl v + (grad /) x v, for
/€C'(R3).
(viii) Prove the identity in Exercise 8.25.(vi). Todoso, start with bi (grad (ni, 112)) =
(doiy^)[)lV2 and apply the homotopy formula. Further, useL„,/ = {Df)v\,
for / € C'(R3), and [^, L„,] = 0.
Exercise 8.40 (Divergence in arbitrary coordinates - sequel to Exercise 3.14).
Using differential forms we give two different proofs of the following formula (•)
from Exercise 3.14:
(*) (div f) o vl/ = y Diif'-'^ det Dvl/).
l<i<n
Here U and V are open subsets of R", while / : t/ —> R" is a C^ vector field and
<if :V ^ UissiC^ diffeomorphism, and / o 4/ = I^,^,-^,, f^''> Dj^ : V -^ R".
(i) Consider b„_i/ € Q"^^(U), and derive from Example 8.8.2 the following
equality of differential forms in ^""^(V):
vl'*(b„_,/) = Y^ (-ly-'f^detD^dyr;
1 <i <n

766 Exercises for Chapter 8: Oriented Integration
deduce
^(vl'*(b„_i/)) = J2 A(/'^detDvI/)rfy€^"(V).
l<i<n
Prove, for g € C(U),
(••) ^*(gdx) = g o 4' det D4' (iy
and verify, as in Example 8.8.2,
4'*(^(b„_i/)) = (div /) o ^ det D^ dy.
Using Theorem 8.6.12 deduce Formula (•).
(ii) Using the homotopy formula from Lemma 8.9.1 deduce that div f dx =
doif dx € Q" (U). Prove, by applying 4'*, Formula (••) and Theorem 8.6.12,
(div /) o 4/ det D^/ dy = ^(4'*(i/ dx)) = d(i^*f^*dx)
= d(detD^i^*fdy).
Here, in the notation of Exercises.14, we have the vectorfield 4'*/ : V —> R"
satisfying (vI^VX}') = Z)vI/0,)-'(/o vl/)(y) = i:i,,,„/«0')^/. Deduce
Formula (•) using
rf(det D^/ i^*f dy) = ^ A (f^ det D^) dy.
l<i<n
Exercise 8.41 (Lie derivative of vector field and differential form - sequel to
Exercise 3.14 - needed for Exercises 8.42,8.43 and 8.46). Let U be open in R".
Suppose X is the vector field on U satisfying X = ^|,_o'I'', for a one-parameter
group of C ^ diffeomorphisms (^')tsR- If <w is a C ^ differential form in Q''(U) and
Xi, ..., Xjt are C^ vector fields on t/, then g := a)(Xi,..., X^) belongs to C^(U).
(i) Verify that Definition 8.6.7 of the pullback (<!>')* acting on differential forms
gives, for X € U,
i^rgix) = aiiXu...,Xk)(^'ix))
= a)i^'ix)){D^'ix)D^'ix)-^Xii^'ix)),..., D^'(x)D^'(x)-^Xki^'(x)))
= {i^roj)ix){i^rxiix),..., i^rx.ix)).
Here we used that, in view of the definition of pullback of a vector field from
Exercise 3.14,
D^'ix)-\X! o ct>')(x) = i^yXiix) (1 < i < k).

Exercises for Chapter 8: Oriented Integration 767
Note that Lxg = Xg := (Dg)X. Now define, for a vector field Y on U,
d
LxY =
dt
(ii) Using Proposition 2.7.6 and the definition of Lie derivative of a differential
form from Formula (8.55), deduce from part (i) the following derivation
property for the Lie derivative Lx'.
XicoiXi,..., Xk)) = (LxcoXXu ...,Xk)
+ J2 o)iXi,...,LxXi,...,Xk).
Next we study LxY, for vector fields X and Y on U. The proper framework for
studying vector fields is that of derivations; it is in this context that one obtains the
correct functorial properties.
(iii) Prove, on account of the chain rule, for any / € C^(U),
(i^'rY)if)ix) = D/(x)((<I.')*F)(x)
= Dfix)D^-'i^'ix))iY o ct>')(x) = Dif o ct>-')((t>'(x))F(<I>'(x))
= i^'nY(i^-'rf))ix).
{i^'TY)/ = ^
at
i^'r{Y{i^-'rf))
t=0
Deduce
d
(LxY)f =-
dt ,^0
= X(Yif))-Y(Xif)),
which implies
LxY = XY-YX = [X, Yl (Lx^)(X,,..., X^)
= XicoiXi,..., Xk)) + i:,<,<, ^(Xi,..., [X,-, X],..., X,).
Exercise 8.42 (Homotopy formula and exterior derivative - sequel to
Exercise 8.41 - needed for Exercise 8.43). Let U be open in R", let co € ^*(t/) be a
C^ form, and let Xi, ..., X^+i be C^ vector fields on U. The homotopy formula
from Lemma 8.9.1 implies
ix.do) = Lx.ct) — dix,M,

768 Exercises for Chapter 8: Oriented Integration
while the derivation property for Lx, from Exercise 8.41 yields
— 2_^ (^(Xi,■ ■ -AXi,Xj],...,Xk+i).
2<j<k+l
By combination of these two formulae derive
dMXi,..., Xk+i) = XiicoiX2,..., Xk+i))
+ J2 i-'^y^'(^i[Xi,XjiX2,...,Tj,...,Xk+i)
l<j<k+l
-diixiO))iX2,...,Xk+i).
More generally, one can tackle the last term, which involves ixiO) € ^*"' (U), by
the same method. Verify the following formula by mathematical induction over
Jfe€No:
d(o(Xr,...,X,+0= Yl (-^y^'X,i(o(Xi,...,Xi,...,X,+r))
\<i<k+i
+ Yl i-iy^'o)aXi,xjixi,...,Xi,...Xj,...,Xk+i).
l<i<j<k+l
Background. In algebraic contexts, for instance in Lie algebra cohomology, the
formula above is often adopted as the definition of the exterior derivative d.
Furthermore, the result from Proposition 8.6.11 is a direct consequence. However, a
direct proof of (i^ = 0 (compare with Theorem 8.7.2) on the basis of this definition
is tedious and not illuminating; therefore we give a different argument in
Exercise 8.43 under a mildly restrictive extra condition. (Using some more theory, one
may get rid of this restriction.)
Exercise 8.43 (Proof by algebra ofd^ = 0- sequel to Exercises 8.41 and 8.42).
Let the notation be as in Exercise 8.42. Furthermore, let <I> : t/ —> t/ be a
diffeomorphism and let <I>* be the corresponding pushforward of vector fields on U
as defined in Exercise 3.15. For a vector field X on U and / € C^(U) we define
Xf € CiU) by Xf = {Df)X.
(i) Verify <I>*((<I>*X)/) = X(<I>*/), and conclude that
(<I>,X)/ = (<I>-')*(X(<I>*/)).
(ii) Let F be a vector field on U. Deduce from part (i) that
<I>,[X,F] = [<I>,X, cD.F].

Exercises for Chapter 8: Oriented Integration 769
(iii) Using Exercise 8.41.(i) prove, for a C^ differential form m € Q^(U) and C'
vector fields Xi, ..., Xj. on U,
(cD^ojXX,, ..., X,) = ^^oii^^Xi,..., ^A)).
(iv) Successively apply part (iii). Exercise 8.42, and parts (i) and (ii) to obtain
^*(dM) = rf(<t>*ftj), in other words [<!>*, d] = 0.
Let X be as in Exercise 8.41 but otherwise arbitrary and deduce [Lx, d] = 0.
(v) Prove the homotopy formula LxO) = d(ixO)) + ix{doi) on the basis of the
formula for LxOi from Exercise 8.41.(iii), and for da> from Exercise 8.42,
respectively. Next conclude that [ix, d^'\ = 0 by means of part (iv) and the
homotopy formula. Finally, use mathematical induction over ^ € No to show
d'^o) = 0, for every o) € OJ^iU).
Exercise 8.44 (Closed but not exact). Supposes > 2. Define, as in Formula (8.73)
a € ^"-^(R"\{0}) by a{x) = ij-^ dx) =-\- T (-1)'-^ d^.
\\\x\\" I ||x||" |^^_
(i) Demonstrate that the closed differential form a is not exact, that is, there is
no ?7 € ^"-2(R" \ {0}) with a = dr].
Hint: Recall that /^„_i a = |S""'| and apply Stokes' Theorem, noting that
dS"-^ = 0.
(ii) Take n = 2 and let 4' : V —> t/ with ^(r, a) = r(cosa, sin a) be the
substitution of polar coordinates from Example 3.1.1. Prove ^*a = da on
V.
Background. The angle function a is multi-valued on R^ \ {0}, and this is the
obstruction why a is not exact on all of R^ \ {0}. On the other hand, the summands
involving multiples of 2;7r are annihilated when one applies d toa.
Exercise 8.45. Let t/ = R^ \ {(0,0, X3) | X3 € R} C R^ and let m be the C°
differential form
X3
CO = — j(X2dXi — Xi dX2) € ^ (U)

