\input fremtex
%\fullsize
\smallprint
\atUEssex
\def\Res{\mathop{\text{Res}}\displaylimits}
\def\Ln{\mathop{\text{Ln}}}
\font\twelverm=cmr12
\def\lectureend#1{\discrversionA{\hfill{\twelverm #1}}{}}
\def\omitted#1{\discrversionA{\hfill{\twelverm omitted #1}}{}}
\Centerline{\bf MA302 Complex Variables}
\Centerline{\smc D.H.Fremlin}
\oldfootnote{}{These notes are made available to students on the
understanding that they have not been fully checked for errors. They
are not intended to provide a substitute for attendance at lectures.
If you notice a mistake, please tell
the lecturer!}
\filename{cv2n6.tex}
\versiondate{14.1.99}
\noindent{\bf 6. Trick Integrals I: trigonometric polynomials and the
unit circle}
\medskip
In the next four chapters I describe methods of calculating certain
integrals and infinite sums using the calculus of residues. The first
gives a (fairly) quick method of dealing with certain integrals between
$0$ and $2\pi$.
\medskip
{\bf 6A Example} Consider the integral
$\int_0^{2\pi}\Bover{\sin\theta}{2+\sin\theta}d\theta$.
We can convert this into an integral around the unit circle, as follows:
$$\int_{\Gamma}{{{1\over{2i}}(z-{1\over z})}\over
{2+{1\over{2i}}(z-{1\over z})}}{{dz}\over{iz}}
=\int_0^{2\pi}{{\sin\theta}\over{2+\sin\theta}}d\theta,$$
\noindent parametrizing $\Gamma$ by $z=e^{i\theta}$ for $0\le\theta\le
2\pi$, $dz=ie^{i\theta}d\theta$, so that
$\Bover1{2i}(z-{1\over z})=\Bover1{2i}(e^{i\theta}-e^{-i\theta})
=\sin\theta$ and $\Bover{dz}{iz}=d\theta$.
Accordingly we have
$$\int_0^{2\pi}{{\sin\theta}\over{2+\sin\theta}}d\theta
=\int_{\Gamma}{{{1\over{2i}}(z-{1\over z})}\over
{2+{1\over{2i}}(z-{1\over z})}}{{dz}\over{iz}}
=\int_{\Gamma}{{z^2-1}\over{iz(z^2+4iz-1)}}dz.$$
\noindent Now this is a suitable case for the calculus of residues.
Setting $f(z)=\Bover{z^2-1}{iz(z^2+4iz-1)}$ we see that $f$ has simple
poles at $0$ and $i(-2\pm\sqrt{3})$, of which only $0$ and
$i(-2+\sqrt{3})$ lie inside the unit circle; so we need to calculate
$$\eqalign{\Res_0f+\Res_{i(-2+\sqrt{3})}f
&={{-1}\over{i(-1)}}+{{(i(-2+\sqrt{3}))^2-1}\over{i(i(-2+\sqrt{3}))
(i(-2+\sqrt{3})+2i+i\sqrt{3})}}\cr
&={1\over i}+{{-7+4\sqrt{3}-1}\over{(2-\sqrt{3})2i\sqrt{3}}}\cr
&=-i+{{4\sqrt{3}-8}\over{2i(2\sqrt{3}-3)}}\cr
&=-i-2i{{(\sqrt{3}-2)(-2\sqrt{3}-3)}\over{-12+9}}\cr
&=-i+{{2i}\over 3}\sqrt{3}
=i({{2\sqrt{3}}\over 3}-1).\cr}$$
\noindent To get the integral, we must multiply by $2\pi i$, getting
$$2\pi(1-{{2\sqrt{3}}\over 3})\bumpeq -0.97.$$
%checked with DERIVE
\medskip
{\bf 6B Remarks (a)} This method can, in principle, be applied to any
integral of the form $\int_0^{2\pi}{\Bover{p(\sin\theta,\cos\theta)}
{q(\sin\theta,\cos\theta)}}d\theta$, where $p$ and $q$ are polynomials,
using the translations
$\cos\theta={\Bover12}(e^{i\theta}+e^{-i\theta})={\Bover12}(z+{\Bover1z})$,
$\sin\theta={\Bover1{2i}}(z-{\Bover1z})$,
$d\theta={\Bover{dz}{iz}}$. Of course the method will work only when $q$ is such that $q(\sin\theta,\cos\theta)$ is never zero, because we must be sure that the function $f$ we integrate has no singularities lying on the unit circle.
\medskip
{\bf (b)} A $\sin k\theta$ or $\cos k\theta$ can be treated as
${\Bover1{2i}}(e^{ik\theta}-e^{-ik\theta})
={\Bover1{2i}}(z^k-{\Bover1{z^k}})$ or
${\Bover1{2}}(e^{ik\theta}+e^{-ik\theta})
={\Bover1{2}}(z^k+{\Bover1{z^k}})$.
\medskip
{\bf (c)} Note however that every integral of this type can also be done
with the substitution $t=\tan{\Bover{\theta}2}$,
$\sin\theta={\Bover{2t}{1+t^2}}$, $\cos\theta={\Bover{1-t^2}{1+t^2}}$,
$d\theta={\Bover2{1+t^2}}dt$, if we first change the range of integration
to $\int_{-\pi}^{\pi}$, so that $\tan{\Bover{\theta}2}$ is well-defined
on the whole (open) interval.
\medskip
{\bf (d)} As suggested in (c), note that all integrals which can be
dealt with by this method at all are of functions which are periodic
with period $2\pi$, so that $\int_0^{2\pi}f(\theta)d\theta
=\int_{-\pi}^{\pi}f(\theta)d\theta=
=\int_{\alpha}^{\alpha+2\pi}f(\theta)d\theta$ for all real $\alpha$.
\medskip
{\bf (e)} It is also well worth while looking for elementary
substitutions or other tricks which will simplify the algebra you will
need. For example:
\medskip
{\bf 6C Example} To calculate $\int_0^{2\pi}{\Bover a{a^2+\sin^2t}}dt$,
where $a>0$.
Start with the substitution $\theta=2t$, so that $\sin^2t={
\Bover{1-\cos
2t}2}={\Bover{1-\cos\theta}2}$, $dt={\Bover{d\theta}2}$ and the
integral becomes
\centerline{$\int_0^{4\pi}{\Bover a{a^2+{1\over 2}(1-\cos\theta)}}
{\Bover{d\theta}2}=2\int_0^{2\pi}{\Bover a{2a^2+1-\cos\theta}}d\theta$.}
\noindent Taking $\Gamma$ to be the unit circle as usual, this is
\centerline{$2\int_{\Gamma}{\Bover a{2a^2+1-{1\over 2}(z+{1\over z})}}
{\Bover{dz}{iz}}
=\int_{\Gamma}{\Bover{4a}{i((4a^2+2)z-z^2-1}}dz
=\int_{\Gamma}{\Bover{4ia}{z^2-(4a^2+2)z+1}}dz$.}
\noindent Setting $f(z)={\Bover{4ia}{z^2-(4a^2+2)z+1}}$, $f$ has simple
poles at $z_0=2a^2+1+2a\sqrt{a^2+1}$, $z_1=2a^2+1-2a\sqrt{a^2+1}$, of
which only
$z_1$ lies within the unit circle; the residue there is
${\Bover{4ia}{z_1-z_0}}
={\Bover{4ia}{-4a\sqrt{a^2+1}}}=-{\Bover i{\sqrt{a^2+1}}}$.
Multiplying by $2\pi i$ we get $\Bover{2\pi}{\sqrt{a^2+1}}$ as the value
of the integral.
%checked with DERIVE
\medskip
{\bf 6D} Another trick worth remembering is the real- or imaginary-part
method:
\medskip
\noindent{\bf Example} To find $\int_0^{2\pi}{\Bover{\cos
n\theta}{3+\cos\theta}}d\theta$.
Express this as
\centerline{$\int_0^{2\pi}
{\Bover{\Real(e^{in\theta})}{3+\cos\theta}}d\theta
=\Real\int_0^{2\pi}{\Bover{e^{in\theta}}{3+\cos\theta}}d\theta$.}
\noindent Now if we have $z=e^{i\theta}$ as usual, $e^{in\theta}$
becomes $z^n$, so we find ourselves with
\centerline{$\Real\int_{\Gamma}{\Bover{z^n}{3+{1\over 2}(z+{1\over
z})}}{\Bover{dz}{iz}}
=\Real\int_{\Gamma}{\Bover{2z^n}{i(z^2+6z+1)}}dz
$.}
\noindent Setting $f(z)={\Bover{2z^n}{i(z^2+6z+1)}}$, then, {\it
provided} $n\ge 0$, $f$ has singularities only at $z_0=-3+2\sqrt{2}$,
$z_1=-3-2\sqrt{2}$, of which only $z_0$ lies within the unit circle, and
the residue at $z_0$ is ${\Bover{2z_0^n}{i(z_0-z_1)}}
={\Bover{-2i}{4\sqrt{2}}}z_0^n=-{\Bover{i\sqrt{2}}4}z_0^n$.
Multiplying by $2\pi i$, the integral is
${\Bover{\pi\sqrt{2}}2}(-3+2\sqrt{2})^n$.
%checked with DERIVE
Of course this is real, so
this is also $\int_0^{2\pi}{\Bover{\cos n\theta}{3+\cos\theta}}d\theta$.
This argument works only for $n\ge 0$. But of course $\cos$ is an even
function, so for general $n$ we have
\centerline{$\int_0^{2\pi}{\Bover{\cos n\theta}{3+\cos\theta}}d\theta
=\int_0^{2\pi}{\Bover{\cos |n|\theta}{3+\cos\theta}}d\theta
={\Bover{\pi\sqrt{2}}2}(-3+2\sqrt{2})^{|n|}$.}
Note that discovering that
\centerline{$\int_{\Gamma}{\Bover{z^n}{3+{1\over 2}(z+{1\over
z})}}{\Bover{dz}{iz}}
=\int_0^{2\pi}{\Bover{e^{in\theta}}{3+\cos\theta}}d\theta$}
\noindent was real implied that its imaginary part was zero, that is,
that
\centerline{$\int_0^{2\pi}{\Bover{\sin n\theta}{3+\cos\theta}}d\theta
=0$}
\noindent for $n\ge 0$ (and therefore for all integers $n$). This is
of course true, because ${\Bover{\sin n\theta}{3+\cos\theta}}$ is an odd
function, so its integral over $[0,2\pi]$, which must be equal to its
integral over $[-\pi,\pi]$, is surely $0$.
Note also that $\lim_{n\to\infty}\int_0^{2\pi}{\Bover{\cos
n\theta}{3+\cos\theta}}d\theta=0$. It is generally true that for
any integrable function $h$ on $[0,2\pi]$,
$\sum_{n=0}^{\infty}|\int_0^{2\pi}h(\theta)\cos n\theta
\,d\theta|^2<\infty$, so that
$\lim_{n\to\infty}\int_0^{2\pi}h(\theta)\cos n\theta\,d\theta=0$.
\discrpage
\filename{cv2n7.tex}
\versiondate{28.1.99}
\noindent{\bf 7. Trick integrals II: Big Semicircle Method}
\medskip
The next method is a very much more powerful and useful technique which
handles integrals of the type
$\int_{-\infty}^{\infty}{\Bover{p(x)}{q(x)}}\cos\lambda x\,dx$,
$\int_{-\infty}^{\infty}{\Bover{p(x)}{q(x)}}\sin\lambda x\,dx$ where $p$
and $q$ are polynomials with real coefficients.
The idea is to express these as the real and imaginary parts of
$f(x)={\Bover{p(x)}{q(x)}}e^{i\lambda x}$, and to integrate $f$ around a
large contour which runs from $-R$ to $R$ along the axis and then back
from $R$ to $-R$ along a semi-circle, centre $0$, in the upper
half-plane. The integral along the straight-line portion will be
$\int_{-R}^Rf(x)dx$, which will approximate the integral
$\int_{-\infty}^{\infty}f(x)dx$ which we seek; the integral along the
whole contour will, we hope, be calculable using the residues of the
complex function $f(z)={\Bover{p(z)}{q(z)}}e^{i\lambda z}$; and it
remains only to deal with the integral along the curved part.
Typically, there will be no hope of getting an exact expression for
this, and the best we can do is to find an adequate approximation as $R\to\infty$;
in fact, in all the cases I shall look at here, the limit as
$R\to\infty$ of the integral along the curved section is zero.
A basic tool in dealing with this is the following.
\medskip
{\bf 7A Lemma (a)} If $\lambda>0$, then
$\int_0^{\pi}e^{-\lambda\sin\theta}d\theta\le{\Bover{\pi}{\lambda}}$.
\quad{\bf (b)} If $\lambda$, $R>0$ and $f$ is a continuous
complex-valued function defined (at least) on the semicircle $\Gamma_R$
running from $R$ to $-R$ in the upper half-plane, then
\centerline{$|\int_{\Gamma_R}f(z)e^{i\lambda
z}dz|\le{\Bover{\pi}{\lambda}}\sup_{0\le\theta\le\pi}|f(Re^{i\theta})|$.}
\medskip
\noindent{\bf proof (a)} Because $\sin\theta$ is symmetric about
${\Bover{\pi}2}$,
\def\Caption{$\sin \theta$ for $0\le\theta\le\pi$}
\picture{cv2n7a1}{52.21pt}
%domain -1./4. scale 1. dash&gap 0.2/0.2
\medskip
\noindent so is $e^{-\lambda\sin \theta}$:
\def\Caption{$e^{-\sin\theta}$ for $0\le\theta\le\pi$}
\picture{cv2n7a2}{55.34pt}
%domain -1./4. scale 1. dash&gap 0.2/0.2
%cv2n7a2.for
\medskip
\noindent Accordingly
\centerline{$\int_0^{\pi}e^{-\lambda\sin\theta}d\theta
=2\int_0^{\pi/2}e^{-\lambda\sin\theta}d\theta$.}
\noindent Now for $0\le\theta\le{\Bover{\pi}2}$,
$\sin\theta\ge {\Bover2{\pi}}\theta$ -- see diagram --
\def\Caption{$\sin\theta$ and ${\Bover2{\pi}}\theta$ for
$0\le\theta\le{\Bover{\pi}2}$}
\picture{cv2n7a3}{49.69pt}
%domain -0.5/2. scale 1. dash&gap 0.1/0.1
\medskip
\noindent so
\centerline{$e^{-\lambda\sin\theta}\le e^{-2\lambda\theta/\pi}$,}
\noindent and
\centerline{$\int_0^{\pi/2}e^{-\lambda\sin\theta}d\theta
\le\int_0^{\pi/2}e^{-2\lambda\theta/\pi}d\theta
=\left[-{\Bover{\pi}{2\lambda}}e^{-2\lambda\theta/\pi}\right]_0^{\pi/2}
={\Bover{\pi}{2\lambda}}(1-e^{-\lambda})
\le{\Bover{\pi}{2\lambda}}$.}
\noindent Multiplying by $2$,
\centerline{$\int_0^{\pi}e^{-\lambda\sin\theta}d\theta
\le{\Bover{\pi}{\lambda}}$.}
\medskip
{\bf (b)} Set $M=\sup_{0\le\theta\le \pi}|f(Re^{i\theta}|$.