770 Exercises for Chapter 8: Oriented Integration
(i) Verify
1
do) = — ^(xi dx2Adx3+X2dxsAdxi) € Q,^(U) and d^co = 0.
Fix T/f € [-f, f ] and define the C°° embedding
(P: D :=]-7t,7t[ x'^f,^[^ U
by (j)(a,6) = (cos a cos 0, sin a cos 0, sin0).
(ii) Prove(f)*a) = —sm9da € ^'(D). Deduce
d((p''a)) = (j)*{dci>) = cosOdaAdO € 0^(0),
and verify the second identity also by a direct computation,
(iii) Check the identity f da> = /g m from Stokes' Theorem by proving
I COS 6 dadO = In(1 — sin rj/) = — sin rj/ I da — I i
Jd J-jt Jtt
JdD
da
J —JT Jtt
sin 6 da.
Background. The oriented line integral above gives the angle of daily rotation of
Foucault's pendulum from Exercise 5.57. See Exercise 8.10 for the same
computation in terms of vector fields.
Exercise 8.46 (Hamiltonian mechanics in terms of differential forms - sequel
to Exercises 3.8,3.15,5.76 and 8.41). In mechanics the cotangent bundle T* Q —
Q X R''* of a submanifold Q of dimension d, see Exercise 5.76, plays an important
role. T*Q arises as momentum phase space of a system: q € Q represents the
generalized coordinates and p € R''* the generalized momenta for the system.
The evolution in time of the system is described by Hamilton's equation in part (v)
below, which is a system of 2d first-order ordinary differential equations. As in
Exercise 5.17.(ii) one proves that T*Q is a submanifold of dimension 2d.
(i) Define ;7r : T* Q —> Q as the projection onto the first factor. Then we have
D7r(q, p) : T(q^p)T*Q —> TqQ, for all {q, p) € T*Q\ and additionally p :
TqQ —> R. Therefore the tautological 1-form t on T*Q may be introduced
by
riq, p) = po Dniq, p) : T(q^p)T*Q -^ R.
Show
T= Y. p,dqi€^\T*Q).
l<i<d

Exercises for Chapter 8: Oriented Integration 111
The following explains the name of t. Let r} € ^'(Q); in other words, let
Y} : Q —> T*Q he a section of the cotangent bundle, that is, ;7r o ?; = /
on Q; then r]*T = r}. Indeed, use Definition 8.6.7 to prove (?;*t)(^) =
-^{q, nil)) o Dr](q) = r](q) o Dniq, r](q)) o Dr](q) = r](q).
(ii) Next introduce the symplectic 2-form a on T* Q hy
a =dT. Verify o- = X! ^^' ^ ^^' ^ ^^(^*2)-
This implies that t is all but closed; now prove that o itself is closed. Verify
for the rf-fold exterior product
o'' = o A ■ • • A o = d\ {—!) 2' dqi A • • ■ A dqj A dp\ A ■ • • A dp J.
In other words, ^~^^.,"—o'' is the Euclidean volume form on T*Q.
(iii) On the strength of Definition 8.6.2 verify, for vector fields v and tJ on T* Q,
aiv,v)= ^ (vj+iVi - VjVj+i) = {v, Jjv),
with
Ji = {^j^ ~^0 ) ^ ^^^^'^' ^^
Prove that cr is a nondegenerate bilinear form.
(iv) A vector field v on T*Q is said to be a Hamilton vector field corresponding
to the Hamiltonian H : T* Q ^> R if i^o = —dH (see (8.57)), that is,
i-oO is an exact differential 1-form on T*Q. Now deduce from (iii) that
{JdV,v) = {grad H, v), and use the nondegeneracy of a to conclude that
V = vh := —Jd grad H.
(v) Prove that a solution curve x = (q, p) : J ^> T* Q of the Hamilton vector
field Vf] satisfies the following, known as Hamilton's equation (compare with
Exercise 7.62), that is, for I < j <n and t € J,
J dpj J dqj
(vi) Show that iv,^cr from part (iv) is exact, at least locally, if and only if iv,^cr
is closed, that is, div^cr = Ly^a = 0, on account of (ii) and the homotopy
formula.

772 Exercises for Chapter 8: Oriented Integration
(vii) Assume that i^')reR is a one-parameter group of C^ diffeomorphisms of
r* Q having a Hamilton vector field v^ as tangent vector field and that a is
the symplectic 2-form. By means of part (vi) show, for f € R,
>'r' - T.
h=0
(vi/'+'')v = i^y—
ah
(vl/")V = (vI/')*L„,,or = 0.
/i=0
Consequently, (^')*a = a; such diffeomorphisms are called canonical
transformations. Using part (ii) deduce Liouville 's Theorem, which asserts
that the Euclidean volume form on T*Q is invariant under (4'')rgR.
(viii) For a canonical transformation 4'prove, for ally € T* Q and n, tJ € Ty{T*Q),
hence
on account of (iii). Here we have written the transpose as ^ instead of, in
order to avoid any confusion with the time variable t. Another way of saying
this is that G = D^ (y) belongs to the symplectic group Sp(rf, R) defined by
Sp(^, R) = {G € GL(2^, R) I G'JdG = Jj}.
Originally this group was called the linear complex group. This terminology
was too confusing, so the Latin roots in com-plex (meaning "plaited together")
were replaced by the Greek roots sym-plectic.
(ix) Prove the mapping Jd € End(r*R'') is canonical, in view of J^ € Sp(rf, R).
(x) A diffeomorphism if '■ Q ^ Q induces the mappings
TQ^TQ with {q, q') ^ {f{q), Df{q)qJ,
^■.T*Q^T*Q with ^{q, p) = {f{q), {Df{q)-'r p).
Prove D^{q,p){Sq,Sp) = {Df{q)Sq, {Dfiq^^Sp), for {Sq,Sp) €
^2rf Verify that the induced mapping 4' : T*Q —> r*(2 is a canonical
transformation.
(xi) Let 4' : r*(2 —> r*(2 be a canonical transformation and 7/ : r*(2 —> R
a Hamiltonian. Suppose x is as in part (v). Using Exercises 3.15.(i) and
3.8.(ii)) show, withx = 4'0') and f € 7,
y'{t) = DHyiOr' JADHyiOr'y grad(vI/*/f)(y(0) = v^.Hiyit)).
Here we have used that G € Sp(^, R) if and only if G'^JdiG'^ = Ja-
In other words, the pullback '^*vh of the vector field vh under the
canonical transformation 4' is the Hamiltonian vector field corresponding to the
pullback 4'*7/ of H under 4', that is, '^*V[j = v^*h. Prove this also via
2**.„o- = i**„„*V = 4'*(/„„o-) = ^*dH = d^*H = iy^,„(r.

Exercises for Chapter 8: Oriented Integration 773
(xii) Define, for functions / and g € C^(T*Q), the Poisson brackets {/, g} €
C(T*Q) of / and g by {/, g} = i^^dg = dg(Vf). Using dg = -i^^a from
part (iv) show
The definition of {/, g} is independent of the choice of coordinates (q, p) =
f € T*Q, in view of the invariance of a under canonical transformations.
(xiii) Prove the relation [^f,'^g] = '^{f,g) between the Lie and the Poisson brackets.
In fact, apply successively the formula from Exercise 8.41.(iii) for the Lie
derivative of a differential form with X = Vfando) = i^o € Q.^{T*Q),^sri
(vi), and Exercise 8.41.(iii) again, to obtain
(L„^0"„^,o-))(t;) = Vf{i^^a{v)) + {i^^a){[v, Vf])
= -iL^^a)iVg, v) + Vf{aiVg, v)) + criVg, [ v, Vf ])
= cr([vf, Vglv) = (i[vf,v,]cr)(v),
and deduce i[vf,vg]Cr = L^^i-dg) = -d{L^^.g) = -d{f, g} = ivy-^^cr from
part (iv), the homotopy formula and part (xii). Show that the Poisson brackets
satisfy Jacobi's identity from Exercise 5.26.(iii), that is
{/l, {/2, /3}} + [fl, {/3, /l}} + {/3, {/l, fl}} = 0.
(xiv) Suppose X : 7 —> r* (2 is a solution curve of the Hamilton vector field vh as
in part (v), and let / € C^iT*Q). Verify by means of Formula (2.12)
(/ o xYit) = {H, /}(x(0), in particular (H o x)' = 0.
That is, the Hamiltonian H is a conserved quantity. Hamilton's equation
itself takes the form
q'j = {H,qj} p'j = {H,pj} (1 < y < rf), x' = {H,x}.
Here we have extended the definition of the Poisson brackets to vector-valued
functions. Assuming convergence deduce that the solution is given, with
ShX = {H,x},hy
xit) = e""xiO) = Y] -AH, ■■•{H, [H,x}} • ■ • }(0) (t € R).
^-^ n\
neNo
This formula shows an analogy with descriptions of quantum physics.

774 Exercises for Chapter 8: Oriented Integration
(xv) The Hamiltonian of a classical point particle in R of mass m under the influ-
2
ence of gravity F(q) = —g, for ^ € R, is given by H(q, p) = j;;^, + tngq.
Prove
{H,q} = ^, {H,{H,q}} = -g, [H, [H, [H, q}}} = 0;
m
and obtain the law of free fall as the solution, where p = mq',
piO)t gt^ , gt^
q(t) = q(0) + ^^^^ - ^ = q(0)+q'(0)t - ^ (t € R).
m Z 2.
Exercise 8.47 (Minimal hypersurface - needed for Exercise 8.48). Consider a
compact oriented C^ submanifold V C R" of codimension 1, and a one-parameter
group of C^ diffeomorphisms (<I>')reR of R" with C^ tangent vector field t; : R" —>
R". Then all the Vf = <I>'(V),forf € R, are compact oriented C^ hypersurfaces too.
Select C' mappings « : R x R" —> R" such that«,(x) = «(f, x), forx € V,, is the
normal to V, at x compatible with the orientation on V,. According to Formula (7.15)
the Euclidean hyperarea form on V, is given by (W, = z„, dx. Note that Vq = V, and
write ftJo = ftj.
(i) Use the homotopy formula from Lemma 8.9.1 to verify
"' lr=o Jvi "' \t=QJv
= I (d oiy + iy o d)a)+ I i J \ dx.
Jv Jv ^l,=o"'
(ii) Deduce firom \\ntf=l,foTt€ R, that
(— n,,no) = 0;
\dt ,_„ /
accordingly
r=o
thus —
dt
nr(x)€ T,V;
r=0
ij\ dx = 0,
Jv si<=o"'
since computing the integral involves evaluation of dx at the points x € V
on n vectors belonging to the (n — l)-dimensional space TxV.
(iii) Use Stokes' Theorem to show
d_
dt
I cOf = I iyCt) + I iy od(i(n)dx) = I iu(d oi(n)dx).
t=aJv, Jdv Jv Jv

Exercises for Chapter 8: Oriented Integration 775
Apply the equality d o /„ dx = div n dx from Example 8.9.2 and prove
d\ f f
— I 0), = I (div n)i^dx.
Background. We call div« = tr Dn : V —> R the mean curvature of V (see
Section 5.7); it depends on the choice of orientation. A hypersurface V with fixed
boundary and having smallest possible hyperarea is called a minimal hypersurface.
The arguments above imply that the mean curvature of a minimal hypersurface
vanishes identically, as one sees by taking the vector field v restricted to V equal to
/«, for arbitrary C' functions /.
Exercise 8.48 (Catenoid and helicoid are minimal surfaces - sequel to
Exercises 4.6 and 8.47). As in Exercise 4.6 we define 0 : R^ —> R^ by x = 0(j, f) =
(cosh s cos t, cosh ^ sin f, s), and we call C = im(0) the catenoid.
(i) Show that the Gauss mapping « : C —> S^ is given by
1
n(x) = (—cost, — sint, siahs) ((s,t) € R^).
cosh J
(ii) Take Di(j)(s, t) and D2^(s, t) as basis vectors for T^C. With respect to this
basis compute, as in Example 5.7.2, the matrix of the Weingarten mapping
Dn(x) € End+(rvC) to be
sh^cU -1 )■
cosh s
Deduce from Exercise 8.47 that C is a minimal surface.
(iii) Fix a € R4-. Compute the area of the subset Ca of C consisting of the x € C
with IX3I < a to be In (a + cosh a sinh a). On the other hand, the area of
the two disks D* = {x € R^ | x^ + x| < cosh^a, X3 = zta} equals
2n cosh^ a. So the minimal surface Ca will not minimize the area among all
surfaces with boundary the two circles 90* if a + cosh a sinh a > cosh^ a,
that is, if 2fl > 1 + e"^, which is satisfied for a sufficiently large.
(iv) Prove that the helicoid from Exercise 4.8 is a minimal surface.
Exercise 8.49 (Special case of Gauss-Bonnet Theorem). Consider a compact
oriented C^ submanifold V C R" of codimension 1. Extending the theory of
Section 5.7 in a straightforward manner we say that the Gaussian curvature K of
V is given by AT = det Dn, where w : V —> S""' is the Gauss mapping. Let

776 Exercises for Chapter 8: Oriented Integration
0) € ^""'(R") be the differential form from Example 8.11.9 that computes the
Euclidean (n — l)-dimensional hyperarea of V. Prove
n*a)^„.i = KcOy, and deduce (•) -—— Kiy)dn-iy = degin).
!'-> I Jv
In particular, if V C R^ is a multi-donut with g holes, prove that deg(«) = I — g
by means of Formula (8.68). The number g is called the genus of the multi-donut.
Background. The equality (•) above is half the assertion of the Gauss-Bonnet
Theorem. The other half identifies deg(«) € Z as an invariant of V. Furthermore,
in R^ the integer 2 deg(«) equals the Euler characteristic x(V) of V: partition V
into a finite number of triangles, then x(V) equals the number of vertices minus
the number of edges plus the number of faces of the triangles, irrespective of the
chosen subdivision.
Exercise 8.50 (Zeros of a holomorphic function). As usual, write C B z =
xi + ix2 -^ X = (xi, X2) € R^. Let f = fi + if2 : C —> C be a holomorphic
function and set f/ = {z € C | f(z) 7^ 0}.
(i) Leto- € ^'(R2\{0}) be as in Formula (8.73). Verify that we have on R^ \ {0}
and U, respectively.
-X2dXi+XidX2 r '^f Iji 11^1,2
2 , 2 ' rflog/ = — = -f'l"""'^!'
xf +xi / 2
cr{x) = ' / ■ rflog/ = 4 = -rflogll/f+ //V.
In complex function theory it is shown that / has only isolated zeros (see
Example 2.2.6) if / 7^ 0.
(ii) Suppose that f'{z) 7^ 0 if f{z) = 0. Use the Cauchy-Riemann equation to
show det Dfix) = |/'(z)P > 0, and deduce that sgn (det(D/(x)) = 1, for
every zero x € R^ for /.
(iii) Let ^ C C be as in Example 8.11.11. By means of parts (i) and (ii) prove
277:/ i
where n(f, Q) is the number of zeros of / that belong to Q.
(iv) In the case of f(a) = /'(a) = 0, for some a € ^, we argue as follows. In
view of Exercise 8.12.(v) we may develop / in a power series about a, hence
fiz) = J2„>n(f,a) ^"(z - '^)"' with nif,a) € N and 0 7^ c„(/,a) € C. This
gives f(z) = (z — fl)"^-^'"^ giz), for z near a, with g holomorphic near a and
g(a) ^ 0. Hence, for z near a,
fiz) nif,a) , g'iz) n{fa)
fiz) z-a giz) z-a

Exercises for Chapter 8: Oriented Integration 111
with h holomorphic near a. If /"' ({0}) n ^ = {a,- | 1 < z < m }, deduce
2^2 Jbu f .f-',..
l<!<ni
Exercise 8.51 (Linking number, Kronecker's integral and Biot-Savart's law
- sequel to Exercise 8.28). Let V; be connected compact and oriented C^ sub-
manifolds of R" of dimension rf,-, for 1 < z < 2, where d\ + d2 = n — \, and
suppose these have no point in common. (Best example: two disjoint closed curves
in R^.) In the notation of Example 8.n.9 define the linking number L(Vi, V2) as
W((l>iVi X V2), 0), where 0 : Vi x V2 ^^ R" \ {0} is given by ^(xi, xz) = xz -xi.
(i) Prove
«>'..w=^X,^,/-(j^H-
(ii) We will compute the integral in (i) for a pair of closed curves V, = im()/,) in
R^, where y,- : [0, 1 ] —> Vi-. Verify, for the standard basis vectors e,- € R^,
Di<t> o (y, X y2)Xtu ^2) e,- = (-l)')A'(f,) (l < i < 2),
and show that this implies
L{Vu V2) = — / / ^—— —— dtidt2
Jo \477: Jo
ll/l(fl)-/2(f2)IP
/2(^2) X (Yiiti) - Y2ih))dh, Y[ih))dh.
Recognize the inner integral as the Biot-Savart law from Exercise 8.28.(v)
describing the magnetic field at Yiih) due to a steady unit electric current
flowing around the closed loop V2. Deduce that L{V\, V2) is precisely the
work done by this magnetic field on a unit magnetic pole which makes one
circuit around Vi.
(iii) The curves yi(f) = (cosf, sinf, 0) and J4-(f) = r{—l + cosf, 0, — sinf),
with r > 1, define two disjoint oriented circles in R^ that are linked. Prove
that Yr converges to 72 with 72 (0 = (0, 0, —t), for r —> 00. Now verify
that L(Vi, V2) = 1 by explicit evaluation of the integral in (ii). Note that
0(Vi X V2) is the cylinder {x € R^ | x^ + x| = 1}, which winds once
around the origin in R^.
Hint: Compute J^ -—\r^ dt2 by means of Exercise 6.50.(iv) or the
substitution t2 = sinhM.