Parametrizing $\Gamma_R$ by $z=Re^{i\theta}$ for $0\le\theta\le\pi$,
$dz=iRe^{i\theta}d\theta$, we get
$$\eqalign{|\int_{\Gamma_R}f(z)e^{i\lambda z}dz|
&=|\int_0^{\pi}f(Re^{i\theta})e^{i\lambda
R(\cos\theta+i\sin\theta)}iRe^{i\theta}d\theta|\cr
&\le\int_0^{\pi}|f(Re^{i\theta})e^{i\lambda
R(\cos\theta+i\sin\theta)}iRe^{i\theta}|d\theta\cr
&=\int_0^{\pi}|f(Re^{i\theta})|
|e^{i\lambda R\cos\theta-\lambda R\sin\theta}||iRe^{i\theta}|d\theta\cr
&=\int_0^{\pi}|f(Re^{i\theta})|
e^{-\lambda R\sin\theta}R\,d\theta\cr
&\le\int_0^{\pi}M
e^{-\lambda R\sin\theta}R\,d\theta\cr
&=MR\int_0^{\pi}e^{-\lambda R\sin\theta}d\theta\cr
&\le MR{{\pi}\over{\lambda R}}\cr
&={{\pi}\over{\lambda}}M,\cr}$$
\noindent as claimed.
\medskip
{\bf 7B Corollary} (\lq{\bf Jordan's Lemma}') Let $f$ be a continuous
complex-valued function such that $\dom f$ includes $\{z:|z|\ge
R_0,\,\Imag z\ge 0\}$ for some $R_0\ge 0$. Set
$M_R=\sup_{0\le\theta\le\pi}|f(Re^{i\theta})|$ for $R\ge R_0$, and
suppose that $\lim_{R\to\infty}M_R=0$. Then, for any $\lambda > 0$,
\centerline{$\lim_{R\to\infty}\int_{\Gamma_R}f(z)e^{i\lambda z}dz=0$,}
\noindent where $\Gamma_R$ is the semicircle running from $R$ to $-R$ in
the upper half-plane.
\medskip
\noindent{\bf proof} By 7Ab,
\centerline{$|\int_{\Gamma_R}f(z)e^{i\lambda
z}dz|\le{\Bover{\pi}{\lambda}}M_R\to 0$.}
\medskip
{\bf 7C Example} To find $\int_{-\infty}^{\infty}{\Bover{x\sin\lambda
x}{x^2+1}}$ for real $\lambda$.
I will try to set the main argument out in a standard form with four sections: (a) the idea; (b) the parametrizations, with a little algebra; (c) residues and the calculation of the whole integral; (d) the assembly of parts, with appropriate estimates of the limits required.
\medskip
{\bf (a)} The idea is to consider the integral of
${\Bover z{z^2+1}}e^{i\lambda z}$ around the
contour consisting of the straight-line segment $[-R,R]$ and the
semicircle $\Gamma_R$ from $R$ to $-R$ in the upper half-plane, where
$R$ is large.
\medskip
{\bf (b)} We can use the standard parametrization of $[-R,R]$ by $z=x$ for $-R\le x\le R$, $dz=dx$ so the first part of the integral is
$$\eqalign{I_1(R)&=\int_{-R}^R{{xe^{i\lambda x}}\over{x^2+1}}dx\cr
&=\int_{-R}^R{{x\cos\lambda x}\over{x^2+1}}dx
+i\int_{-R}^R{{x\sin\lambda x}\over{x^2+1}}dx\cr
&=i\int_{-R}^R{{x\sin\lambda x}\over{x^2+1}}dx\cr}$$
\noindent because ${\Bover{x\cos\lambda x}{x^2+1}}$ is an odd function.
Writing $I_2(R)$ for the other part we have
$$I_2(R)=\int_{\Gamma_R}{{ze^{i\lambda z}}\over{z^2+1}}dz.$$
\lectureend{98/3}
\medskip
{\bf (c)} The singularities of ${\Bover z{z^2+1}}e^{i\lambda z}$
are when $z^2+1=0$,
that is, $z=\pm i$; of these, only $i$ lies within the contour (taking
$R>1$), so we get
$$I_1(R)+I_2(R)=2\pi i\Res_{z=i}{z\over{z^2+1}}e^{i\lambda z}.$$
\noindent Now ${\Bover z{z^2+1}}={\Bover z{(z+i)(z-i)}}$, so
$$\eqalign{\Res_{z=i}{z\over{z^2+1}}e^{i\lambda z}
&=\lim_{z\to i}(z-i){z\over{(z+i)(z-i)}}e^{i\lambda z}\cr
&=\lim_{z\to i}{z\over{z+i}}e^{i\lambda z}\cr
&={i\over{2i}}e^{-\lambda}\cr
&={1\over 2}e^{-\lambda},\cr}$$
\noindent and
$$I_1(R)+I_2(R)=\pi ie^{-\lambda}.$$
\medskip
{\bf (d)} Next, if $|z|=R>1$, then
$$|{z\over{z^2+1}}|={{|z|}\over{|z^2+1|}}\le{{|z|}\over{|z^2|-1}}
={R\over{R^2-1}}\to 0\text{ as }R\to\infty,$$
\noindent so by Jordan's Lemma
$$\lim_{R\to\infty}I_2(R)=0$$
\noindent {\it provided} $\lambda>0$. For the moment, therefore, let
us take it that $\lambda >0$. In this case
$$\eqalign{i\int_{-\infty}^{\infty}{{x\sin\lambda x}\over{x^2+1}}
&=\lim_{R\to\infty}I_1(R)\cr
&=\pi ie^{-\lambda},\cr}$$
\noindent that is,
$$\int_{-\infty}^{\infty}{{x\sin \lambda x}\over{x^2+1}}dx
=\pi e^{-\lambda}.$$
\medskip
{\bf (e)} The argument in (d) required us to assume $\lambda >0$. What about other
$\lambda$? If $\lambda=0$ then we have
$$\int_{-\infty}^{\infty}{{x\sin \lambda x}\over{x^2+1}}dx
=\int_{-\infty}^{\infty}0\,dx=0,$$
\noindent while if $\lambda<0$ then
$$\int_{-\infty}^{\infty}{{x\sin \lambda x}\over{x^2+1}}dx
=-\int_{-\infty}^{\infty}{{x\sin (-\lambda) x}\over{x^2+1}}dx
=-\pi e^{-(-\lambda)}=-\pi e^{\lambda}.$$
\medskip
{\bf 7D Notes (a)} Observe that for $\lambda\le 0$ the integral
$\int_{-\infty}^{\infty}{\Bover{x\sin\lambda x}{x^2+1}}$ is no longer
$2\pi i\Res_{z=i}{\Bover z{z^2+1}}e^{i\lambda z}$; this is because
$\lim_{R\to\infty}\int_{\Gamma_R}{\Bover z{z^2+1}}e^{i\lambda z}dz$ is no
longer $0$. When using Jordan's Lemma you {\it must} remember to check
that you have $\lambda > 0$.
\medskip
{\bf (b)} Because ${\Bover{x\sin\lambda x}{x^2+1}}$ is an even function of
$x$, we have
$$\eqalign{\int_0^{\infty}{{x\sin\lambda x}\over{x^2+1}}dx
&={1\over 2}\int_{-\infty}^{\infty}{{x\sin\lambda x}\over {x^2+1}}dx\cr
&={{\pi}\over 2}e^{-\lambda}\text{ if }\lambda>0,\cr
&=0\text{ if }\lambda=0,\cr
&=-{{\pi}\over 2}e^{\lambda}\text{ if }\lambda<0.\cr}$$
\noindent Plotting this as a function of $\lambda$, we have
\def\Caption{$\int_0^{\infty}{\Bover{x\sin\lambda x}{x^2+1}}dx$}
\picture{cv2n7d}{88pt}
%domain -4./4. scale 1
\medskip
\noindent (This is of course an odd function of $\lambda$.)
\medskip
{\bf (c)} The method calculated $\int_{-\infty}^{\infty}{
\Bover{xe^{i\lambda
x}}{x^2+1}}dx$ (for $\lambda>0$). The fact that the integral came
out purely imaginary shows that $\int_{-\infty}^{\infty}{
\Bover{x\cos\lambda
x}{x^2+1}}dx=0$. This is obvious, because
${\Bover{x\cos\lambda x}{x^2+1}}$ is an odd function of $x$.
\medskip
{\bf (d)} To my surprise, MAPLE doesn't seem to be able to do this
one. It can do $\int_0^{\infty}\bover{\sin x}{x}dx$ (below).
\medskip
{\bf (e)} In the form above, the method works for integrals of the form
$\int_{-\infty}^{\infty}f(x)\sin\lambda x\,dx$,
$\int_{-\infty}^{\infty}f(x)\cos\lambda x\,dx$ where $f(x)$ is of the
form ${\Bover{p(x)}{q(x)}}$, both $p$ and $q$ being real polynomials and the
degree of $q$ being greater than the degree of $p$ (so that
$\lim_{R\to\infty}\sup_{0\le\theta\le\pi}|f(Re^{i\theta})|=0$); we also
need to have $q(x)\ne 0$ for all real $x$, as otherwise we can have a
singularity arising on the contours we wish to integrate around. All
other singularities can be avoided by taking the semicircle
large enough to enclose all those in the upper half-plane.
To actually calculate the residues we need, of course, we normally have
to be able to factorize $q$.
Variants of the method, however, can give further information. I show how
(sometimes!) we can cope with a singularity on the real axis.
\medskip
{\bf 7E Example} To find $\int_0^{\infty}{\Bover{\sin x}x}dx$.
\medskip
{\bf (a)} Begin by noting that the singularity at $0$ is not necessarily
disastrous, because $\lim_{x\to 0}{\Bover{\sin x}x}$ is finite;
accordingly this will not give rise to any difficulty in defining the
real integral, though it will certainly affect the complex integral we
shall need to consider.
This time, we integrate ${\Bover{e^{iz}}z}$ ({\it not} ${\Bover{\sin z}z}$!)
around a more complex contour, designed to avoid the singularity
at $0$:
\def\Caption{Dodging $0$}
\picture{cv2n7e}{85.05pt}
%domain -3.5/3.5 scale 1 delete annotations
\medskip
\noindent In this contour, we first proceed along the straight line
segment $[-R,-\epsilon]$ (where $0<\epsilon\ll 1\ll R$), next along
the semicircle $\Delta_{\epsilon}$ from $-\epsilon$ to $\epsilon$ in the
upper half-plane, then along the straight-line segment $[\epsilon,R]$,
and finally return along the semicircle $\Gamma_R$ in the upper
half-plane from $R$ to $-R$.
\medskip
{\bf (b)} Now we consider the
four pieces:
\quad(i) ($[-R,-\epsilon]$) Parametrize with $z=x$ for $-R\le x\le-\epsilon$, $dz=dx$,
$$\eqalignno{I_1(\epsilon,R)
&=\int_{-R}^{-\epsilon}{{e^{ix}}\over x}dx\cr
&=\int_R^{\epsilon}{{e^{-iy}}\over{-y}}(-dy)\cr
\noalign{\noindent (substituting $x=-y$, $dx=-dy$)}
&=-\int_{\epsilon}^R{{e^{-iy}}\over y}dy\cr
&=-\int_{\epsilon}^R{{e^{-ix}}\over x}dx.\cr}$$
\quad(ii) ($\Delta_{\epsilon}$) For reasons which will appear later, it's not yet helpful to parametrize this, so let us just write
$$I_2(\epsilon)=\int_{\Delta_{\epsilon}}{{e^{iz}}\over z}dz.$$
\quad(iii) ($[\epsilon,R]$) Parametrize with $z=x$ for $\epsilon\le x\le R$, $dz=dx$,
$$I_3(\epsilon,R)=\int_{\epsilon}^R{{e^{ix}}\over x}dx.$$
\quad(iv) ($\Gamma_R$)
$$I_4(R)=\int_{\Gamma_R}{{e^{iz}}\over z}dz.$$
\medskip
{\bf (c)} The integrand ${\Bover{e^{iz}}z}$
has no singularities inside the contour, so by Cauchy's theorem
$$I_1(\epsilon,R)+I_2(\epsilon)+I_3(\epsilon,R)+I_4(R)=0.$$
\medskip
{\bf (d)} Now consider the various parts of this sum.
We have
$$\eqalign{I_1(\epsilon,R)+I_3(\epsilon,R)
&=-\int_{\epsilon}^R{{e^{-ix}}\over x}dx
+\int_{\epsilon}^R{{e^{ix}}\over x}dx\cr
&=\int_{\epsilon}^R{{e^{ix}-e^{-ix}}\over x}dx\cr
&=2i\int_{\epsilon}^R{{\sin x}\over x}dx.\cr}$$
\noindent Next,
$$\lim_{R\to\infty}I_4(R)=0$$
\noindent by Jordan's Lemma, because
$\lim_{R\to\infty}\sup_{0\le\theta\le\pi}|{\Bover1{Re^{i\theta}}}|=
\lim_{R\to\infty}{\Bover1R}=0$. Letting $R\to\infty$, we have
$$2i\int_{\epsilon}^{\infty}{{\sin x}\over x}dx+I_2(\epsilon)=0.$$
\lectureend{98/4}
Accordingly the next stage is to find $\lim_{\epsilon\downarrow
0}I_2(\epsilon)$. There are important general theorems to handle such
problems; they are difficult; I recommend the following
approach. The point is that we have a Laurent series for
${\Bover{e^{iz}}z}$:
$$\eqalign{{{e^{iz}}\over z}
&={1\over z}(1+iz-{{z^2}\over{2!}}-i{{z^3}\over{3!}}+\ldots)\cr
&={1\over z}+(i-{1\over 2}z-{i\over 6}z^2+\ldots)\cr
&={1\over z}+g(z)\text{ say,}\cr}$$
\noindent where $g$ is an analytic function defined at and near $0$ (in
fact, everywhere). Now because $0\in\dom g$, we know that $\lim_{z\to
0}|g(z)|=|\lim_{z\to 0}g(z)|=|g(0)|$ exists, so for all $z$ small enough
we shall have $|g(z)|\le|g(0)|+1=M$ say. Accordingly, for all
$\epsilon$ small enough,
$$|\int_{\Delta_{\epsilon}}g(z)dz|\le M\times\text{length of
}\Delta_{\epsilon}=M\pi\epsilon,$$
\noindent and
$$\lim_{\epsilon\downarrow 0}\int_{\Delta_{\epsilon}}g(z)dz=0.$$
\noindent Thus
$$\lim_{\epsilon\downarrow 0}\int_{\Delta_{\epsilon}}{{e^{iz}}\over z}dz
=\lim_{\epsilon\downarrow 0}\int_{\Delta_{\epsilon}}{1\over z}dz.$$
\noindent But this last expression is easy to calculate; parametrizing
$\Delta_{\epsilon}$ with $z=\epsilon e^{-i\theta}$ for $-\pi\le\theta\le
0$, $dz=-i\epsilon e^{-i\theta}d\theta$, we have
$$\int_{\Delta_{\epsilon}}{1\over z}dz
=\int_{-\pi}^0{1\over{\epsilon e^{-i\theta}}}(-i\epsilon e^{-i\theta})
d\theta
=\int_{-\pi}^0(-i)d\theta=-i\pi.$$
\noindent Accordingly
$$\lim_{\epsilon\downarrow 0}I_2(\epsilon)=\lim_{\epsilon\downarrow 0}
(-i\pi)=-i\pi.$$
\noindent Putting this together with the results above,
$$\lim_{\epsilon\downarrow 0}2i\int_{\epsilon}^{\infty}{{\sin x}\over
x}dx-i\pi=0,$$
\noindent and
$$\int_0^{\infty}{{\sin x}\over x}dx={{\pi}\over 2}.$$
\medskip
{\bf 7F Notes (a)} This is a basic formula for the theory of Fourier
transforms.