Notation
c
■"^
*
0
H
b
a
dvA
d
X
A
/\'^y*
A
V
Vx
D
11-11
II ■ IIeuc.
<-,■>
{-,■}
[-,■]
Ia
I
I
L
/^^
!y fix) ddX
fv fix)p(x)dx
Iv{f,v){y)dy
fy{f(^),dis)
fs.isix),d2x)
fan /(^) '^^
Wk
complement
Fourier transform
convolution
composition
closure
mapping from vector fields to differential forms
boundary
boundary of A in y
derivative in direction of outer normal
cross product
exterior multiplication
A^-th exterior power of y *
Laplace operator
nabla or del
covariant derivative in direction of X
wave operator
Euclidean norm on R"
Euclidean norm on Lin(R", R/")
standard inner product
Poisson brackets
Lie brackets
characteristic function of A
integral
upper Riemann integral
lower Riemann integral
integral of differential fonn co along mapping ip
integral of / over V w.r.t. Euclidean density
integral of / over V w.r.t. density p
flux of / across V w.r.t. v
oriented line integral of / along y
oriented surface integral of vector field g over 3
complex line integral of / along 9 f2
shifted factorial
8
466
468
17
8
583
9
11
534
147
577
568
470, 528
59, 528
407
755
3
39
2
773
169
34
426
426
426
572
495
490
530
537
560
555
180
779

?0
r'
^ jk
V(A-)
r
O : f/ -*■ y
o*
^ -.v ^ u
Xfl*
A'
A^
A^
Af
A(n,R)
ad
Ad
Ad
arg
Aut(R")
S = {Bi\i el]
B{a; S)
c'
Cc(R")
codim
curl
d
Dj
Df
D^fia)
d{;-)
deg
div
dom
End(R")
End+(R")
End-(R")
/-'{{■})
GL(H, R)
grad
graph
H(H, C)
hyperarea
/
iv
im
inf
int
/
Christoffel symbol
outer normal on 9f2 at x
pullback under ip
C diffeomorphism of open subsets U and V of R"
pushforward under diffeomorphism O
inverse mapping of O : f/ —>■ y
pullback under ^
adjoint or transpose of matrix A
complementary matrix of A
closure of A in y
antisymmetric part of Df
linear subspace of antisymmetric matrices
in Mat(n, R)
inner derivation
adjoint mapping
conjugation mapping
argiunent function
group of bijective linear mappings
partition of rectangle
open ball of center a and radius S
k times continuously differentiable
R" -^ R"
mapping
linear space of continuous functions on R" with
compact support
codimension
curl
exterior differentiation
7-th partial derivative
derivative of mapping /
Hessian of / at a
Euclidean distance
degree of mapping
divergence
domain space of mapping
linear space of linear mappings R"
linear subspace in End(R") of self-
linear subspace in End(R") of anti-
inverse image under /
^ R"
adjoint operators
-adjoint operators
general linear group, of invertible matrices
in Mat(n, R)
gradient
graph of mapping
linear subspace of Hennitian matrices in Mat(n, C)
hyperarea
identity mapping
contraction with vector (field) v
image under mapping
infimum
interior
complex structure on R^
Notation
408
515
571
88
270
88
88
39
41
11
539
159
168
168
168
88
28,38
424
6
65
434
112
540, 541
574
47
43
71
5
591
166, 528
108
38
41
41
12
38
59
107
385
507
44
570
108
21
7
555

Notation 781
${A) collection of compact and Jordan measurable 431
subsets of A
L-*- orthocomplement of linear subspace L 201
Lu Lie derivative in direction of vector field v 585
lim limit 6, 12
Lin(R",R'') linear space of linear mappings R"—>■ R/" 38
Lin*(R",R'') linear space of Winear mappings R"—>■ R'' 63
Lo(« + 1, R) Lorentz group of R"+^ 756
Lo°(4, R) proper Lorentz group 386
Mat(n, R) linear space of « x n matrices with coefficients in R 38
Mat(/? X n, R) linear space of p x n matrices with coefficients in R 38
N{c) level set of c 15
O = {Oi \ i & 1} open covering of set 30
0(R") orthogonal group, of orthogonal operators in End(R") 73
0(n, R) orthogonal group, of orthogonal matrices in Mat(n, R) 124
R" Euclidean space of dimension n 2
31{W) linear space of Riemann integrable functions on R" 428
with compact support
S upper sum 425
S_ lower sum 425
Sf symmetric part of Df 539
S"" ^ unit sphere in R" 124
■^CR") linear space of Schwartz functions on R" 466
5l{n, C) Lie algebra of SL(n, C) 385
SL(«,R) special linear group, of matrices in Mat(n, R) with 303
detenninant 1
SO(3, R) special orthogonal group, of orthogonal matrices 219
in Mat(3, R) with detenninant 1
SO(n, R) special orthogonal group, of orthogonal matrices 302
in Mat(n, R) with detenninant 1
Sp(«, R) symplectic group of R" 772
5U(2) Lie algebra of SU(2) 385
SU(2) special unitary group, of unitary matrices in Mat(2, C) 377
with determinant 1
sup supremum 21
supp support of function 427
Tx V tangent space of submanifold V at point x 134
T*V cotangent bundle of submanifold y 399
T*V cotangent space of submanifold V at point x 399
tr trace 39
V{a; S) closed ball of center a and radius S 8
vol,, n-dimensional volume 429

Index
Abel's integral equation 670
— partial summation formula 189
Abel-Ruffini Theorem 102
Abelian group 169
acceleration 159
action 238, 366
—, group 162
—, induced 163
addition 2
— fonnula for lemniscatic sine 288
adjoint linear operator 39, 398
— mapping 168, 380
affine function 216
— transformation, bijective 479
Airy function 256
Airy's differential equation 256
algebraic topology 307, 483, 550, 563
amplitude 469
— function 656
Ampere—Maxwell law 748
analog formula 395
analogs of Delambre-Gauss 396
angle 148, 219, 301, 498
— of spherical triangle 334
angle-preserving 336
angular momentum operator 230, 264,
272
anti-adjoint operator 41
anticommutative 169, 577
antiderivation 580
antiderivative 539, 547
antisymmetric matrix 41
— multilinear mapping 568
approximation, best affine 44
arc length 498
arc-length function 501
Archimedes' law 718
— Theorem 604
area, Euclidean 502
argument function 88
Arzela's Dominated Convergence
Theorem 472
associated Legendre function 177
associativity 2
astroid 324, 729
asymptotic expansion 197, 626
— expansion, Stirling's 197
automorphism 28
autonomous vector field 163
axis of rotation 219, 301
ball, closed 8
—, open 6
basic linear mapping 477
basis, orthonormal 3
—, standard 3
Bernoulli function 190
— number 186, 607
— polynomial 187
Bernoulli's summation formula 188,
194
Bessel function 634, 640
Bessel's equation 634
best affine approximation 44
Beta function 620
— function, generalized 688, 690
bifurcation set 289
bijective affine transfonnation 479
Binet—Cauchy's, formula 677
binomial coefficient, generalized 181
— series 181
binormal 159
Biot—Savart law 751
biregular 160
bitangent 327
black body radiation 623
boost 390, 756, 758
boundary 9
— condition, at infinity 721, 750
— condition, Dirichlet 535
— condition, Neumann 726
783

784
Index
— in subset 11
—, C*, of open set 515
—, lying at one side of 515
bounded convergence 471
— sequence 21
Brouwer's Fixed-point Theorem 583,
764
— Theorem 20
bundle, tangent 322
—, unit tangent 323
C mapping 51
Cagnoli's formula 395
canonical 58
— transfonnation 772
Cantor's Theorem 211
cardioid286,299, 501
Cartan decomposition 247, 385
— decomposition, (infinitesimal) 539
Casimir operator 231, 265, 370
Catalan's constant 613
catenary 295
catenoid 295
Cauchy distribution 660
— sequence 22
Cauchy's integral formula 734, 737
— integral formula, generalized 739
— Integral Theorem 557
— Minimum Theorem 206
Cauchy—Riemann equation 556, 738
Cauchy—Schwarz inequality 4, 642
— inequality, generalized 245
caustic 351, 761
Cayley's surface 309
Cayley—Klein parameter 376, 380, 391
centrifugal 604
centripetal 604
chain rule 51
change of coordinates, (regular) C* 88
Change of Variables Theorem 444
characteristic function 34, 428
— polynomial 39, 239, 527
—, Euler 776
charge (density) 747
— density 491, 696
chart 111
Christoffel symbol 408
circle, osculating 361
circles, Villarceau's 327
circulation 538, 553
cissoid. Diodes' 325
Clausen function 614
Clifford algebra 383
— multiplication 379
closed ball 8
— differential form 544, 574
— in subset 10
— mapping 20
— set 8
closure 8
— in subset 11
cluster point 8
— point in subset 10
codimension of submanifold 112
coefficient of matrix 38
—, Fourier 190
cofactor matrix 233
cohomology, de Rham 590
commutativity 2
commutator 169, 217, 231, 407
— relation 231, 367
compactness 25, 30
—, sequential 25
complement 8
complementary matrix 41
completeness 22, 204
complex analysis 737
— line integral 555
— structure 555
complex-analytic 735
complex-differentiable 105, 556, 738
component function 2
— of vector 2
composition of mappings 17
computer graphics 385
conchoid, Nicomedes' 325
cone 447
—, unrolling of 684
confluent hypergeometric differential
equation 641
— hypergeometric function 639
conformal 336
conjugate axis 330
connected component 35
connectedness 33
connection 408
conservation of charge, law of 748
— of energy 290
— of energy, law of 749
conservative 566

Index
785
constant, Catalan's 613
continuity 13
— equation 747
continuity. Holder 213
Continuity Theorem 77
continuous mapping 13
continuously differentiable 51
contractible 589
contraction 23
— factor 23
Contraction Lemma 23
contraction with vector 570
— with vector field 585
—, FitzGerald—Lorentz 758
convergence 6
—, uniform 82
convex 57
convolution 468, 663
— equation 672
coordinate function 19
— of vector 2
—, generalized 770
coordinates, change of, (regular) C*
88
—, confocal 267
—, cylindrical 260
—, polar 88
—, simplex 691
—, spherical 261
coordinatization 111
cosine rule 331
cotangent bundle 399
— space 399
Coulomb gauge condition 750
Coulomb's law 750
covariance matrix 617, 655
covariant 754
— derivative 407
— differentiation 407
covering group 383, 386, 389
—, open 30
Cramer's rule 41
critical point 60
— point of diffeomorphism 92
— point of function 128
— point, nondegenerate 75
— value 60
cross product 147
— product operator 363
curl, of vector field 540, 541
curl-free 546
current (density) 747
curvature form 363, 410
— of curve 159
—, Gaussian 157
—, mean 775
—, nonnal 162
—, principal 157
curve 110
—, differentiable (space) 134
—, space-filling 211
cusp, ordinary 144
cycloid 141, 293, 499
cylindrical coordinates 260
D'Alembert's fonnula 277
D'Alembertian 755, 756
Darboux's equation 707, 708
de I'Hopital's rule 311
decomposition, (infinitesimal) Cartan
539
—, Cartan 247, 385
—, Helmholtz-Weyl 751
—, Hodge 753
—, Iwasawa 356
—, polar 247
—, Stokes 539
degree of mapping 591, 594
del 59, 528
Delambre-Gauss, analogs of 396
DeMorgan's laws 9
dense in subset 203
density, continuous (f-dimensional 489
—, Euclidean 495
—, Gel'fand-Leray 700
—, positive 491
derivation 404, 585
— at point 402
—, inner 169
derivative 43
—, directional 47
—, exterior 544, 574
—, partial 47
derived mapping 43
Descartes' folium 142, 554
de Rham cohomology 590
diagonal 722
diameter 355
diffeomorphism, C^ 88
—, orthogonal 271

786
Index
—, volume-preserving C* 445
differentiability 43
differentiable mapping 43
—, continuously 51
—, partially 47
differential 400
— at point 399
— equation, Airy's 256
— equation, Bessel's 634
— equation, confluent hypergeometric
641
— equation, Darboux's 707, 708
— equation, for rotation 165
— equation, Hamilton's 717, 771
— equation, Helmholtz' 636
— equation, hypergeometric 638
— equation, Legendre's 177
— equation, Newton's 290
— equation, ordinary 55, 163, 177,
269
— equation, partial 470, 534
— equation, Poisson's 721
— equation, Whittaker's 641
— fonii400,544,571
— fonn, closed 544, 574
— fonn, exact 544, 574
— operator, linear partial 241
— operator, partial 164
Differentiation Theorem 78, 84, 473
differentiation, fractional 669
dilatation 759
dilogarithm 613
—, functional equation for 613
dimension of submanifold 109
Dini's Theorem 31
Diodes' cissoid 325
directional derivative 47
Dirichlet problem 535
Dirichlet's formula 632, 633, 701
— principle 719
— series 664
— test 189
disconnectedness 33
discrete 396
discriminant 347, 348
— locus 289
distance, Euclidean 5
distribution 617, 763
— theory 669
—, X-622
—, X^- 657
—, Cauchy 660
—, Gamma 657
—, nonnal 617, 618, 655
—, one-sided stable 659
distributivity 2
divergence in arbitrary coordinates
268,714
— with respect to volume form 586
—, of vector field 166, 268, 528
divergence-free 546
dodecahedron, rhombic 616
dual basis for spherical triangle 334
— vector space 398
duality, definition by 404, 406
Duhamel's principle 763
duplication formula for (lemniscatic)
sine 180
— formula, Legendre's 621, 623
eccentricity 330, 499
— vector 330
Egregium, Theorema 409
eigenvalue 72, 224
eigenvector 72, 224
Einstein summation convention 583
Eisenstein series 658
electromagnetic energy 748
— waves in vacuum 749
electromagnetism 683, 747
elimination 125
— theory 344
elliptic curve 185
— integral of first kind 679
— integral of second kind 499
— paraboloid 297
embedding 111
endomorphism 38
energy of wave 761
— surface 716
—, electromagnetic 748
—, kinetic 290
—, potential 290
—, total 290
entry of matrix 38
envelope 348
epicycloid 342
equality of mixed partial derivatives 62
equation 97
—, Cauchy—Riemann 556, 738

Index
787
—, continuity 747
—, Kepler's 635
—, Poisson's 750
—, Schrodinger's 728
equations. Maxwell's 747
—, Maxwell's, in covariant fonu 753
—, Maxwell's, in vacuum 749
—, Maxwell's, time-independent case
749
equivariant 384
Euclid's parallel postulate 686
Euclidean area 502
— density 495
— distance 5
— hyperarea 507
— norm 3
— norm of linear mapping 39
— space 2
Euler characteristic 776
— number A{n, k) 703
— number £„ 607
— operator 231, 265
Euler's Beta function 620
— constant 627
— foniiula300, 364
— Gamma function 620
— generalized Beta function 688, 690
— identity 228
— series 615
— Theorem 219
Euler—MacLaurin summation fonnula
194
evolute 350
exact differential fonn 544, 574
excess of spherical triangle 335
expectation 617
— vector 617, 645, 655
exponential 55
exterior derivative 544, 574
— differentiation 544, 574
— multiplication 577
— power 568
Faraday fonn 754
Faraday's law 748
Feynman's formula 693
fiber bundle 124
— of mapping 124
field 696
—, electric 747
—, magnetic 747
finite intersection property 32
— part of integral 199
FitzGerald—Lorentz contraction 758
fixed point 23
Fixed-point Theorem, Brouwer's 583,
764
flow of vector field 163
flux 530, 560
focus 329
focusing effect 761
folium, Descartes' 142, 554
formal power series 606
formula, analog 395
—, Binet—Cauchy's 677
—, Cagnoli's 395
—, D'Alembert's 277
—, Dirichlet's 632, 633, 701
—, Euler's 300, 364
—, Feynman's 693
—, Gegenbauer's 694
—, Girard's 681
—, Hankel's 694
—, Heron's 330
—, Kirchhoff's 760
—, Laplace's 744
—, Leibniz—Honnander's 243
—, Lipschitz' 658
—, Mellin's 662
—, Pizzetti's 689, 708
—, Poisson's 693
—, Rodrigues' 177
—, Schlafli's 743
—, Sonine's 670
—, Stirling's 624
—, Taylor's 68, 250
formulae, Frenet—Serret's 160
forward (light) cone 386
Foucault's pendulum 362, 733, 770
four-square identity 377
Fourier analysis 191
— coefficient 190
— Inversion Theorem 468
— series 190
— transform 466
— transformation 467
fractional differentiation 669
— integration 668
— linear transformation 384
Frenet—Serret's formulae 160