\medskip
{\bf (b)} We could have worked through the whole thing with
$\int_{0}^{\infty}{\Bover{\sin\lambda x}x}dx$, where $\lambda>0$, and I
suppose it would be good for you to do this. However the substitution
$y=\lambda x$ reduces this at once to the case $\lambda = 1$.
(Question: what happens if $\lambda<0$?)
\medskip
{\bf (c)} The method is good, in principle, for other singularities on
the real axis. But note that the reason for the convergence of
$I_2(\epsilon)$ was the fact that $0$ was a simple pole of
${\Bover{e^{iz}}z}$, so that we could reduce
$\lim\int_{\Delta_{\epsilon}}{\Bover{e^{iz}}z}dz$ to
$\lim\int_{\Delta_{\epsilon}}{\Bover{d_1}z}dz=-i\pi d_1$. If it had been a
pole of order $2$, or worse, we should have had to worry about terms of
the form
$$\eqalign{\int_{\Delta_{\epsilon}}{1\over {z^2}}dz
&=\int_{-\pi}^0{1\over{\epsilon^2e^{-2i\theta}}}
(-i\epsilon e^{-i\theta})d\theta\cr
&=-{i\over{\epsilon}}\int_{-\pi}^0e^{i\theta}d\theta\cr
&=-{2\over{\epsilon}},\cr}$$
\noindent which has no finite limit.
\medskip
{\bf (d)} There is also the point that what we find ourselves
calculating, at best, is
$$\lim_{\epsilon\downarrow 0}\lim_{R\to\infty}
\bigl(\int_{-R}^{-\epsilon}f(x)e^{i\lambda x}dx
+\int_{\epsilon}^Rf(x)e^{i\lambda x}dx\bigr).$$
\noindent There are cases in which this limit can exist even though
neither of the limits
\centerline{$\lim_{\epsilon\downarrow 0}\lim_{R\to\infty}
\int_{-R}^{-\epsilon}f(x)e^{i\lambda x}dx$,
\quad$\lim_{\epsilon\downarrow
0}\lim_{R\to\infty}\int_{\epsilon}^Rf(x)e^{i\lambda x}dx$}
\noindent exists, and in such cases we should not like to speak of
$\int_0^{\infty}f(x)e^{i\lambda x}dx$.
\medskip
{\bf (d)} To summarise: we can hope to use this singularity-dodging
technique to find
$\int_{-\infty}^{\infty}{\Bover{p(x)}{q(x)}}\sin\lambda
x$ or $\int_{-\infty}^{\infty}{\Bover{p(x)}{q(x)}}\cos\lambda x$ in cases
where $p$ and $q$ are real polynomials, the degree of $q$ is greater than
that of $p$, $\lambda>0$, any singularities of ${\Bover{p(x)}{q(x)}}$ on
the real axis are either removable or simple poles, and the integrand
${\Bover{p(x)}{q(x)}}\sin\lambda x$ or
${\Bover{p(x)}{q(x)}}\cos\lambda x$ has a finite limit at each of these
singularities.
\discrpage
\filename{cv2n8.tex}
\versiondate{14.1.99}
\def\Lnstar{\mathop{\text{Ln}^*}}
\def\degree{\mathop{\text{degree}}}
\noindent{\bf 8. Trick integrals III: Logarithms and Fractional Powers}
Further versions of big-semicircle methods give us a variety of
integrals of the form $\int_0^{\infty}f(x)\ln x\,dx$,
$\int_0^{\infty}x^{\alpha}f(x)dx$.
\medskip
{\bf 8A Complex Logarithms} Recall the following properties of the
complex function $\Ln$:
\quad $\dom(\Ln)=\{z:z\text{ is not real and }\le 0\}$;
\quad $\Ln(z)=\ln(|z|)+i\arg z$;
\quad $\exp\Ln z=z$ for every $z\in\dom(\Ln)$;
\quad $\Ln\exp z=x$ iff $|\Imag(z)|<\pi$.
\noindent Note also that there is something arbitrary about this. We
choose to cut the plane along the negative real axis from $-\infty$ to
$0$, and a very wide variety of alternative cuts are possible. Indeed,
if $\theta$ is any real number, we can define a
function $\Lnstar$ by setting
$$\Lnstar(z)=\Ln(e^{-i\theta}z)+i\theta$$
\noindent whenever this is defined; that is,
$\Lnstar(z)=\ln(|z|)+i\arg({e^{-i\theta}z})+i\theta$
except when ${e^{-i\theta}z}$ is real and
negative; and $\Lnstar$ will have many of the properties of $\Ln$ --
for instance, $\Lnstar(x)=\ln x$ for real $x>0$ (at least when
$-\pi<\theta<\pi$), $\exp\Lnstar z=z$ for
every $z\in\dom(\Lnstar)$, and $\Lnstar'(z)={\Bover1z}$.
But it will be undefined along a different cut -- this time, the \lq
bad' line will be $\{z:e^{-i\theta z}\text{ is real }\le 0\}=\{z:\arg z=
\theta w\pm\pi\}$ if $-\pi<\theta<\pi$
-- and this may be more convenient. A typical choice might
be $\theta=\Bover{\pi}2$;
then $\Lnstar$ is defined except on the negative imaginary
axis.
\medskip
{\bf 8B Example} To find $\int_0^{\infty}{\Bover{\ln x}{1+x^2+x^4}}dx$.
\medskip
{\bf (a)} The idea is to integrate ${\Bover{\Ln z}{1+z^2+z^4}}$ along a
curve
consisting of a large semicircle with a small pimple at $0$, as in 7E;
that is, a straight-line segment $[\epsilon,R]$, where $0<\epsilon\ll
1\ll R$, a semicircle $\Gamma_R$ from $R$ to $-R$ in the upper
half-plane, a straight-line segment $[-R,-\epsilon]$ and a semicircle
$\Delta_{\epsilon}$, again in the upper half-plane, from $-\epsilon$ to
$\epsilon$. Of course there is an immediate difficulty: the function
$\Ln$ is not defined on the segment $[-R,-\epsilon]$. There are two
possible methods of dealing with this: (i) to replace $\Ln$ by
$\Lnstar$, where $\Lnstar z=\Ln(z/i)+\Ln(i)=\Ln(-iz)+{\Bover{i\pi}2}$,
so that $\Lnstar$ agrees with $\Ln$ on the upper half-plane and in
addition sets $\Lnstar z=\ln(|z|)+i\pi$ if $z$ is real and negative;
(ii) to regard the interval $[-R,-\epsilon]$ as replaced by one lying
infinitesimally above the axis, so that again we can treat $\Ln z$ as
being $\ln(|z|)+i\pi$. Of these, (i) seems easier to justify in terms
of the methods I have used in the course so far. On the other hand,
(ii) is easier to use in practice, and is a technique which will be
useful later in this chapter. Thus I suggest that you regard the
left-hand straight section as running from $-\sqrt{R^2-\eta^2}+\eta i$
to $-\sqrt{\epsilon^2-\eta^2}+\eta i$, where $0<\eta\ll\epsilon$, so
that as a rule we take the limit as $\eta\downarrow 0$ in formulae
before even writing them down.
\def\Caption{Keeping clear of trouble}
\picture{cv2n8b}{94.03pt}
%domain -3.5/3.5 scale 1 annotations deleted
\medskip
{\bf (b)} Now for the value of the integral, calculated by using
residues. The singularities of the integrand ${\Bover{\Ln
z}{1+z^2+z^4}}$ are at the solutions of $1+z^2+z^4=0$, that is,
when
$$z^2={{-1\pm\sqrt{1-4}}\over 2}=-{1\over 2}\pm i{{\sqrt{3}}\over
2}=e^{\pm 2\pi i/3},$$
\noindent so that
we have four simple poles at
$$e^{\pi i/3},\quad -e^{\pi i/3}=e^{-2\pi i/3},\quad e^{-\pi i/3},
\quad e^{2\pi i/3}.$$
\noindent Of these, the first and last lie within the contour, so these
are the residues we need to calculate. Now if $a$ is any zero of
$1+z^2+z^4$,
$$\Res_{z=a}{{\Ln z}\over{1+z^2+z^4}}=\lim_{z\to a}(z-a){{\Ln
z}\over{1+z^2+z^4}}=\Ln a\lim_{z\to a}{1\over{4z^3+2z}}
={{\Ln a}\over{4a^3+2a}}.$$
\noindent Taking $a=z_1=e^{i\pi/3}={\Bover12}+i{\Bover{\sqrt{3}}2}$, we
get
$$\eqalign{\Res_{z=z_1}{{\Ln z}\over{1+z^2+z^4}}
&={{\Ln z_1}\over{4z_1^3+2z_1}}\cr
&={{{i\pi}\over 3}\over{-4+1+i\sqrt{3}}}\cr
&={{i\pi(-3-i\sqrt{3})}\over{3.12}}\cr
&={{\pi(\sqrt{3}-3i)}\over{36}}.\cr}$$
\noindent Similarly, taking $a=z_2=e^{2i\pi/3}
=-{\Bover12}+i{\Bover{\sqrt{3}}2}$, we get
$$\eqalign{\Res_{z=z_2}{{\Ln z}\over{1+z^2+z^4}}
&={{\Ln z_2}\over{4z_2^3+2z_2}}\cr
&={{{2i\pi}\over 3}\over{4-1+i\sqrt{3}}}\cr
&={{2i\pi(3-i\sqrt{3})}\over{3.12}}\cr
&={{2\pi(\sqrt{3}+3i)}\over{36}}.\cr}$$
\noindent Adding, the sum of the residues is
$${{\pi(\sqrt{3}-3i+2\sqrt{3}+6i)}\over{36}}
={{\pi(3\sqrt{3}+3i)}\over{36}}
={{\pi}\over {12}}(i+\sqrt{3});$$
\noindent multiplying by $2\pi i$, the integral round the whole contour
is
$${{\pi^2}\over 6}(i\sqrt{3}-1).$$
\medskip
{\bf (c)} Now let us consider the four sections of our contour:
\quad(i) ($[\epsilon,R]$) Parametrize with $z=x$ for $\epsilon\le x\le
R$, $dz=dx$ so
$$I_1(\epsilon,R)=\int_{\epsilon}^R{{\ln x}\over{1+x^2+x^4}}dx.$$
\quad(ii) ($\Gamma_R$) We shall not need to parametrize this part of the
curve; write
$$I_2(R)=\int_{\Gamma_R}{{\Ln z}\over{1+z^2+z^4}}dz.$$
\quad(iii) (\lq$[-R,-\epsilon]$') Strictly speaking, parametrize with
$z=x+i\eta$ for $-\sqrt{R^2-\eta^2}\le x\le-\sqrt{\epsilon^2-\eta^2}$;
but it is safe to move directly to the limit, so that we have $z=x$ for
$-R\le x\le-\epsilon$, $dz=dx$ and
$$\eqalignno{I_3(\epsilon,R)
&=\int_{-R}^{-\epsilon}{{\ln(|x|)+i\pi}\over{1+x^2+x^4}}dx\cr
&=\int_R^{\epsilon}{{\ln y+i\pi}\over{1+y+y^2}}(-dy)\cr
\noalign{\noindent (substituting $x=-y$ for $\epsilon\le y\le x$)}
&=\int_{\epsilon}^R{{\ln x+i\pi}\over{1+x^2+x^4}}dx.\cr}$$
\quad(iv) ($\Delta_{\epsilon}$) Again, there is no virtue in writing
down a parametrization, so set
$$I_4(\epsilon)=\int_{\Delta_{\epsilon}}{{\Ln z}\over{1+z^2+z^4}}dz.$$
We have
$$I_1(\epsilon,R)+I_2(R)+I_3(\epsilon,R)+I_4(\epsilon)=
{{\pi^2}\over 6}(i\sqrt{3}-1).$$
\medskip
{\bf (d)} From the expressions above,
$$\eqalign{I_1(\epsilon,R)+I_3(\epsilon,R)
&=\int_{\epsilon}^R{{\ln x}\over{1+x^2+x^4}}dx+\int_{\epsilon}^R{{\ln
x+i\pi}\over{1+x^2+x^4}}dx\cr
&=2\int_{\epsilon}^R{{\ln x}\over{1+x^2+x^4}}dx
+i\pi\int_{\epsilon}^R{1\over{1+x^2+x^4}}dx.\cr}$$
\noindent Now consider the two curved sections.
(i) To estimate $I_2(R)$, observe that the length of $\Gamma_R$ is $\pi
R$ (strictly speaking, slightly less), while the greatest value
of ${\Bover{\Ln z}{1+z^2+z^4}}$ on $\Gamma_R$ is at most
$$M_R={{\ln R+\pi}\over{R^4-R^2-1}};$$
\noindent this is because any $z$ on $\Gamma_R$ is of the form
$Re^{i\theta}$ for some $\theta\in\coint{0,\pi}$, so that
$$|\Ln z|=|\ln(|z|)+i\arg z|=|\ln R+i\arg z|\le\ln R+|\arg z|
\le\ln R+\pi,$$
\noindent while $|z^4+z^2+1|\ge|z^4|-|z^2+1|\ge|z^4|-|z^2|-1=R^4-R^2-1$.
Accordingly
$$\eqalign{|I_2(R)|
&=|\int_{\Gamma_R}{{\Ln z}\over{1+z^2+z^4}}dz|\cr
&\le \pi RM_R\cr
&={{\pi R(\ln R+\pi)}\over{R^4-R^2-1}}\cr
&\to 0\text{ as }R\to\infty.\cr}$$
(ii) To estimate $I_4(\epsilon)$, observe that the length of
$\Delta_{\epsilon}$ is $\pi\epsilon$ (strictly speaking, slightly
less), while the greatest value of ${\Bover{\Ln z}{1+z^2+z^4}}$ on
$\Delta_{\epsilon}$ is at most
$$M_\epsilon={{|\ln\epsilon|+\pi}\over{1-\epsilon^2-\epsilon^4}};$$
\noindent this is because any $z$ on $\Delta_{\epsilon}$ is of the form
$\epsilon e^{i\theta}$ for some $\theta\in\coint{0,\pi}$, so that
$$|\Ln z|=|\ln(|z|)+i\arg z|=|\ln \epsilon+i\arg
z|\le|\ln\epsilon|+|\arg z|
\le|\ln\epsilon|+\pi,$$
\noindent while $|z^4+z^2+1|\ge1-|z^4+z^2|\ge1-|z^2|-|z^4|
=1-\epsilon^2-\epsilon^4$.