788
Index
frequency 469
Fresnel's integral 630
Frobenius norm 39
Frullani's integral 81
Fubini's Theorem 476
function of several real variables 12
—, C°°, on subset 399
—, affine216
—, Airy 256
—, Bernoulli 190
—, Bessel 634, 640
—, characteristic 34, 428
—, Clausen 614
—, complex-analytic 735
—, complex-differentiable 105, 556,
738
—, confluent hypergeometric 639
—, continuous with compact support
434
—, coordinate 19
—, Green's 722
—, holomorphic 105, 556, 738
—, hypergeometric 637
—, Kummer 639
—, Lagrange 149
—, Lerch 658
—, Lobachevsky 614
—, monomial 19
—, Morse 244
—, piecewise affine 216
—, polynomial 19
—, positively homogeneous 203, 228
—, rational 19
—, real-analytic 70
—, reciprocal 17
—, scalar 12
—, Schwartz 466
—, spherical 271
—, spherical harmonic 272
—, step 435
—, vector-valued 12
—, zonal spherical 271
functional dependence 312, 611
— equation 175, 313
— equation for dilogarithm 613
— equation for zeta function 631, 652
— equation of Jacobi 651
fundamental solution 763
Fundamental Theorem of Algebra
291, 593,735
Gamma distribution 657
— function 620
— function, product fonnula for 627
gauge condition, Coulomb 750
— condition, Lorentz 755
Gauss mapping 156
Gauss' Divergence Theorem 529
— law 748
Gauss—Bonnet Theorem 776
Gaussian curvature 157
Gegenbauer's fonnula 694
Gel'fand—Leray density 700
general linear group 38
generalized Beta function 688, 690
— binomial coefficient 181
— coordinate 770
— momentum 770
generator, infinitesimal 163
genus 776
geodesic 361
geometric tangent space 135
(geometric) tangent vector 322
geometric—arithmetic inequality 351
Girard's fonnula 681
Global Inverse Function Theorem 93
gradient 59
— operator 59
— vector field 59, 527, 539
Gram's matrix 39
Gram—Schmidt orthononnalization
356
graph 107,203
Grassmann's identity 331, 364
great circle 333
greatest lower bound 21
Green's first identity 534
— function 722
— Integral Theorem 554
— second identity 534
group action 162, 163, 238
—, covering 383, 386, 389
—, general linear 38
—, Lorentz 386, 756
—, orthogonal 73, 124, 219
—, permutation 66
—, proper Lorentz 386
—, special linear 303
—, special orthogonal 302
—, special orthogonal, in R^ 219
—, special unitary 377

Index
789
—, spinor 383
—, symplectic 772
Hadamard's inequality 152, 356
— Leiiima 45
Hamilton vector field 771
Hamilton's equation 717, 771
— Theorem 375
Hamilton-Cayley, Theorem of 240
Hamiltonian 716, 771
— vector field 717
Hankel transfonn 665
Hankel's formula 665, 694
Hardy's inequality 647
harmonic function 265, 535
— vector field 546
heat equation 469, 726
Heine—Borel Theorem 30
Heisenberg's uncertainty relations 645
helicoid 296
helix 107, 138, 296
Helmholtz' equation 636
Helmholtz—Weyl decomposition 751
Hermitian inner product 645
— matrix 385
Heron's fonnula 330
Hessian 71
— matrix 71
hexagon 703
highest weight of representation 371
Hilbert's inequality 643
— Nullstellensatz311
Hilbert—Schmidt norm 39
Hodge decomposition 753
— operator 753, 754
hodograph 605
Holder continuity 213
Holder's inequality 353, 642
holomorphic 105, 556, 738
holonomy 363
homeomorphic 19
homeomorphism 19
homogeneous function 203, 228
homographic transfonnation 384
homology theory 563
homomorphism of Lie algebras 170
— of rings, induced 401
homotopic 587
homotopy 550, 587
— fonnula 586
Homotopy Lemma 587
homotopy operator 587
— theory 483, 550
Hopffibration307, 383
— mapping 306
hydrostatic pressure 718
hyperarea, Euclidean 507
hyperbolic reflection 391
— screw 390, 756, 758
hyperboloid of one sheet 298
— of two sheets 297
hypercube 702
hyperfunction 662
hypergeometric differential equation
638
— function 637
— series 637
hypersurface 110, 145, 507
— integral, oriented 563
—, minimal 775
hypocycloid 342
—, Steiner's 343, 346, 731
icosahedron 504
identity mapping 44
—, Euler's 228
—, four-square 377
—, Grassmann's 331, 364
—, Green's first 534
—, Green's second 534
—, Jacobi's 332
—, Jacobi's, for minors 584
—, Lagrange's 332
—, parallelogram 201
—, Parseval—Plancherel's 650
—, polarization 3
—, symmetry 201
image, inverse 12
immersion 111
— at point 111
Immersion Theorem 114
implicit definition of function 97
— differentiation 103
Implicit Function Theorem 100
— Function Theorem over C 106
incircle of Steiner's hypocycloid 344
incompressible 546
indefinite 71
index set 424
—, of function at point 73

790
Index
—, of operator 73
induced action 163
— homomorphism of rings 401
inequality, Cauchy—Schwarz' 4, 642
—, Cauchy—Schwarz', generalized
245
—, geometric—arithmetic 351
—, Hadamard's 152, 356
—, Hardy's 647
—, Hubert's 643
—, Holder's 353, 642
—, isoperimetric 709
—, isoperimetric for triangle 352
—, Kaiitorovich's 352
—, Minkowski's 353, 643
—, Poincare's 644
—, reverse triangle 4
—, Sobolev's 645
—, triangle 4
—, Young's 353
infimum 21
infinitesimal generator 56, 163, 239,
303, 365, 366
inhomogeneous wave equation 761
initial condition 55, 163, 277
— value problem, for heat equation
470
— value problem, for wave equation
759
inner derivation 169
— measure 429
— product, Hermitian 645
integrability conditions 539, 546
integral 539, 547
— equation, Abel's 670
— fonnula for remainder in Taylor's
formula 67, 68
— fonnula, Cauchy's 734, 737
— formula, Cauchy's, generalized 739
— fonnula, Poisson's 725, 745
— of differential form 572
— of differential form over
submanifold 573, 594
— of gradient 519
— of total derivative 518
— over rectangle 426
— over submanifold w.r.t. density 490
— over submanifold w.r.t. Euclidean
density 495
— over subset 429
— over tubular neighborhood 698
Integral Theorem, Stokes' 560
integral, (n — l)-dimensional w.r.t.
Gel'fand—Leray density 700
—, Fresnel's 630
—, Frullani's 81
—, Kronecker's 595
—, oscillatory 656
—, Poisson's 661, 725
integrals, Laplace's 254, 661
integration, fractional 668
—, Lebesgue 476
interior 7
— point 7
intennediate value property 33
interval 33
intrinsic property 408
invariance of dimension 20
— of Laplacian under orthogonal
transfonnations 230
inverse image 12
— mapping 55
inverse-square law 604
Inversion Theorem, Fourier 468
inversion w.r.t. sphere 724
involution 685
irreducible representation 371
irrotational 546
isolated zero 45
isometry 685
Isomorphism Theorem for groups 380
isoperimetric inequality 709
— inequality for triangle 352
isotopy 587
Isotopy Lemma 592
iteration method, Newton's 235
Iwasawa decomposition 356
Jacobi matrix 48, 445
Jacobi's functional equation 651
— identity 169,332
— identity for minors 584
— notation for partial derivative 48
Jacobi an 445
Jordan measurable set 429
— measurable, (f-dimensional 492
— measure of set 429
— measure, (f-dimensional 492
Jordan—Brouwer Separation Theorem
596