Accordingly
$$\eqalign{|I_2(\epsilon)|
&=|\int_{\Delta_{\epsilon}}{{\Ln z}\over{1+z^2+z^4}}dz|\cr
&\le \pi \epsilon M_\epsilon\cr
&={{\pi \epsilon(|\ln \epsilon|+\pi)}\over{1-\epsilon^2-\epsilon^4}}\cr
&\to 0\text{ as }\epsilon\downarrow 0\cr}$$
\noindent because $\lim_{\epsilon\downarrow 0}\epsilon\ln\epsilon = 0$.
This means that, taking $R\to\infty$,
$$2\int_{\epsilon}^{\infty}{{\ln x}\over{1+x^2+x^4}}dx
+i\pi\int_{\epsilon}^{\infty}{1\over{1+x^2+x^4}}dx
+I_4(\epsilon)={{\pi^2}\over 6}(i\sqrt{3}-1),$$
\noindent and taking $\epsilon\downarrow 0$
$$2\int_0^{\infty}{{\ln x}\over{1+x^2+x^4}}dx
+i\pi\int_0^{\infty}{1\over{1+x^2+x^4}}dx
={{\pi^2}\over 6}(i\sqrt{3}-1).$$
\noindent Taking the real parts, we have
$$2\int_0^{\infty}{{\ln x}\over{1+x^2+x^4}}dx
=-{{\pi^2}\over 6},$$
\noindent that is,
$$\int_0^{\infty}{{\ln x}\over{1+x^2+x^4}}dx
=-{{\pi^2}\over {12}}\bumpeq 0.82.$$
\noindent Taking imaginary parts, we also get
$$\pi\int_0^{\infty}{1\over{1+x^2+x^4}}dx
={{\pi^2}\over 6}\sqrt{3},$$
\noindent that is
$$\int_0^{\infty}{1\over{1+x^2+x^4}}dx
={{\pi\sqrt{3}}\over 6}\bumpeq 0.91.$$
%checked with DERIVE
\medskip
{\bf 8C Notes (a)} This is not of course an efficient method of
calculating $\int_0^{\infty}{\Bover1{1+x^2+x^4}}dx$; but it is valuable
as a check.
\medskip
{\bf (b)} The method will work, in principle, for integrals of the form
$\int_0^{\infty}{\Bover{p(x)}{q(x)}}\ln x\,dx$ where $p$ and $q$ are {\it
even} real polynomials, the degree of $p$ is less than that of $q$, and
$q$ has no real zeroes. We need to have even functions because the
integral $I_3$ depends on ${\Bover{p(x)}{q(x)}}$ for negative $x$ and we
need to combine it with the integral $I_1$. We need to have the degree
of $p$ less than that of $q$ in order to make
$\lim_{R\to\infty}I_2(R)=0$; when we estimate
$$|I_2(R)|\le\pi RM_R \le\pi R(\ln
R+\pi)\sup_{|z|=R}{{|p(z)|}\over{|q(z)|}},$$
\noindent we need an extra $R^2$ on the bottom to control the $R\ln R$
on the top, and because $p$ and $q$ are both supposed to be even this
will work whenever the degree of $p$ is less than that of $q$.
\medskip
{\bf (c)} As in 7E-7F, we can cope with a zero of $q$ on the real axis
if it is compensated for by a zero of $\ln$; that is, if it is a simple
zero at $1$. So, for instance, the method would give an expression for
$\int_0^{\infty}{\Bover{\ln x}{x^2-1}}dx$.
\medskip
{\bf (d)} In fact we can cope with integrals of the form
$\int_0^{\infty}{\Bover{p(x)}{q(x)}}\ln x\,dx$ in further cases, in which
$p$ and $q$ are not even, by using different contours.
\medskip
{\bf $^*$8D Example} To find $\int_0^{\infty}{\Bover{\ln x}{1+x^3}}dx$.
\medskip
{\bf (a)} This time we take the integral of ${\Bover{\Ln z}{1+z^3}}$
around a contour consisting of a straight-line segment $[\epsilon,R]$,
one third of a circle (centre 0) from $R$ to $Re^{2\pi i/3}$, a straight
line from $Re^{2\pi i/3}$ to $\epsilon e^{2\pi i/3}$, and one third of a
circle (again with centre $0$) back from $\epsilon e^{2\pi i/3}$ to
$\epsilon$.
\def\Caption{A third of a cake}
\picture{cv2n8d}{85.05pt}
%domain -3.5/3.5 scale 1. annotations deleted
\medskip
\noindent The reason why this will work is that the two
straight-line segments again match up, and this is so because our
multiplier ${\Bover1{1+z^3}}$ is a function of $z^3$.
\medskip
{\bf (b)} The singularities of the integrand ${\Bover{\Ln z}{1+z^3}}$ are at the cube roots of $-1$, that is, at $-1$ and ${\Bover12}
\pm{\Bover{\sqrt{3}}2}i$. Of these, only $z_1={\Bover12}
+{\Bover{\sqrt{3}}2}i=e^{\pi i/3}$ lies inside the contour, so that is
the residue we need. Now
$$\eqalign{\Res_{z=z_1}{{\Ln z}\over{1+z^3}}
&=\lim_{z\to z_1}(z-z_1){{\Ln z}\over{1+z^3}}\cr
&=\Ln z_1\lim_{z\to z_1}{1\over{3z^2}}\cr
&={{\Ln z_1}\over{3z_1^2}}\cr
&={{i\pi}\over 3}{1\over 3}e^{-i2\pi/3}\cr
&={{\pi}\over 9}e^{-i\pi/6}.\cr}$$
\noindent Multiplying by $2\pi i$, the integral is
$${{2\pi^2}\over 9}e^{\pi i/3}={{\pi^2}\over 9}(1+i\sqrt{3}).$$
\medskip
{\bf (c)} Working through the parametrizations, we have
$$\eqalignno{I_1(\epsilon,R)&=\int_{\epsilon}^R{{\ln x}\over
{1+x^3}}dx;\cr
I_2(R)&=\int_{\Gamma_R}{{\Ln z}\over{1+z^3}}dz\cr
\noalign{\noindent (this time writing $\Gamma_R$ for one third of the circle,
rather than one half);}
I_3(\epsilon,R)
&=\int_{-R}^{-\epsilon}{{\ln|x|+{{2\pi i}\over 3}}\over{1-x^3}}
(-e^{2\pi i/3})dx\cr
\noalign{\noindent(parametrizing the second straight line with
$z=-e^{2\pi i/3}x$ for $-R\le x\le-\epsilon$, $dz=-e^{2\pi i/3}dx$)}
&=-e^{2\pi i/3}\int_R^{\epsilon}{{\ln y+{{2\pi i}\over 3}}\over{1+y^3}}(-dy)\cr
\noalign{\noindent(substituting $x=-y$)}
&={1\over 2}(1-i\sqrt{3})\int_{\epsilon}^R{{\ln x}\over{1+x^3}}dx
+{{\pi}\over 3}(\sqrt{3}+i)\int_{\epsilon}^R{1\over{1+x^3}}dx;\cr
I_4(\epsilon)&=\int_{\Delta_{\epsilon}}{{\Ln z}\over{1+z^3}}dz,\cr}$$
\noindent writing $\Delta_{\epsilon}$ for the final one-third circle.
From (b), we have
$$I_1(\epsilon,R)+I_2(R)+I_3(\epsilon,R)+I_4(\epsilon)
={{\pi^2}\over 9}(1+i\sqrt{3}).$$
\medskip
{\bf (d)} Using the expressions above
$$I_1(\epsilon,R)+I_3(\epsilon,R)
=({1\over 2}(3-i\sqrt{3})\int_{\epsilon}^R{{\ln x}\over{1+x^3}}dx
+{{\pi}\over 3}(\sqrt{3}+i)\int_{\epsilon}^R{1\over{1+x^3}}dx.$$
(i) To estimate $I_2(R)$, observe that the length of $\Gamma_R$ is $2\pi
R/3$, while the greatest value of $|{\Bover{\Ln z}{1+z^3}}|$ on
$\Gamma_R$ is at most
$$M_R={{\ln R+{{2\pi}\over 3}}\over{R^3-1}};$$
\noindent so that
$$|I_2(R)|\le{{2\pi R}\over 3}\,{{\ln R+{{2\pi}\over 3}}\over{R^3-1}}
\to 0$$
\noindent as $R\to\infty$.
(ii) To estimate $I_4(\epsilon)$, observe that the length of
$\Delta_{\epsilon}$ is $2\pi\epsilon/3$, while the greatest value of
$|{\Bover{\Ln z}{1+z^3}}|$ on $\Delta_{\epsilon}$ is at most
$$M_{\epsilon}={{|\ln\epsilon|+{{2\pi}\over 3}}\over{1-\epsilon^3}};$$
\noindent so that
$$|I_4(\epsilon)|\le{{2\pi \epsilon}\over 3}\,{{|\ln\epsilon|+{{2\pi}\over
3}}\over{1-\epsilon^3}}
\to 0$$
\noindent as $\epsilon\downarrow 0$.
Taking the limit as $R\to\infty$,
$${1\over 2}(3-i\sqrt{3})
\int_{\epsilon}^{\infty}{{\ln x}\over{1+x^3}}dx
+{{\pi}\over 3}(\sqrt{3}+i)\int_{\epsilon}^{\infty}{1\over{1+x^3}}dx
+I_4(\epsilon)={{\pi^2}\over 9}(1+i\sqrt{3});$$
\noindent taking the limit as $\epsilon\downarrow 0$,
$${1\over 2}(3-i\sqrt{3})\int_{0}^{\infty}{{\ln x}\over{1+x^3}}dx
+{{\pi}\over 3}(\sqrt{3}+i)\int_{0}^{\infty}{1\over{1+x^3}}dx
={{\pi^2}\over 9}(1+i\sqrt{3}).$$
\noindent Setting $J=\int_0^{\infty}{\Bover{\ln x}{1+x^3}}dx$, $K=\int_0
^{\infty}{\Bover1{1+x^3}}dx$, and taking real and imaginary parts, we have
$$\eqalign{{3\over 2}J+{{\pi\sqrt{3}}\over 3}K&={{\pi^2}\over 9},\cr
-{{\sqrt{3}}\over 2}J+{{\pi}\over 3}K&={{\pi^2\sqrt{3}}\over 9}.\cr}$$
\noindent Solving these, we get
$$\eqalign{\int_0^{\infty}{{\ln x}\over{1+x^3}}dx&=J=-{{2\pi^2}\over {27}}
\bumpeq -0.73,\cr
\int_0^{\infty}{1\over{1+x^3}}dx&=K={{2\pi\sqrt{3}}\over 9}
\bumpeq 1.21.\cr}$$
\omitted{98}
\medskip
{\bf $^*$8E Note} The point of the method is that the emphasis on \lq even' functions in 8B-8C is slightly misplaced; really what we need is a function ${\Bover{p(z^n)}{q(z^n)}}$ where $n\ge 2$, $p$ and $q$ are real polynomials, the degree of $q$ is greater than that of $p$, and $q$ has no zeroes on $\coint{0,\infty}$ except perhaps a simple zero at $1$.
We can then use a contour around ${\Bover1n}$th of a cake, returning along the line $\{z:\arg z={\Bover{2\pi i}n}\}$.
\omitted{98}
\medskip
{\bf 8F Fractional powers} Recall that for any $z\in\dom\Ln$ and any $a\in\Bbb C$ we can define
$$z^a=\exp(a\Ln z).$$
\noindent For real $\alpha$, this becomes
$$z^{\alpha}=\exp(\alpha(\ln|z|+i\arg z))=e^{\alpha\ln|z|}e^{i\alpha\arg z}=|z|^{\alpha}e^{i\alpha\arg z}.$$
\noindent This is defined except along a slit consisting of the negative real axis. If $z$ is close to this slit, then
$$\eqalign{z^{\alpha}
&\bumpeq|z|^{\alpha}e^{i\alpha\pi}\text{ if }\Imag z>0,\cr
&\bumpeq|z|^{\alpha}e^{-i\alpha\pi}\text{ if }\Imag z<0.\cr}$$
\noindent If $\alpha$ is an integer, then $e^{i\alpha\pi}
=e^{-i\alpha\pi}=(-1)^{\alpha}$, so these two expressions match up
to give $z^{\alpha}=(-1)^{\alpha}|z|^{\alpha}$ along the negative real axis (there is still a problem at $0$ if $\alpha<0$); but if $\alpha$ is not an integer, they differ, and $z^{\alpha}$ jumps by a factor of $e^{2i\alpha\pi}$ as $z$ crosses the slit.
As with $\Ln$, it is essentially arbitrary where we set the slit. For any
real number $\theta$, we could examine
$$e^{i\theta a}(e^{-i\theta}z)^a$$
\noindent and get a version of $z^a$ undefined along a different slit.
The fact that (for non-integer $\alpha$) $z^{\alpha}$ jumps across the negative real axis gives us a chance to calculate integrals of the form $\int_0^{\infty}x^{\alpha}f(x)dx$ for certain functions $f$.
\medskip
{\bf 8G Example} To find $\int_0^{\infty}{\Bover{x^{\alpha}}{1+x}}dx$.
\medskip
{\bf (a)} The idea is to integrate ${\Bover{z^{\alpha}}{1-z}}$
around a contour which proceeds from (nearly) $-R$ to (nearly)
$-\epsilon$ in the upper half-plane, circumnavigates the origin
along the circle of radius $\epsilon$, returns to (nearly) $-R$
in the lower half-plane, and then goes round a big circle to get
back to $-R$ above the axis:
\def\Caption{Going both ways}
\picture{cv2n8g}{170.10pt}
%domain -3.5/3.5 scale 1. annotations deleted
\medskip
\noindent We think of the straight-line sections as being so close to the axis that we can treat them as if they actually lay on it, but interpreting $z^{\alpha}$ as $|z|^{\alpha}e^{\alpha\pi i}$ for the upper section and as $|z|^{\alpha}e^{-\alpha\pi i}$ for the lower section.
\medskip
{\bf (b)} The only singularity of the integrand
${\Bover{z^{\alpha}}{1-z}}$ is at $1$, where it has a simple pole with
residue $-1^{\alpha}=-1$; so the integral is $-2\pi i$.
\medskip
{\bf (c)} Now for the four parts of the contour:
\quad(i) (\lq upper $[-R,-\epsilon]$') Parametrizing as $z=x$ for $-R\le x\le-\epsilon$, we have
$$\eqalign{I_1(\epsilon,R)
&=\int_{-R}^{-\epsilon}{{|x|^{\alpha}e^{\alpha\pi i}}\over{1-x}}dx\cr
&=e^{\alpha\pi i}\int_R^{\epsilon}{{y^{\alpha}}\over{1+y}}(-dy)\cr
&=e^{\alpha\pi i}\int_{\epsilon}^R{{y^{\alpha}}\over{1+y}}dy\cr
&=e^{\alpha\pi i}\int_{\epsilon}^R{{x^{\alpha}}\over{1+x}}dx.\cr}$$
\quad(ii) (inner circle) Write this part of the contour as $\Delta_{\epsilon}$, and
$$I_2(\epsilon)=\int_{\Delta_{\epsilon}}{{z^{\alpha}}\over{1-z}}dz.$$
\quad(iii) (\lq lower$[-R,-\epsilon]$') Parametrize as $z=-x$ for $\epsilon\le x\le R$, $dz=-dx$, to get
$$\eqalign{I_3(\epsilon,R)
&=\int_{\epsilon}^R{{x^{\alpha}e^{-\alpha\pi i}}\over{1+x}}(-dx)\cr
&=-e^{-\alpha\pi i}\int_{\epsilon}^R{{x^{\alpha}}\over{1+x}}dx.\cr}$$
\quad(iv) (outer circle) Write this part of the contour as $\Gamma_R$,
and
$$I_4(R)=\int_{\Gamma_{R}}{{z^{\alpha}}\over{1-z}}dz.$$
From (b), we have
$$I_1(\epsilon,R)+I_2(\epsilon)+I_3(\epsilon,R)+I_4(R)=-2\pi i.$$
\medskip
{\bf (d)(i)} Putting the expressions for $I_1$ and $I_3$ together, we have
$$\eqalign{I_1(\epsilon,R)+I_3(\epsilon,R)
&=(e^{\alpha\pi i}-e^{-\alpha\pi i})
\int_{\epsilon}^R{{x^{\alpha}}\over{1+x}}dx\cr
&=2i\sin\alpha\pi\int_{\epsilon}^R{{x^{\alpha}}\over{1+x}}dx.\cr}$$
\medskip
\quad{\bf (ii)} To estimate $I_2(\epsilon)$, we observe that the length of $\Delta_{\epsilon}$ is slightly less than $2\pi\epsilon$, while the greatest value of $|{{z^{\alpha}}\over{1-z}}|$ on $\Delta_{\epsilon}$ is at most $M_{\epsilon}={{\epsilon^{\alpha}}\over{1-\epsilon}}$. So
$$|I_2(\epsilon)|\le 2\pi\epsilon M_{\epsilon}={{2\pi\epsilon^{\alpha+1}}
\over{1-\epsilon}}\to 0$$
\noindent as $\epsilon\downarrow 0$ {\it provided} that $\alpha+1>0$, that is, that $\alpha>-1$.
\medskip
\quad{\bf (iii)} To estimate $I_4(R)$, we observe that the length of $\Gamma_{R}$ is slightly less than $2\pi R$, while the greatest value of $|{
\Bover{z^{\alpha}}{1-z}}|$ on $\Gamma_{R}$ is at most $M_{R}={
\Bover{R^{\alpha}}{R-1}}$. So
$$|I_2(R)|\le 2\pi RM_{R}={{2\pi R^{\alpha+1}}
\over{R-1}}\to 0$$
\noindent as $R\to\infty$ {\it provided} that $\alpha+1<1$, that is, that $\alpha<0$.
\medskip
\quad{\bf (iv)} Supposing that $-1<\alpha<0$, therefore, we may take the limit as $R\to\infty$ to get
$$2i\sin\alpha\pi\int_{\epsilon}^{\infty}{{x^{\alpha}}\over{1+x}}dx
+I_2(\epsilon)=-2\pi i,$$
\noindent and now the limit as $\epsilon\downarrow 0$ to get
$$2i\sin\alpha\pi\int_{0}^{\infty}{{x^{\alpha}}\over{1+x}}dx=-2\pi i,$$
\noindent so that
$$\int_{0}^{\infty}{{x^{\alpha}}\over{1+x}}dx=-{{\pi}\over {\sin\alpha\pi}},$$
\noindent provided that $-1<\alpha<0$. You may think this looks prettier if we write it as
$$\int_{0}^{\infty}{{x^{-\gamma}}\over{1+x}}dx={{\pi}\over {\sin\gamma\pi}}\text{ if }0<\gamma<1.$$
\medskip
{\bf 8H Notes (a)} In this case, the requirement $-1<\alpha<0$
imposed by the proof corresponds to a real restriction on the admissible values of $\alpha$. For if $\alpha\ge 0$ then $\int_1^{\infty}{
\Bover{x^{\alpha}}{1+x}}dx=\infty$ (compare with $\int_1^{\infty}{\Bover{1}
{2x}}dx=\infty$), while if $\alpha\le -1$ then $\int_0^1{
\Bover{x^{\alpha}}{1+x}}dx=\infty$ (compare with $\int_0^1{\Bover12}x^{\alpha}dx=\infty$).
\medskip
{\bf (b)} The method is applicable to any integral of the form $\int_0^{\infty}{\Bover{p(x)}{q(x)}}x^{\alpha}dx$ where $p$ and $q$ are real polymomials, $q$ has no zeroes in $\coint{0,\infty}$,
\centerline{$-1<\alpha<\degree(q)-\degree(p)-1,$}
\noindent and $\alpha$ is not an integer.
We need $\alpha>-1$ to show that $\lim_{\epsilon\downarrow 0}I_2(\epsilon)=0$, and $\alpha<\degree(q)-\degree(p)-1$ to show that $\lim_{R\to\infty}I_4(R)=0$; moreover, just as in 8G, these are exactly the limits needed to ensure that $\int_0^{\infty}{
\Bover{p(x)}{q(x)}}x^{\alpha}dx$ is finite. We need $\alpha\notin\Bbb Z$ to ensure that the two integrals $I_1(\epsilon,R)$, $I_3(\epsilon,R)$ don't cancel out; in fact their sum will always be
$$2i\sin\alpha\pi\int_{\epsilon}^R{{p(x)}\over{q(x}}x^{\alpha}dx,$$
\noindent so integral $\alpha$, corresponding to zero $\sin\alpha\pi$, must involve a zero sum of the residues.
\medskip
{\bf (c)} Of course if the degree of $q$ is two or more above the degree of $p$, there will be $\alpha\in \Bbb Z$ for which the integral is finite. The integral cannot be calculated directly by this method (and, as always, there will be better methods of calculating the integral of a rational function ${
\Bover{p(x)}{q(x)}}x^n$), but if $0\le n<\degree(q)-\degree(p)$ we shall have
$$\int_0^{\infty}{{p(x)}\over{q(x)}}x^ndx
=\lim_{\alpha\to n}\int_0^{\infty}{{p(x)}\over{q(x)}}x^{\alpha}dx,$$
\noindent so that we can find the exceptional values quickly if we need them.
\discrpage
\filename{cv2n9.tex}
\versiondate{25.2.99}
\def\degree{\mathop{\text{degree}}}
\noindent{\bf 9. Infinite sums: the cotangent trick}
I describe here a method of calculating certain infinite sums. It
depends on special properties of the cotangent function.
\medskip
{\bf 9A Cotangents (a)} Recall that
$$\cot z
={{\cos z}\over{\sin z}}
={{e^{iz}+e^{-iz}}\over 2}\bigg/{{e^{iz}-e^{-iz}}\over{2i}}
=i{{e^{iz}+e^{-iz}}\over{e^{iz}-e^{-iz}}}
=i{{1+e^{-2iz}}\over{1-e^{-2iz}}}$$
\noindent is an analytic function defined except when $e^{-2iz}=1$, that
is, except when $-2iz$ is an integral multiple of $2\pi i$, that is,
except when $z=n\pi$ for some $n\in\Bbb Z$.
Because $\exp$ is periodic with period $2\pi i$, $\cot$ is periodic with
period $\pi$. At each singularity,
$$\Res_{z=n\pi}\cot z=\lim_{z\to n\pi}(z-n\pi)\cot z
=\lim_{z\to n\pi}\cos z{{z-n\pi}\over{\sin z}}
=\cos n\pi\lim_{z\to n\pi}{1\over{\cos z}}
=1.$$
This shows also that $\cot$ has a simple pole at each $n\in\Bbb Z$. It
is necessary sometimes to know the early terms of its Laurent series
near $0$. Because $\cot$ has a simple pole with residue $1$ at $0$, it
is expressible in the form
$$\cot z={1\over z}+c_0+c_1z+c_2z^2+\ldots$$
\noindent for $0<|z|<\pi$. Because $\sin$ is odd and $\cos$ is even,
$\cot$ is odd, so $c_0=c_2=\ldots=0$, and we have
$$\cot z={1\over z}+c_1z+c_3z^3+\ldots.$$
\noindent Multiplying this by the series for $\sin$ we get
$$({1\over z}+c_1z+c_3z^3+\ldots)(z-{{z^3}\over{3!}}
+{{z^5}\over{5!}}-\ldots)
=1-{{z^2}\over{2!}}+{{z^4}\over{4!}}-\ldots;$$
\noindent equating coefficients, we get
\quad (coefficient of $z^2$)\quad $c_1-{\Bover1{3!}}=-{\Bover1{2!}}$ so
$c_1=-{\Bover13}$;
\quad (coefficient of $z^4$)
\quad $c_3-{\Bover1{3!}}c_1+{\Bover1{5!}}={\Bover1{4!}}$ so
$c_3=-{\Bover1{45}}$;
\quad (coefficient of $z^6$)
\quad $c_5-{\Bover1{3!}}c_3+{\Bover1{5!}}c_1-{\Bover1{7!}}
=-{\Bover1{6!}}$ so $c_5=-{\Bover2{945}}$.
\medskip
{\bf (b)} It will in fact be convenient to do calculations with $\cot\pi
z$ rather than with $z$. This has a simple pole at $z=n$ for every
$n\in\Bbb Z$; the residue is
$$\lim_{z\to n}(z-n)\cot\pi z
=\lim_{z\to n}\cos\pi z{{z-n}\over{\sin\pi z}}
=\cos n\pi\lim_{z\to n}{1\over{\pi\cos\pi z}}
={1\over\pi}.$$
\noindent Generally, if $n\in\Bbb Z$ and $f$ is any analytic function
defined at $n$,
$$\eqalign{\Res_{z=n}f(z)\cot\pi z=\lim_{z\to n}(z-n)f(z)\cot\pi z
&=\lim_{z\to n}f(z)\cos\pi z{{z-n}\over{\sin\pi z}}\cr
&=f(n)\cos n\pi\lim_{z\to n}{1\over{\pi\cos\pi z}}
={1\over\pi}f(n).\cr}$$
\lectureend{98/8}
We can also compute the Laurent series of $\cot\pi z$ for $0<|z|<1$;
this is
$${1\over{\pi z}}
-{{\pi}\over 3}z-{{\pi^3}\over{45}}z^3-{{2\pi^5}\over{945}}z^5-\ldots.$$
\noindent This gives us the possibility of computing the residue of
$f(z)\cot\pi z$ at $0$ when $f$ has a pole at $0$; thus
$$\Res_{z=0}{1\over z}\cot\pi z
=\Res_{z=0}{1\over{\pi z^2}}
-{{\pi}\over 3}-{{\pi^3}\over{45}}z^2-{{2\pi^5}\over{945}}z^4
-\ldots=0;$$
$$\Res_{z=0}{1\over{z^2}}\cot\pi z
=\Res_{z=0}{1\over{\pi z^3}}
-{{\pi}\over 3z}-{{\pi^3}\over{45}}z-{{2\pi^5}\over{945}}z^3
-\ldots=-{{\pi}\over 3};$$
$$\Res_{z=0}{1\over z^3}\cot\pi z
=\Res_{z=0}{1\over{\pi z^4}}
-{{\pi}\over{3z^2}}-{{\pi^3}\over{45}}-{{2\pi^5}\over{945}}z^2
-\ldots=0;$$
$$\Res_{z=0}{1\over{z^4}}\cot\pi z
=\Res_{z=0}{1\over{\pi z^5}}
-{{\pi}\over{3z^3}}-{{\pi^3}\over{45z}}-{{2\pi^5}\over{945}}z
-\ldots=-{{\pi^3}\over{45}};$$
$$\Res_{z=0}{1\over{z^5}}\cot\pi z
=\Res_{z=0}{1\over{\pi z^6}}
-{{\pi}\over{3z^4}}-{{\pi^3}\over{45z^2}}-{{2\pi^5}\over{945}}
-\ldots=0;$$
\noindent and so on. If $f$ has a pole of order at most $5$ at $0$, so
that it is expressible as ${\Bover{d_5}{z^5}}+\ldots+{\Bover{d_1}z}
+c_0+c_1z+\ldots$ for $z$ near to $0$, then
$$\Res_{z=0}f(z)\cot\pi z=-{{\pi^3}\over{45}}d_4-{{\pi}\over
3}d_2+c_0.$$
Because $\cot$ is periodic with period $1$, we can use the same ideas to
calculate residues at $n$ for non-zero integers $n$;
if $f$ has a pole of order at most $5$ at $n$, so that it is expressible
as ${\Bover{d_5}{(z-n)^5}}+\ldots+{\Bover{d_1}{z-n}}+c_0+c_1(z-n)+\ldots$
for $z$ near to $n$, then
$$\Res_{z=n}f(z)\cot\pi z=-{{\pi^3}\over{45}}d_4-{{\pi}\over
3}d_2+c_0.$$
\medskip
{\bf 9B Lemma} If {\it either} $|\Imag z|\ge 1$ {\it or} $\Real
z=n+{\Bover12}$ for some integer $n$, then
$$|\cot\pi z|\le{{1+e^{-2\pi}}\over{1-e^{-2\pi}}}\le 2.$$
\medskip
\noindent{\bf proof}
$$|\cot\pi z|=\big|{{e^{i\pi z}+e^{-i\pi z}}\over
{e^{i\pi z}-e^{-i\pi z}}}\big|
=\big|{{1+e^{2\pi iz}}\over{1-e^{2\pi iz}}}\big|
=\big|{{1+e^{-2\pi iz}}\over{1-e^{-2\pi iz}}}\big|.$$
\noindent Now if $z=x+iy$, then $e^{2\pi iz}=e^{-2\pi y}e^{2\pi ix}$, so
$|e^{2\pi iz}|=e^{-2\pi y}$; similarly, $|e^{-2\pi iz}|=e^{2\pi y}$.
{\bf case 1} If $y=\Imag z\ge 1$, then
$$|\cot\pi z|=\big|{{1+e^{2\pi iz}}\over{1-e^{2\pi iz}}}\big|
\le{{1+|e^{2\pi iz}|}\over{1-|e^{2\pi iz}|}}
={{1+e^{-2\pi y}}\over{1-e^{-2\pi y}}}
\le{{1+e^{-2\pi}}\over{1-e^{-2\pi}}}.$$
{\bf case 2} If $y=\Imag z\le -1$, then
$$|\cot\pi z|=\big|{{1+e^{-2\pi iz}}\over{1-e^{-2\pi iz}}}\big|
\le{{1+|e^{-2\pi iz}|}\over{1-|e^{-2\pi iz}|}}
={{1+e^{2\pi y}}\over{1-e^{2\pi y}}}
\le{{1+e^{-2\pi}}\over{1-e^{-2\pi}}}.$$
{\bf case 3} If $x=n+{\Bover12}$, where $n\in\Bbb Z$, then $e^{2\pi i
z}=e^{-2\pi y}e^{2\pi ix}=e^{-2\pi y}e^{2\pi in+\pi i}=-e^{-2\pi y}$, so
$$|\cot\pi z|=\big|{{1+e^{2\pi iz}}\over{1-e^{2\pi iz}}}\big|
=\big|{{1-e^{-2\pi y}}\over{1+e^{-2\pi y}}}\big|<1.$$
Thus in all cases
$$|\cot\pi z|\le
{{1+e^{-2\pi}}\over{1-e^{-2\pi}}};$$
\noindent now $e\ge 1+1+{\Bover12}={\Bover52}$ and $\pi\ge 3$, so
$e^{2\pi}\ge e^6\ge({\Bover52})^6\ge 6^3=216$, and
$${{1+e^{-2\pi}}\over{1-e^{-2\pi}}}
\le{{1+{1\over{216}}}\over{1-{1\over{216}}}}
={{217}\over{215}}=1{2\over{215}}\le 2.$$
\medskip
{\bf 9C Example} To find $\sum_{n=1}^{\infty}{\Bover1{n^2}}$.