Index
791
A^-lineai- mapping 63
Kakeya's needle problem 731
Kantorovich's inequality 352
Kepler's equation 635
— first law 605
— second law 450, 451
— third law 605
kinetic gas theory 622
Kirchhoff's formula 760
kissing number 504
Kronecker's integral 595
Kummer function 639
Lagrange function 149
— multipliers 150
Lagrange's identity 332
— Theorem 377
Laplace operator 229, 263, 265, 270,
470, 528
Laplace's formula 744
— integrals 254, 661
Laplacian 229, 263, 265, 270, 470,
528
— in arbitrary coordinates 715
— in cylindrical coordinates 263
— in polar coordinates 712
— in spherical coordinates 264
latus rectum 330
law of conservation of charge 748
— of conservation of energy 749
— of free fall 774
— of gravitation, Newton's 450
—, Ampere—Maxwell's 748
—, Archimedes' 718
—, Biot-Savart's 751
—, Coulomb's 750
—, Faraday's 748
—, Gauss' 748
—, inverse-square 604
—, Kepler's first 605
—, Kepler's second 450, 451
—, Kepler's third 605
—, Pascal's 718
laws, DeMorgan's 9
least upper bound 21
Lebesgue integration 476
— number of covering 210
Legendre function, associated 177
— polynomial 177
Legendre's duplication formula 621,
623
— equation 177
Leibniz' rule 240
Leibniz—Honnander's formula 243
Lemma, Hadamard's 45
—, Homotopy 587
—, Morse's 131
—, Poincare's 548
—, Poincare's, for differential forms
589
—, Rank 113
—, Urysohn's 608
lemniscate 323, 447, 678
lemniscatic sine 180, 287
— sine, addition fonnula for 288, 313
length of vector 3
Lerch function 658
level set 15
Lie algebra 169, 231, 332
— algebra cohomology 768
— bracket 169
— derivative 404, 585
— derivative at point 402
— group, linear 166
Lie's product formula 171
light cone 386
limit of mapping 12
— of sequence 6
line integral, complex 555
— integral, oriented 537, 552
linear Lie group 166
— mapping 38
— mapping, basic 477
— operator, adjoint 39
— partial differential operator 241
— regression 228
— space 2
linearized problem 98
linking number 777
Liouville's Theorem 708, 726, 772
Lipschitz constant 13
— continuous 13
Lipschitz' formula 658
Lobachevsky function 614
Local Inverse Function Theorem 92
— Inverse Function Theorem over C
105
locally isometric 341
Lorentz gauge condition 755

792
Index
— group 386, 756
— group, proper 386
— transfonnation 386, 756
— transfonnation, proper 386
lower Riemann integral 426
— sum 425
lowering operator 371
loxodrome 338
lying at one side of boundary 515
MacLaurin's series 70
magnetic monopole 748
manifold 108, 109
— at point 109
mapping 12
—, C^ 51
—, k times continuously differentiable
65
—, A^-linear 63
—, adjoint 380
—, closed 20
—, continuous 13
—, derived 43
—, differentiable 43
—, identity 44
—, inverse 55
—, linear 38
—, of manifolds 128,594
—, open 20
—, partial 12
—, proper 26, 206
—, tangent 225
—, uniformly continuous 28
—, Weingarten 157
mass 290
— density 491
matrix 38
—, antisymmetric 41
—, cofactor 233
—, Hermitian 385
—, Hessian 71
—, Jacobi 48
—, orthogonal 219
—, symmetric 41
—, transpose 39
Maxwell's equations 747
— equations, in covariant form 753
— equations, in vacuum 749
— equations, time-independent case
749
mean curvature 775
— value property 707
Mean Value Theorem 57
— Value Theorem for harmonic
functions 707, 723
mean, spherical 689, 706, 708
mechanics, statistical 622
Mellin transform 662
— transformation 663
Mellin's fonnula 662
Mercator projection 339
method of least squares 248
metric 685
— space 5
metric-preserving 685
minimal hypersurface 775
minimax principle 246
Minkowski's inequality 353, 643
minor of matrix 41
Mobius strip 564
moment 649
— of inertia 690
momentum operator 644
— phase space 770
—, generalized 770
monomial function 19
Morse function 244
Morse's Lemma 131
Morsification 244
moving frame 267
multi-index 69
multilinear algebra 568
Multinomial Theorem 241
multipliers, Lagrange 150
musical isomorphism 583
nabla 59, 528
needle problem, Kakeya's 731
negative (semi)definite 71
negligible, (f-dimensional 492
—, ^-dimensional 429
—, (n — l)-dimensional 525
neighborhood 8
— in subset 10
—, open 8
nephroid 343
Neiunann boundary condition 726
— problem 752
Newton vector field 529
Newton's Binomial Theorem 240

Index
793
— equation 290
— iteration method 235
— law of gravitation 450
— potential 728
— potential of function 721
— potential of point 720
— potential of set 448
Nicomedes' conchoid 325
non-Euclidean geometry 685
nondegenerate critical point 75
norm 27
—, Euclidean 3
—, Euclidean, of linear mapping 39
—, Frobenius 39
—, Hilbert-Schmidt 39
—, operator 40
normal 146
— curvature 162
— distribution 617, 618, 655
— plane 159
— section 162
— space 139
—, inner 515
—, outer 515
nullity, of function at point 73
—, of operator 73
Nullstellensatz, Hilbert's 311
numerical mathematics 675
octahedron 703
one-parameter family of lines 299
— group 669
— group of diffeomorphisms 162
— group of invertible linear mappings
56
— group of rotations 303
one-sided stable distribution 659
open ball 6
— covering 30
— in subset 10
— mapping 20
— neighborhood 8
— neighborhood in subset 10
— set 7
operator norm 40
—, anti-adjoint 41
—, self-adjoint 41
—, unitary 663
orbit 163
orbital angular momentum quantum
number 272
ordinary cusp 144
— differential equation 55, 163, 177,
269
orientation 563
—, positive 552
orthocomplement 201
orthogonal group 73, 124, 219
— matrix 219
— projection 201
— transformation 218
— vectors 2
orthononnal basis 3
orthononnalization. Gram—Schmidt
356
oscillatory integral 656
osculating circle 361
— plane 159
outer measure 429
p function, Weierstrass' 185
paraboloid, elliptic 297
parallel postulate, Euclid's 686
— translation 363
parallelepiped 446
parallelogram identity 201
parameter 97
—, Cayley-Klein 376, 380, 391
parametrization 111
—, of orthogonal matrix 364, 375
—, positive 552
parametrized set 108
Parseval—Plancherel's identity 650
partial derivative 47
— derivative, second-order 61
— differential equation 470, 534
— differential operator 164
— differential operator, linear 241
— mapping 12
— summation fonnula of Abel 189
partial-fraction decomposition 183,
653
partially differentiable 47
particle without spin 272
partition 424
— of unity 452
Pascal's law 718
Pearson's x^ distribution 657
pendulum, Foucault's 362, 733, 770

794
Index
periapse 330
period 185
— lattice 184, 185
periodic 190
permutation group 66
perpendicular 2
phase function 656
physics, quantum 230, 272, 366, 644,
728
piecewise affine function 216
Pizzetti's formula 689, 708
planar curve 160
plane, normal 159
—, osculating 159
—, rectifying 159
Pochhammer symbol 180
Poincare's inequality 644
— Lemma 548
— Lemma, for differential fonns 589
point of function, critical 128
— of function, singular 128
—, cluster 8
—, critical 60
—, interior 7
—, saddle 74
—, stationary 60
Poisson brackets 242, 773
— integral 745
Poisson's equation 721, 750
— fonnula 693
— integral 661, 725
— integral formula 725, 745
— kernel 660, 722
— summation formula 650, 651
polar coordinates 88
— decomposition 247
— part 247
— triangle 334
polarization identity 3
polygon 274
polyhedron 274
polynomial function 19
—, Bernoulli 187
—, characteristic 39, 239, 527
—, Legendre 177
—, Taylor 68
polytope 274
position operator 644
positive (semi)definite 71
— orientation 552
— parametrization 552
positively homogeneous function 203,
228
potential 697
— difference 547, 566
—, Newton's 728
—, Newton's, of function 721
—, Newton's, of point 720
—, retarded 755, 762
—, scalar 547, 750
—, vector 547, 750
Poynting vector field 748
principal curvature 157
— nonnal 159
principle of stationary phase 656
—, Dirichlet's719
—, Duhamel's 763
—, minimax 246
probability density 644
— density of x-distribution 622
— density of Cauchy distribution 660
— density of distribution 617
— density of Gamma distribution 657
— density of nonnal distribution 617,
618,655
— density of one-sided stable
distribution 659
— density of Pearson's x distribution
657
product formula for Gamma function
627
— of mappings 17
—, cross 147
—,Wallis' 184,627
projection, Mercator 339
—, stereographic 336
proper 591
— Lorentz group 386
— Lorentz transformation 386
— mapping 26, 206
property, global 29
—, local 29
pseudosphere 358, 685
pullback 406
— of differential form 571
— under diffeomorphism 88, 268, 404
pushforward under diffeomorphism
270, 404
Pythagoras' Theorem 678
Pythagorean property 3