\medskip
{\bf (a)} The idea is to integrate ${\Bover1{z^2}}\cot\pi z$ around a
square with corners at $(\pm1\pm i)(m+{\Bover12})$, where $m\ge 1$ is a
large
integer. We shall find that for large $m$ the integral is necessarily
small, so that the sum of the residues inside the contour must also be
small, and this will give the result.
\medskip
{\bf (b)} Let us calculate the residues. There is a singularity at
$0$, from both the ${\Bover1{z^2}}$ and the $\cot\pi z$, and also at $\pm
n$ for integers $n$ between $1$ and $m$; larger integers $n$ lie
outside the contour and don't contribute. The residue at $0$, as
computed in 9A above, is $-{\Bover{\pi}3}$; at each other $n$ with
$-m\le n\le m$ it is ${\Bover1{\pi n^2}}$. Accordingly the sum of the
residues is
$$-{{\pi}\over 3}+2\sum_{n=1}^m{1\over{\pi n^2}},$$
\noindent and the integral will be
$$2\pi i(-{{\pi}\over 3}+2\sum_{n=1}^m{1\over{\pi n^2}}).$$
\medskip
{\bf (c)} To estimate the integral, we observe that the length of the
contour is $8(m+{\Bover12})$, while the maximum value of
$|{\Bover1{z^2}}\cot\pi z|$ on the contour is at most
${\Bover2{m^2}}$, because (by 9B) $|\cot\pi z|\le 2$ for every $z$ on
the contour, while also $|z|\ge m+{\Bover12}$ so
$|{\Bover1{z^2}}|\le{\Bover1{m^2}}$. Accordingly the integral is at most
$$8(m+{1\over 2}){2\over{m^2}}\to 0\text{ as }m\to\infty.$$
\medskip
{\bf (d)} Accordingly
$$\lim_{m\to\infty}2\pi i(-{{\pi}\over 3}+2\sum_{n=1}^m{1\over{\pi
n^2}})=0$$
\noindent and
$$\lim_{m\to\infty}\sum_{n=1}^m{1\over{\pi n^2}}={{\pi}\over 3},$$
\noindent that is,
$$\sum_{n=1}^{\infty}{1\over{n^2}}={{\pi^2}\over 6}\bumpeq 1.64.$$
\medskip
{\bf 9D Note} The method can be adapted to find
$$\sum_{n=k}^{\infty}{{p(-n)}\over{q(-n)}}
+\sum_{n=k}^{\infty}{{p(n)}\over{q(n)}}$$
\noindent whenever $p$ and $q$ are (real or complex) polynomials, the
degree of $q$ is at least $2$ more than the degree of $p$, and $q(n)\ne
0$ for any integer $n$ with $|n|\ge k$, provided (as usual) that we can
factorize $q$, so as to identify the residues of $\cot\pi
z{\Bover{p(z)}{q(z)}}$. The condition
\centerline{$\degree(q)\ge 2+\degree(p)$}
\noindent is what is needed to ensure that the sums are summable, as
well as to ensure that the integral around the square with corners
$(m+{\Bover12})(\pm 1\pm i)$ is small for large $m$. If
${\Bover{p(x)}{q(x)}}$ is even, of course, we can get a sum
$\sum_{n=k}^{\infty}{\Bover{p(n)}{q(n)}}$, as in 9C. But this method
does not (for instance) give us an expression for
$\sum_{n=1}^{\infty}{\Bover1{n^3}}$.
\lectureend{98/9}
\discrpage
\filename{cv2n10.tex}
\versiondate{14.1.99}
\def\Caption{}
\noindent{\bf 10. Conformal Mappings}
I turn now to a quite different subject: the problem of understanding
analytic functions as geometrical transformations.
\medskip
{\bf 10A Definition} If $D\subseteq\Bbb R^2$ is an open set, a function
$f:D\to\Bbb R^2$ is {\bf conformal} if for small objects $A\subseteq D$
the image $f[A]$ of $A$ is approximately the same shape as $A$;
possibly rotated and/or expanded or contracted, but not reflected or
significantly distorted.
\medskip
{\bf 10B Theorem} If $f$ is an analytic function and $D$ is an open
subset of $\{z:z\in\dom f,\,f'(z)\ne 0\}$, then $f$ is conformal on $D$.
\medskip
\noindent{\bf remark} Here I identify $\Bbb C$ with the plane $\Bbb
R^2$.
\medskip
\noindent{\bf idea of proof} Take any $z_0\in D$. Then we have
$$f(z)=f(z_0)+(z-z_0)f'(z_0)+{1\over 2}(z-z_0)^2f''(z_0)+\ldots$$
\noindent for $z$ near $z_0$. If $z$ is very near $z_0$ then
$$f(z)\bumpeq f(z_0)+f'(z_0)(z-z_0)=g(z)\text{ say},$$
\noindent with error small compared with $z-z_0$. So if $A\subseteq D$
is an object lying very close to $z_0$, the image $f[A]$ is closely
approximated by $g[A]$, with errors small compared with the diameters of
$A$ and $f[A]$. But the map $g:\Bbb C\to\Bbb C$ is a
composition of the maps
$$\eqalign{z&\mapsto z-z_0,\cr
z&\mapsto f'(z_0)z,\cr
z&\mapsto f(z_0)+z.\cr}$$
\noindent Of these, the first and last are simple translations, so
certainly leave shapes unaffected, while the middle one is an expansion
or contraction (by a factor of $|f'(z_0)|\ne 0$) with a rotation
(through $\arg f'(z_0)$), so also leaves shapes unchanged. Thus $g$ is
certainly conformal, and $f[A]$ will also be very nearly the same shape
as $A$.
\medskip
{\bf 10C Remarks (a)} Conversely, any conformal map on an open subset of
$\Bbb R^2$ or $\Bbb C$ corresponds to an analytic function with
non-zero derivative.
\medskip
{\bf (b)} Note that if $f:D\to\Bbb C$ is conformal, then near any
$z_0\in D$ the transformation is locally approximated by $z\mapsto
f(z_0)+f'(z_0)(z-z_0)$. In particular, the scale is changed by a
factor of $|f'(z_0)|$ and figures are rotated through an angle of $\arg
f'(z_0)$.
\medskip
{\bf (c)} One of the reasons for taking conformal maps seriously is
their use in
solving partial differential equations, using the following theorem. I
star this because we shall not be going further in this direction, but
it is worth knowing about.
\medskip
{\bf $^*$10D Theorem} Suppose that $D$, $E$ are open sets in $\Bbb R^2$
and that $f:D\to E$ is conformal, $\phi:E\to\Bbb R$ is twice
continuously differentiable. Set $\psi=\phi\circ f$. Then
$${{\partial^2\psi}\over{\partial
x^2}}(a,b)+{{\partial^2\psi}\over{\partial y^2}}(a,b)
=|f'(a+ib)|^2\big({{\partial^2\phi}\over{\partial
x^2}}(f(a,b))+{{\partial^2\phi}\over{\partial y^2}}(f(a,b))\big)$$
\noindent for every $(a,b)\in D$, where $f'$ is the derivative of $f$
when regarded as an analytic function.
In particular, if $\phi$ is {\bf harmonic} (i.e.,
${\Bover{\partial^2\phi}{\partial
x^2}}+{\Bover{\partial^2\phi}{\partial y^2}}=0$), so is $\psi$.
\medskip
{\bf 10E M\"obius transformations} Consider the formula
$$f(z) = {{az+b}\over{cz+d}}$$
\noindent where $a$, $b$, $c$ and $d$ are complex numbers and $c$, $d$
are not both $0$. This defines an analytic function, defined
everywhere if $c=0$ (so $d\ne 0$), and everywhere except at $-d/c$ if
$c\ne 0$. If $ad=bc$ then $f$ is constant, with value ${\Bover ac}
={\Bover bd}$. Otherwise, it is invertible, with inverse given by
$$\eqalign{&w={{az+b}\over{cz+d}}\cr
&\iff w(cz+d)=az+b\cr
&\iff z(cw-a)=b-wd\cr
&\iff z={{b-wd}\over{cw-a}}\cr}$$
\noindent (we cannot have either $cz+d=az+b=0$ or $b-wd=cw-a=0$ because
$ad\ne bc$). Observe that the inverse transformation
$$f^{-1}(w)={{b-wd}\over{cw-a}}$$
\noindent is another transformation of the same kind, this time defined
except at $a/c$ (unless $c=0$, in which case it is defined everywhere).
\lectureend{98/10}
If we look at the derivative of $f$, we get
$$f'(z)={{a(cz+d)-c(az+b)}\over{(cz+d)^2}}={{ad-bc}\over{(cz+d)^2}},$$
\noindent which (still assuming that $ad\ne bc$) is never $0$. So $f$
defines a conformal transformation.
Transformations of this type are called {\bf M\"obius} or {\bf linear
fractional} transformations.
\medskip
{\bf 10F Structure of M\"obius transformations} Let
$f(z)={\Bover{az+b}{cz+d}}$ be a M\"obius transformation. (Remember
that $ad\ne bc$.)
\medskip
\quad{\bf case 1} If $c=0$ then $a\ne 0$, $d\ne 0$ and $f(z)={\Bover
ad}z
+{\Bover bd}$. We can regard this as the composition of the maps
$z\mapsto{\Bover ad}z$ and $z\mapsto z+{\Bover bd}$. The first is a
rotation together with a contraction or expansion; the second is a
translation. Both preserve all shapes; in particular, circles are
taken to circles and straight lines are taken to straight lines.
The transformation is a bijection from $\Bbb C$ to itself.
\medskip
\quad{\bf case 2} Before considering the general case of $c\ne 0$, let
us examine the special case $f(z)={\Bover1z}$ (that is, $a=c=1$,
$b=d=0$). In this case, consider the general equation of a straight
line in the plane
$$2Bx+2Cy+D=0,$$
\noindent where $B$, $C$, $D$ are real and $B$, $C$ are not both $0$,
and the general equation of a circle in the plane
$$(x+B)^2+(y+C)^2=r^2,$$
\noindent where $B$, $C$ are real and $r>0$. These can both be put in
the form
$$A(x^2+y^2)+2Bx+2Cy+D=0,$$
\noindent where $A$, $B$ and $C$ are not all $0$, and if $A\ne 0$ then
$({\Bover B{A}})^2+({\Bover C{A}})^2>{\Bover DA}$, that is,
$B^2+C^2>AD$,
while if $A=0$ then $B^2+C^2>0$ -- thus the condition $B^2+C^2>AD$
ensures a non-trivial figure in all cases, which will be a circle if
$A\ne 0$ and a straight line if $A=0$.
In terms of a complex variable $z$, the equation becomes
$$Az\overline{z}+2B{{z+\overline{z}}\over 2}+2C{{z-\overline{z}}\over
{2i}}+D=0.$$
\noindent Setting $w={\Bover 1z}$ this becomes
$${A\over{w\overline{w}}}+{B}\big({1\over
w}+{1\over{\overline{w}}}\big)
+{C\over{i}}\big({1\over w}-{1\over{\overline{w}}}\big)+D=0,$$
\noindent that is,
$$A+2B\bover{w+\overline{w}}2
-2C\bover{w-\overline{w}}{2i}+Dw\overline{w}=0,$$
\noindent which is an equation of the same sort as before (because
$B^2+(-C)^2>DA$), so again corresponds to either a straight line or a
circle. Moreover, the transformed figure is a straight line iff
$D=0$, that is, iff the original figure passed through $0$; just as the
original figure was a straight line iff $A=0$, that is, iff the
transformed figure passes through $0$.
Summarising, we have
\qquad a straight line through $0$ becomes a straight line through $0$,
\qquad a straight line not through $0$ becomes a circle through $0$,
\qquad a circle through $0$ becomes a straight line not through $0$,
\qquad a circle not through $0$ becomes a circle not through $0$.
\noindent Both the transformation and its inverse are bijections from
$\Bbb C\setminus\{0\}$ to itself.
\medskip
\quad{\bf case 3} Now let us turn to the general case $c\ne 0$. In
this case
$$w={{az+b}\over{cz+d}}={{{a \over c}z+{b\over c}}\over{z+{d\over c}}}
={a\over c}+{{{bc-ad}\over{c^2}}\over{z+{d\over c}}},$$
\noindent so we have the composition of the transformations
\qquad $z\mapsto z+{\Bover dc}$ (translation)
\qquad $z\mapsto{\Bover 1z}$ (inversion, as considered in case 2)
\qquad $z\mapsto{\Bover{bc-ad}{c^2}}z$ (rotation \& magnification)
\qquad $z\mapsto{\Bover ac}+z$ (translation).
\noindent Since each of these transform straight lines and circles into
straight lines and circles, so does the composition.
The {\it pole} of the transformation ${\Bover{az+b}{cz+d}}$ is the point
$-{\Bover dc}$ where it is undefined; straight lines through this point
are transformed into straight lines through ${\Bover ac}$, while circles
through the pole are transformed into straight lines not through
${\Bover ac}$, and other straight lines are transformed into circles
through ${\Bover ac}$. (${\Bover ac}$ is of course the pole of the
inverse transformation $z={\Bover{b-dw}{cw-a}}$.)
The transformation is a bijection from $\Bbb C\setminus\{-\Bover dc\}$
to $\Bbb C\setminus\{-\Bover ac\}$.
\medskip
{\bf Remark} It sometimes helps to think of a M\"obius transformation
$w=\bover{az+b}{cz+d}$ as a map from $\Bbb C\cup\{\infty\}$ to
itself; $z=\infty$ corresponds to $w=\bover{a}{c}$ (or to $w=\infty$
if $c=0$), $w=\infty$ corresponds to $z=-\bover{d}{c}$ (or to
$z=\infty$ if $c=0$). Under this convention, a straight line
behaves like a `circle through $\infty$', so the rule becomes just
`generalized circles are transformed into generalized circles',
and the rest is a matter of checking what $z=\infty$, $w=\infty$ mean.
\medskip
{\bf 10G Techniques for finding particular M\"obius transformations (a)}
If $z_1$, $z_2$, $z_3$ are three distinct points in $\Bbb C$ and $w_1$,
$w_2$, $w_3$ are three distinct points in $\Bbb C$, then the formula
$${{(w-w_1)(w_2-w_3)}\over{(w-w_3)(w_1-w_2)}}
={{(z-z_1)(z_2-z_3)}\over{(z-z_3)(z_1-z_2)}}
$$
\noindent defines a M\"obius transformation which moves $z_1$ to $w_1$,
$z_2$ to $w_2$ and $z_3$ to $w_3$.
\lectureend{98/11}
\medskip
{\bf (b)} In this formula, we can allow \lq$\infty$' as a value for one
of the $w_j$ or one of the $z_j$ (or both), subject to the convention
that ${\Bover{\infty}{\infty}}=1$. To say that \lq$z_1$ moves to
$\infty$' declares that $z_1$ is to be the pole of the transformation;
to say that \lq$\infty$ moves to $w_1$' declares that
$\lim_{|z|\to\infty}w=w_1$; to say that \lq$\infty$ moves to $\infty$'
declares that the transformation is to be of the form $w=az+b$, with no
pole.
Generally, we can think of a M\"obius transformation
$w=\Bover{az+b}{cz+d}$
as a bijection from $\Bbb C\cup\{\infty\}$ to itself; if $c=0$ then
$\infty$ is taken to itself; if $c\ne 0$ then $z=\infty$ corresponds
to $w=\Bover bd$, and $z=-\Bover dc$ corresponds to $w=\infty$.
A straight line is now a \lq circle through $\infty$', so a straight
line in the $z$-plane gets transformed into either a straight line or
a circle through $-\Bover dc$.
\medskip
{\bf (c)} To move a given line $L$ to the unit circle, a point $z_1$ on
$L$ to $1$, and a point $z_0$ not on $L$ to $0$, let $z_{\infty}$ be the
reflection of $z_0$ in $L$, and use the transformation
$$w={{z-z_0}\over{z-z_{\infty}}}\times
{{z_1-z_{\infty}}\over{z_1-z_0}};$$
\noindent note that this is just the formula in (a) with $w_0=0$,
$w_1=1$ and $w_{\infty}=\infty$, so that
$$w={{(w-0)(w_1-\infty)}\over{(1-\infty)(1-0)}}
={{(w-w_0)(w_1-w_{\infty})}\over{(w-w_{\infty})(w_1-w_0)}}
={{(z-z_0)(z_1-z_{\infty})}\over{(z-z_{\infty})(z_1-z_0)}}.
$$
\noindent This transformation can be run backwards to transform the unit
circle into a given line.
\medskip
{\bf (d)} Now to transform lines into other circles, we can use the
transformation in (c) and then a suitable rotation, magnification and
translation.
\medskip
{\bf 10H Examples} I will now present some diagrams which will I hope
help to give a notion of the geometric nature of conformal
transformations.
\medskip
\vfill
\vbox{
{\bf (a)} I begin with $w=f(z)=1/z$. This transforms
\picture{cv10ha1}{228pt}
%confor98.for on f1overz.for, cv2n10ha.dat
\noindent into
\picture{cv10ha2}{225pt}
}%end of vbox
\noindent The letters {\tt A}, {\tt B}, {\tt C} and {\tt D}, all more or
less on the
unit circle in the $z$-diagram, are moved again to the unit circle in
the $w$-diagram; {\tt A} and {\tt C} are inverted
(because $f'(1)=f'(-1)=-1$),
while {\tt B} and {\tt D} remain upright (because $f'(i)=f'(-i)=1$).
{\tt C} and
{\tt E} start inside the circle and finish outside the circle, magnified
as
well as rotated, because $|f'|>1$ inside the circle; while {\tt A} and
{\tt F}
start outside and finish inside, contracted because $|f'|<1$ outside the
unit circle. The axis lines look as if they haven't moved, but in fact
they've been turned inside out, and the upper and lower sections of the
imaginary axis have been swapped.
\vfill
\vbox{
{\bf (b)} Now consider $w={\Bover{3z-2}{2z+2}}$. This transforms
\picture{cv10hb1}{300pt}
%cv2n10hb.for, cv2n10hb.dat
\noindent into
\picture{cv10hb2}{300pt}
}%end of vbox
\noindent Note that $f'(z)={\Bover5{2(z+1)^2}}$. In particular, this
is real and positive for any real $z\ne-1$, so that all points on the
real axis are left right way up.
The real axis remains the real axis, but the two sections (above and
below $-1$) are exchanged left-to-right. The imaginary axis, a line
not through the pole, turns into a circle through the pole $w=\bover32$
of the new diagram. Of the three circles in the $z$-diagram, the
left-hand one, a circle centered on the pole $z=-1$, becomes a circle
centered on the pole $w=\bover32$; the middle one, passing through the
pole $z=-1$, becomes a straight line; and the third, which doesn't
enclose the pole $z=-1$, becomes a circle not enclosing the pole
$w=\bover32$.
\vfill
\vbox{
{\bf (c)} Now consider $w=z^2$. This transforms the grid
\picture{cv10hc1}{220pt}
%fzsq.for, cv2n10hc.dat
\noindent into
\picture{cv10hc2}{248pt}
}%end of vbox
\noindent We have $f(1)=f(-1)$, so {\tt B} and {\tt D} are brought
together; similarly, the letters {\tt A} and {\tt C} are brought
together. $|f'(z)|=2$ for $|z|=1$, so all four are approximately
doubled in size. $f'(1)=2$, so {\tt B} stays the right way up;
$f'(-1)=-2$, so {\tt D} is rotated through $180^{\smallcirc}$;
$f'(i)=2i$, so {\tt C} is rotated
anticlockwise; and $f'(-i)=-2i$, so {\tt A} is rotated clockwise. The
real axis is folded on itself and becomes the positive real axis; the
imaginary axis is also folded to become the negative real axis. {\it
The circle surrounding $0$ now winds round $0$ twice\/}; this is almost
the most characteristic behaviour of the function $z\mapsto z^2$; and
something of this sort will happen with any analytic function $g$ near a
point $z_0$ such that $g'(z_0)=0$, $g''(z_0)\ne 0$, so that $g(z)\bumpeq
a+b(z-z_0)^2$ for $z$ near $z_0$. The vertical line at the left of the
original diagram is bent into a parabola.
At $0$, we have $f'(0)=0$, so {\it the mapping is not conformal}; you
can see that not only has an axis line apparently disappeared, but the
{\tt X} has been dramatically distorted. It has also, of course, been
shrunk.
\lectureend{98/12}
\vfill
\vbox{
{\bf (d)} Now consider $w=\exp z$. This transforms
\picture{cv10hd1}{249pt}
%fexp.for, cv2n10hd.dat
\noindent into
\picture{cv10hd2}{249pt}
}%end of vbox
\noindent Observe that $z=0$ goes to $w=1$; $f'(0)=1$ so the spokes are
not rotated. $z=1$ (the letter {\tt A}) goes to $w=e$; no rotation,
magnification by a
factor of $e$. The line $\Imag z=0$ (letters {\tt C}, {\tt O}, {\tt
A}) goes to the half-line
$\{w:w\text{ is real }>0\}$; the line $\Real z=0$ (letters {\tt G},
{\tt D}, {\tt O}, {\tt B}, {\tt F}) becomes the circle
$|w|=1$, rotating around it, no magnification or contraction (because
$|f'(z)|=1$). The line $\Real z=1$ (past {\tt A} and {\tt E}) becomes
the circle $|w|=e$; the line $\Imag z=1$ (past {\tt B} and {\tt E})
becomes the line $\arg w=1$. The circles $|z|=r$, for $00\}$,
\qquad $z\mapsto z+1$ to transform $D_2$ into $D_3=\{z:\Real z>1\}$,
\qquad $z\mapsto {\Bover1z}$ to transform $D_3$ into $D_4=
\{z:|z-{\Bover12}|<{\Bover12}\}$,
\qquad $z\mapsto 2z-1$ to transform $D_4$ into the open unit disk.
\noindent Putting these together,
$$f(z)=2{1\over{1+(\exp z)^{1/2}}}-1
={{1-\exp{z\over 2}}\over{1+\exp{z\over 2}}}=-\tanh{z\over 4}.$$
\noindent Evidently $z\mapsto f(z)=\tanh{\Bover z4}$ will also map $D$
bijectively onto the unit circle, and the picture will be a little
easier to interpret, because $f'(0)={\Bover14}$ is real and positive, so
the transformed grid is right way up near $0$:
\vbox{
\picture{cv2n10j1}{206pt}
\noindent is transformed to
\picture{cv2n10j2}{285pt}
}%end of vbox
\noindent Note that the domain $\{z:|\Imag z|<\pi\}$ chosen for the
function is comfortably inside the natural domain of
$\tanh{\bover{z}4}$,
the nearest singularities being at $\pm 2\pi i$; so that the curves
remain
reasonably smooth right up to the boundary, except near $w=\pm 1$,
corresponding to $z$ far to the right or left. The lines $\Imag
z=\pm\pi$ correspond respectively to the upper and lower parts of the
circle $|w|=1$.
\medskip
{\bf 10K} It is natural to ask whether the condition \lq$D\ne\Bbb C$' in
the Riemann Mapping Theorem (10I) is
really necessary. Indeed it is, as the following theorem shows:
\medskip
\noindent{\bf Liouville's Theorem} If $f$ is a bounded analytic function
with domain $\Bbb C$, it is constant. In particular, there can be no
analytic surjection $f:\Bbb C\to\{z:|z|<1\}$.
\medskip
\noindent{\bf proof} Let $f:\Bbb C\to\Bbb C$ be a bounded analytic
function; suppose that $|f(z)|\le M$ for every $z\in\Bbb C$. If
$z_0\in\Bbb C$, let $\Gamma_R$ be the circle centre $z_0$ radius $R$,
taken once anticlockwise. Then, by Cauchy's Integral Formula (4K),
$$\eqalign{|f'(z_0)|&=\big|{1\over{2\pi
i}}\int_{\Gamma_R}{{f(\zeta)}\over{(\zeta-z_0)^2}}d\zeta\big|\cr
&\le\Bover1{2\pi}2\pi R\Bover{M}{R^2}
=\Bover{M}{R}.\cr}$$
\noindent As $R$ is arbitrary, $|f'(z_0)|=0$. As $z_0$ is arbitrary,
$f$ is constant.
\medskip
{\bf 10L Example}
Consider the function $f(z)={\Bover12}(z+{\Bover1z})$ with domain
restricted to the punctured disk $\{z:0<|z|<1\}$.
\lectureend{98/14}
\medskip
{\bf (a)} $f$ {is injective}. \Prf\ If $f(z_1)=f(z_2)$ then
$z_1-z_2={\Bover1{z_2}}-{\Bover1{z_1}}=(z_1-z_2)/z_1z_2$, so either
$z_1=z_2$ or $z_1z_2=1$; but if $z_1$, $z_2\in\dom(f)$ then
$|z_1z_2|=|z_1||z_2|<1$, so the latter never happens, and we always have
$z_1=z_2$. \Qed
\medskip
{\bf (b)} The set of values of $f$ is $\{w:\,either\,w\text{ is not real
}or\,|w|>1\}$. \Prf\ If $z=r(\cos\theta+i\sin\theta)$, where $0R$ and not for
$|z|R\}$, $A$ will
be an
analytic function; its derivative will be
$$A'(z)=-\sum_{n=0}^{\infty}{{na_n}\over{z^{n+1}}}
=-\sum_{n=1}^{\infty}{{(n-1)a_{n-1}}\over{z^n}}
=-\sum_{n=2}^{\infty}{{(n-1)a_{n-1}}\over{z^n}}.$$
\medskip
{\bf (c)} Note that for any Z-transform with finite $R$, we must have
$a_0=\lim_{|z|\to\infty}A(z)$,
$a_1=\lim_{z\to\infty}z^2A'(z)=\lim_{|z|\to\infty}z(A(z)-a_0)$; these
formulae give useful checks on our calculations.
\medskip
{\bf 11C Basic rules for calculating Z-transforms (a)} If $\langle
a_n\rangle_{n\ge 0}$,
$\langle b_n\rangle_{n\ge 0}$ have Z-transforms $A$, $B$ defined on
$\{z:|z|>R_1\}$, $\{z:|z|>R_2\}$ then $\sequenceon{a_n+b_n}$ has
Z-transform $A+B$ defined (at least) on $\{z:|z|>\max(R_1,R_2)\}$.
\medskip
{\bf (b)} If $\sequenceon{a_n}$ has Z-transform $A$ defined on
$\{z:|z|>R\}$, and $c\in\Bbb C$, then $\sequenceon{ca_n}$ has
Z-transform
$cA$ defined (at least) on $\{z:|z|>R\}$.
\medskip
{\bf (c)} If $\sequenceon{a_n}$ has Z-transform $A(z)$ defined on
$\{s:\Real
s>R\}$, and $c\in\Bbb C\setminus\{0\}$, then $\sequenceon{c^na_n}$ has
Z-transform
$A({\Bover zc})$ defined on $\{z:|z|>R|c|\}$. \Prf\
$\sum_{n=0}^{\infty}{\Bover{a_nc^n}{z^n}}
=\sum_{n=0}^{\infty}{\Bover{a_n}{(z/c)^n}}$. \Qed
\medskip
{\bf (d)} If $\sequenceon{a_n}$
has Z-transform $A(z)$ defined on $\{z:|z|>R\}$, then
$\sequenceon{a_{n+1}}$ has Z-transform $z(A(z)-a_0)$ defined on
$\{z:|z|>R\}$.
\centerline{\Prf\ $\sum_{n=0}^{\infty}{\Bover{a_{n+1}}{z_n}}
=z\sum_{n=0}^{\infty}{\Bover{a_{n+1}}{z^{n+1}}}
=z\sum_{n=1}^{\infty}{\Bover{a_n}{z^n}}
=z(A(z)-a_0)$. \Qed}
\medskip
{\bf (e)} If $\sequenceon{a_n}$ has Z-transform $A(z)$ defined for
$|z|>R$, then $\sequenceon{na_n}$ has Z-transform $-zA'(z)$ defined for
$|z|>R$. \Prf\ Using 11Bb,
\centerline{$-zA'(z)=\sum_{n=1}^{\infty}{\Bover{(n-1)a_{n-1}}{z^{n-1}}}
=\sum_{n=0}^{\infty}{\Bover{na_n}{z^n}}$. \Qed}
\medskip
{\bf 11D Basic transforms: three to memorise (a)} The Z-transform of
the constant sequence $1$ is
$$\sum_{n=0}^{\infty}{1\over{z^n}}={1\over{1-1/z}}={z\over{z-1}}.$$
\noindent whenever $|z|>1$.
\medskip
{\bf (b)} For real $\omega$, the Z-transform of $\sequenceon{\cos\omega
n}$ is ${\Bover{z^2-z\cos\omega}{z^2-2z\cos\omega+1}}$; the Z-transform
of $\sequenceon{\sin\omega n}$ is
${\Bover{z\sin\omega}{z^2-2z\cos\omega+1}}$, both defined for $|z|>1$.