Index
795
quadrature of pai'abola 730
quadric, nondegenerate 297
quantum number, magnetic 272
— physics 230, 272, 366, 644, 728,
773
quaternion 382
radial part 247
raising operator 371
Rank Lemma 113
rank of operator 113
Rank Theorem 314
rapidity 756,757, 758
rational function 19
— parametrization 390
Rayleigh quotient 72
real-analytic function 70
reciprocal function 17
rectangle 30, 423
rectifying plane 159
recursion relation 625
refinement 424
reflection 374
— fonnula for Gamma function 628,
629
regression fine 228
regular value 591
regularity of mapping R'' D-^ R" 116
— of mapping R" >^ R"-'' 124
— of mapping R" >^ R" 92
regularization 199
relative topology 10
relativistic law for addition of
velocities 757
remainder 67
reparametrization 538, 572
representation, highest weight of 371
—, irreducible 371
—, spinor 383
residue 737
Residue Theorem 737
resultant 346
retarded potential 755, 762
reverse triangle inequality 4
revolution, surface of 294
Riemann integrable function with
compact support 428
— integrable over rectangle 426
— integrable over submanifold 490
— integrable over submanifold,
absolutely 492
— integrable over subset 429
— integrable, absolutely 463
— integrable, locally 461
— integral over subset 429
— integral, lower 426
— integral, upper 426
Riemann's zeta function 191, 197,
612, 622
Rodrigues' fonnula 177
— Theorem 382
Rolle's Theorem 228
rose, with four petals 500
rotation group 303
—, in R3 219, 300
—, in R" 303
—, infinitesimal 303
nile, chain 51
—, cosine 331
—, Cramer's 41
—, de THopital's 311
—, Leibniz' 240
—, sine 331
—, spherical, of cosines 334
—, spherical, of sines 334
saddle point 74
Sard's Theorem 610
scalar function 12
— multiple of mapping 17
— multiplication 2
— potential 539, 547
Schlafli's formula 743
Schrodinger operator 728
Schrodinger's equation 728
Schur complement 304
Schwartz function 466
Schwarz' Theorem 725, 746
screw, hyperbolic 756
second derivative test 74
second-order derivative 63
— partial derivative 61
section 400, 404
—, nonnal 162
segment of great circle 333
self-adjoint operator 41
self-adjointness of Laplacian 718
semicubic parabola 144, 309, 678
semigroup property 661

796
Index
semimajor axis 330
semiminor axis 330
Separation Theorem, Jordan-Brouwer
596
sequence, bounded 21
sequential compactness 25
series, binomial 181
—, Dirichlet's 664
—, Eisenstein's 658
—, Euler's 615
—, hypergeometric 637
—, MacLaurin's 70
—, Taylor's 70
set, closed 8
—, open 7
shifted factorial 180
shuffle 578
side of spherical triangle 334
signature, of function at point 73
—, of operator 73
simple zero 101
simplex coordinates 691
—, standard (n - 1)- 690, 691
simply connected 550
sine rule 331
singular point of diffeomorphism 92
— point of function 128
singularity of mapping R" D-^ R" 92
slerp 385
Sobolev's inequality 645
solenoidal 566
solid of revolution, volume of 604
—, Viviani's 681
Sonine's fonnula 670
source-free 546
space, linear 2
—, metric 5
—, vector 2
space-filling curve 211, 214
special linear group 303
— orthogonal group 302
— orthogonal group in R^ 219
— relativity, theory of 756
— unitary group 377
Spectral Theorem 72, 245, 355
sphere 10
spherical coordinates 261
— diangle 681
— function 271
— harmonic function 272
— mean 689, 706, 708
— rule of cosines 334
— rule of sines 334
— triangle 333, 681
spinor 389
— group 383
— representation 383, 389
spiral 107, 138, 296
—, logarithmic 350
spline 675
stability, of atom 728
standard basis 3
— deviation 617, 645
— embedding 114
— inner product 2
— projection 121
star-shaped 548
stationary phase, principle of 656
— point 60
Steiner's hypocycloid 343, 346, 731
— Roman surface 307
step function 435
stereographic projection 306, 336
Stirling's asymptotic expansion 197
— formula 624
stochastics 617
Stokes decomposition 539
Stokes' Integral Theorem 560
— Theorem 575
stratification 304
stratum 304
structure, complex 555
subcovering 30
—, finite 30
subimmersion 315
submanifold 109
— at point 109
—, affine algebraic 128
—, algebraic 128
submersion 112
— at point 112
Submersion Theorem 121
sum of mappings 17
summation convention, Einstein 583
— formula of Bernoulli 188, 194
— formula of Euler—MacLaurin 194
— formula of Poisson 650, 651
superposition 469
support 427
supremum 21

Index
797
surface 110
— integral 560
— of revolution 294
— of revolution, area of 603
—, Cayley's 309
—, Steiner's Roman 307
Sylvester's law of inertia 73
— Theorem 376
symbol, total 241
symmetric matrix 41
symmetry identity 201
symplectic form 771
— group 772
tangent bundle 322, 403
— cluster 684
— mapping 137, 225
— space 134
— space, algebraic description 400
— space, geometric 135
— space, "intrinsic" description 397
— sweep 684
— vector 134
— vector field 163
— vector, geometric 322
tautological fonn 770
Taylor polynomial 68
Taylor's fonnula 68, 250
— series 70
test, Dirichlet's 189
tetrahedron 274, 692
Theorem, Abel-Ruffini's 102
—, Archimedes' 604
—, Arzela's Dominated Convergence
472
—, Brouwer's 20
—, Brouwer's Fixed-point 583, 764
—, Cantor's 211
—, Cauchy's Integral 557
—, Cauchy's Minimum 206
—, Change of Variables 444
—, Continuity 77
—, Differentiation 78, 84, 473
—, Dini's 31
—, Divergence, Gauss' 529
—,Euler's219
—, Fourier's Inversion 468
—, Fubini's 476
—, Fundamental, of Algebra 291, 593,
735
, Gauss—Bonnet's 776
, Global Inverse Function 93
, Green's Integral 554
Hamilton's 375
Hamilton-Cayley's 240
, Heine—Borel's 30
, Immersion 114
Implicit Function 100
Implicit Function, over C 106
Integral, Cauchy's 557
Integral, Green's 554
Isomorphism, for groups 380
, Jordan—Brouwer's Separation 596
, Lagrange's 377
, Liouville's 708, 726, 772
, Local Inverse Function 92
, Local Inverse Function, over C 105
, Mean Value 57
, Mean Value, for harmonic
functions 707, 723
, Multinomial 241
, Newton's Binomial 240
, Pythagoras' 678
, Rank 314
, Residue 737
, Rodrigues' 382
, Rolle's 228
.Sard's 610
, Schwarz' 725, 746
, Spectral 72, 245, 355
, Stokes' 575
, Stokes' Integral 560
, Submersion 121
, Sylvester's 376
, Tietze's Extension 608
, Weierstrass' Approximation 216,
667
Theorema Egregium 409
theory of special relativity 756
thermodynamics 290
Tietze's Extension Theorem 608
tope, standard (« + 1)- 274
topology 10
—, algebraic 307, 483, 550, 563
—, relative 10
toroid 349
torsion 159
torus, ^-dimensional 696
total derivative 43
— symbol 241

798
Index
totally bounded 209
trace 39, 527
tractrix 357, 685
transfonuation, canonical 772
—, orthogonal 218
transition mapping 118
transport equation 451
transpose matrix 39
transversal 172
transverse axis 330
— intersection 396
triangle inequality 4
— inequality, reverse 4
trisection 326
tubular neighborhood 319
— neighborhood, integration over 698
umbrella, Whitney's 309
uncertainty 645
— relations, Heisenberg's 645
uniform continuity 28
— convergence 82
uniformly continuous mapping 28
unit hyperboloid 390
— sphere 124
— tangent bundle 323
unitary operator 663
unknown 97
unrolling of cone 684
upper Riemann integral 426
— sum 425
Uiysohn's Lemma 608
value, critical 60
vaiiance 617
vector analysis 527
— field 268, 404, 527
— field, gradient 59, 527
— field, harmonic 546
— field, Newton 529
— field, Poynting 748
— potential 547
— space 2
vector-valued function 12
ViUarceau's circles 327
Viviani's solid 681
volume, (f-dimensional 492
—, of Jordan measurable set 429
—, of rectangle 423
volume-preserving C*
diffeomorphism 445
Wallis'product 184, 627
wave equation 276, 749
— equation, inhomogeneous 761
— function 644
— operator 755, 756
Weierstrass' p function 185
— Approximation Theorem 216,
667
Weingarten mapping 157
Whitney's umbrella 309
Whittaker's equation 641
winding number 595, 736
Wronskian 233
Young's inequality 353
Zeeman effect 272
zero, simple 101
zero-set 108
zeta function, Riemann's 191, 197,
612, 622
zonal spherical function 271