\vthsp\Prf\ Consider the Z-transform of $\sequenceon{e^{i\omega n}}$. By
11Cc, with $c=e^{i\omega}$ and ${a_n}=1$, $A(z)={\Bover z{z-1}}$ this is
$A(z/c)=A(e^{-i\omega}z)
={\Bover{e^{-i\omega}z}{e^{-i\omega}z-1}}={\Bover z{z-e^{i\omega}}}$ for
$|z|>1$. Similarly, the Z-transform of $\sequenceon{e^{-i\omega n}}$
is ${\Bover{z}{z-e^{-i\omega}}}$. By 11Ca-b we get the
Z-transform of $\sequenceon{\cos\omega n}=\sequenceon{{\Bover12}
(e^{i\omega n}+e^{-i\omega n})}$ to be
$$\eqalign{{1\over 2}\bigl({{z}\over{z-e^{i\omega}}}
+{{z}\over{z-e^{-i\omega}}}\bigr)
&={{z(z-e^{-i\omega})+z
(z-e^{i\omega})}
\over{2(z-e^{i\omega})(z-e^{-i\omega})}}\cr
&={{z^2-z\cos\omega}\over{z^2-2z\cos\omega+1}},\cr}$$
\noindent while the Z-transform of $\sequenceon{\sin\omega n}=
\sequenceon{{\Bover1{2i}}(e^{i\omega n}-e^{-i\omega n})}$ is
$${1\over {2i}}\bigl({z\over{z-e^{i\omega}}}
-{z\over{z-e^{-i\omega}}}\bigr)
={{z\sin\omega}\over{z^2-2z\cos\omega+1}}. \text{ \Qed}$$
\medskip
{\bf 11E Derived transforms (a)}
Putting 11Da and 11Cc together, the Z-transform of the sequence
$\sequenceon{c^n}$ is ${{z\over c}\over{{z\over c}-1}}={\Bover z{z-c}}$,
defined for $|z|>|c|$, whenever $c\ne 0$.
\medskip
{\bf (b)} Combining this with 11Ce, the Z-transform of
$\sequenceon{nc^n}$ is
$$-z{d\over{dz}}{z\over{z-c}}={cz\over{(z-c)^2}};$$
\noindent now the Z-transform of $\sequenceon{n^2c^n}$ is
$$-z{d\over{dz}}{cz\over{(z-c)^2}}={{cz(z+c)}\over{(z-c)^3}}.$$
\noindent In particular, the Z-transforms of $\sequenceon{n}$ and
$\sequenceon{n^2}$ are ${\Bover z{(z-1)^2}}$, ${\Bover{z(z+1)}{(z-1)^3}}$.
\medskip
{\bf (c)} Putting 11Db and 11Cc together, the Z-transforms of
$\sequenceon{c^n\cos\omega n}$, $\sequencen{c^n\sin\omega n}$ are
${\Bover{z^2-cz\cos\omega}{z^2-2cz\cos\omega+c^2}}$,
${\Bover{2cz\sin\omega}{z^2-2cz\cos\omega+c^2}}$.
Adding 11Ce, we get the Z-transforms of $\sequenceon{nc^n\cos\omega n}$
and $\sequenceon{nc^n\sin\omega n}$ to be
$$-z{d\over{dz}}{{z^2-cz\cos\omega}\over{z^2-2cz\cos\omega+c^2}}
={{cz(z^2\cos\omega-2cz+c^2\cos\omega)}
\over{(z^2-2cz\cos\omega+c^2)^2}},$$
$$-z{d\over{dz}}{{cz\sin\omega}\over{z^2-2cz\cos\omega+c^2}}
={{cz\sin\omega(z^2-c^2)}\over{(z^2-2cz\cos\omega+c^2)^2}};$$
\noindent always taking $c\ne 0$ and $|z|>|c|$.
\medskip
{\bf (d)} Using 11Cd twice, we see that if $\sequenceon{a_n}$ is a
sequence with Z-transform $A(z)$ defined on $\{z:|z|>R\}$, then
$\sequenceon{a_{n+2}}$ has Z-transform $z^2A(z)-a_0z^2-a_1z$, defined on
$\{z:|z|>R\}$.
\medskip
{\bf 11F Uniqueness Theorem} Suppose $\sequenceon{a_n}$,
$\sequenceon{b_n}$ are two sequences with Z-transforms $A(z)$ and
$B(z)$, and suppose that $A(z)=B(z)$ at least whenever $|z|$ is large
enough. Then in fact $A(z)=B(z)$ wherever either is defined, and
$a_n=b_n$ for every $n$.
\medskip
{\bf 11G Solution of difference equations} One of the uses of the
Z-transform is in the solution of difference equations. Consider, for
instance, the equation
$$a_{n+2}+4a_n=1,\,a_0=0,\,a_1=1.$$
\noindent We proceed by {\it assuming} that there is a solution in which
$\sequenceon{a_n}$ has a Z-transform defined for $|z|>R$ for some finite
$R$. In this case, if the Z-transform of $\sequenceon{a_n}$ is $A$,
the Z-transform of $\sequenceon{a_{n+2}}$ will be
$z^2A-z^2a_0-za_1=z^2A-z$; on the other side, the Z-transform of the
constant sequence $1$ is ${\Bover z{z-1}}$, so (using 11Ca-b to find the
Z-transform of $\sequenceon{a_{n+2}+4a_n}$) we get
$$z^2A - z + 4A = {z\over{z-1}}.$$
\noindent Now this is easy to solve for $A$; we have
$$A={{z^2}\over{(z-1)(z^2+4)}}.$$
\noindent We seek a sequence $\sequenceon{a_n}$ with this Z-transform.
There are various approaches to this: the one I favour is to say that
we know some relevant ones, being
$$\eqalign{1\quad&:\quad{z\over{z-1}}\cr
\cos {{\pi}\over 2}n\quad&:\quad{{z^2}\over{z^2+1}}\cr
\sin {{\pi}\over 2}n\quad&:\quad{z\over{z^2+1}}.\cr}$$
\noindent These aren't quite enough, because our denominator has a
factor $z^2+4$, not $z^2+1$; but try
$$\eqalign{{2}^n\cos{{\pi}\over 2}n\quad&:\quad{{(z/2)^2}\over{(z/2)^2+1}}
={{z^2}\over{z^2+4}},\cr
{2}^n\sin{{\pi}\over 2}n\quad&:\quad{{z/2}\over{(z/2)^2+1}}
={{2z}\over{z^2+4}}.\cr}$$
\noindent This gives us a target to aim at: if we can express $A(z)$ as
$\alpha{\Bover z{z-1}}+\beta{\Bover{z^2}{z^2+4}}+\gamma{\Bover{2z}{z^2+4}}$,
then we shall have $a_n=\alpha+2^n\beta\cos{\Bover{\pi}2}n
+2^n\gamma\sin{\Bover{\pi}2}n$. Now we want
$$\alpha{z\over{z-1}}+\beta{{z^2}\over{z^2+4}}+\gamma{{2z}\over{z^2+4}}
={{z^2}\over{(z-1)(z^2+4)}},$$
\noindent that is,
$$\alpha z(z^2+4)+\beta z^2(z-1) + 2\gamma z(z-1)=z^2.$$
\noindent Equating coefficients, this becomes
$$\vbox{
\halign{#1\hfil&\quad\hfil#2&=#3\hfil\cr
(coefficient of $z^3$)&$\alpha+\beta$&$0$,\cr
(coefficient of $z^2$)&$-\beta+2\gamma$&$1$,\cr
(coefficient of $z$)&4$\alpha-2\gamma$&$0$,\cr}}$$
\noindent giving $\alpha={\Bover15}$, $\beta=-{\Bover15}$ and
$\gamma={\Bover25}$, so that
$$a_n={1\over 5}-{{2^{n}}\over 5}\cos{{\pi}\over 2}n+{{2^{n+1}}\over
5}\sin{{\pi}\over 2}n.$$
\medskip
{\bf 11H Notes (a)} We can use this method on a wide variety of linear
difference equations with constant coefficients. In principle, it will
work on equations of all orders, because we can find the Z-transform of
all the sequences $\sequenceon{a_{n+k}}$ in terms of the Z-transform $A$
of $\sequenceon{a_n}$ -- for instance, the Z-transform of
$\sequenceon{a_{n+3}}$ is $z^3A -z^3a_0 - z^2a_1 - za_2$. We do need
to be able to find the Z-transform of the sequence on the right-hand
side of the equation, and we do need to be able to calculate, or guess,
the sequence whose Z-transform is the $A$ we finally extract from the
equation.
\medskip
{\bf (b)} An equation with constant coefficients, of the form
$$c_ka_{n+k}+\ldots+c_1a_{n+1}+c_0a_n=d_n,$$
\noindent will always turn into an equation of the form
$$(c_kz^k+\ldots+c_1z+c_0)A(z) + p(z) = D(z),$$
\noindent where $p$ is a polynomial depending on the initial values
given for $a_0,\ldots,a_{k-1}$, and $D$ is the Z-transform of
$\sequenceon{d_n}$. The polynomial
$c_kz^k+\ldots+c_1z+c_0$ is the characteristic polynomial of the
equation, and the method depends on factorising it, at least into
quadratic factors. It may well happen, of course, that the transform
$D(z)$ introduces further factors into the denominator of the final
expression for $A(z)$.
Given this factorisation, a suitable combination of transforms of the
form
$${z\over{z-c}},
{{cz}\over{(z-c)^2}},
{{cz(z+c)}\over{(z-c)^3}},
\ldots,$$
$${{(z^2-cz\cos\omega)}\over{z^2-2cz\cos\omega+c^2}},
{{cz\sin\omega}\over{z^2-2cz\sin\omega+c^2}},
{{cz(z^2\cos\omega-2cz+c^2\cos\omega)}
\over{(z^2-2cz\cos\omega+c^2)^2}}
\ldots$$
\noindent corresponding to the sequences
$$c^n,nc^n,n^2c^n,\ldots,$$
$$c^n\cos\omega n,c^n\sin\omega n,nc^n\cos\omega
n,\ldots$$
\noindent ought to work, at least when $D(z)$ is a rational function of
$z$.
\medskip
{\bf (c)} In the example above, we can identify a \lq particular
solution' as the constant sequence ${\Bover15}$, while the
terms in $2^{n}\cos{\Bover{\pi}2}n$ and $2^n\sin{\Bover{\pi}2}n$
are solutions of the homogeneous equation $a_{n+2}+4a_n=0$. (Note that
of course these can be rearranged, if we allow ourselves complex
coefficients, as combinations of $(2i)^n=2^n\exp(i{\Bover{\pi}2}n)$ and
$(-2i)^n=2^n\exp(-i{\Bover{\pi}2}n)$; the point being that $\pm 2i$ are
the two complex roots of the characteristic polynomial $z^2+4$.)
\medskip
{\bf (d)} Note the use of the Uniqueness Theorem here. We started by
assuming that there was a solution, and that it would have a
Z-transform. We worked out what that Z-transform would have to be, and
found a sequence which had that transform. I then cheerfully asserted
that that sequence did indeed solve the equation. The point is that I
-- if not you -- could have proved, from the general theory of
difference equations, that there was a solution, and that it would have
a Z-transform defined at least for $z$ with large modulus. So the
first part of the argument was in fact sound, even if it relied on
deeper results than I wish to go into here. Now, having found a
sequence with the right Z-transform, the uniqueness theorem tells me
that it must be the right sequence.
Of course any prudent person does not rely entirely on such abstract
considerations, but substitutes the supposed solution into the original
equation (and especially the initial conditions) as a check.
\medskip
{\bf (e)} Of course in this form the method cannot solve any equation
not soluble by the basically more elementary method of trying suitable
combinations of the sequences on the list just given, choosing $c$ and
$\omega$ among those parameters occuring in the sequence
$\sequenceon{d_n}$, or from roots of the characteristic polynomial in
the form $c\pm i\omega$.
For the real power of the method, we need some more theorems.
\medskip
{\bf 11I Delay switch} Suppose a sequence $\sequenceon{a_n}$ has
Z-transform $A$ defined for $|z|>R$. Suppose that $k\ge 0$ and that we
define a new sequence $\sequenceon{b_n}$ by setting
\centerline{$b_n=0$ for $0\le nR$.
\medskip
\noindent{\bf proof}
$${\sum_{n=0}^{\infty}{{b_n}\over{z^n}}
=\sum_{n=k}^{\infty}{{b_n}\over{z^n}}
=\sum_{n=k}^{\infty}{{a_{n-k}}\over{z^n}}
=\sum_{n=0}^{\infty}{{a_n}\over{z^{n+k}}}
={{A(z)}\over{z^k}}}$$
\noindent whenever $A(z)$ is defined.
\medskip
{\bf 11J Example} Consider the equation $a_{n+1}+a_n=d_n$, where
$d_n=2^n$ for $0\le nk$.
In other cases (and in particular with
higher-order equations and sequences $d_n$ subject to more than one
delayed switch) the method here is quicker and more reliable.
\medskip
{\bf 11L Convolutions} The next theorem which gives further scope to the
Z-transform method is the {\bf convolution theorem}, as follows:
\medskip
\noindent{\bf Theorem} Let $\sequenceon{a_n}$, $\sequenceon{b_n}$ be two
sequences with Z-transforms $A$, $B$, defined respectively for $|z|>R$
and $|z|>S$. Set
$$c_n=\sum_{k=0}^na_kb_{n-k}$$
\noindent for $n\ge 0$. Then $\sequenceon{c_n}$ has Z-transform
$A(z)B(z)$, at least for $|z|>\max(R,S)$.
\medskip
{\bf 11M Example} Consider the recurrence relation
$$a_0=1,\quad a_{n+1}=\sum_{j=0}^na_ja_{n-j}.$$
\noindent Suppose, optimistically, that the solution to this has a
Z-transform $A(z)$ defined for large $|z|$. Then the Z-transform of
$\sequenceon{a_{n+1}}$ is $z(A(z)-1)$, while the right-hand side is the
convolution of $\sequenceon{a_n}$ with itself, so has Z-transform
$A(z)^2$. Thus we have
$$z(A-1)=A^2,$$
\noindent yielding
$$A={{z\pm\sqrt{z^2-4z}}\over 2}.$$
Which square root should we take? We want
$\lim_{|z|\to\infty}A(z)=a_0=1$. If we express $A$ as
$${z\over 2}(1\pm\sqrt{1-{4\over z}}),$$
\noindent then plainly we must take
$$A={z\over 2}(1-(1-{4\over z})^{1\over 2})$$
\noindent to have a chance. Now we know that
$$\eqalign{(1+w)^{1\over 2}=1+{1\over 2}w
+{1\over {2!}}({1\over 2})(-{1\over 2})w^2
&+{1\over{3!}}({1\over 2})(-{1\over 2})(-{3\over 2})w^3\cr
&+{1\over{4!}}({1\over 2})(-{1\over 2})(-{3\over 2})(-{5\over
2})w^4+\ldots,\cr}$$
\noindent so
$$\eqalign{(1-{4\over z})^{1\over 2}
&=1-{2\over z}
-{1\over{2!}}{4\over{z^2}}
-{1\over{3!}}.3.{8\over{z^3}}
-{1\over{4!}}.3.5.{{16}\over{z^4}}
-{1\over{5!}}.3.5.7.{{32}\over{z^5}}
-\ldots\cr
&=1-{2\over z}
-{1\over{2!}}{{1!}\over 1}{{2^2}\over{z^2}}
-{1\over{3!}}{{3!}\over {2}}{{2^3}\over{z^3}}
-{1\over{4!}}{{5!}\over{2.4}}{{2^4}\over{z^4}}
-{1\over{5!}}{{7!}\over{2.4.6}}{{2^5}\over{z^5}}
-\ldots\cr
&=1-{2\over z}
-{1\over{2!}}{{1!}\over 1}{{2^2}\over{z^2}}
-{1\over{3!}}{{3!}\over 2}{{2^3}\over{z^3}}
-{1\over{4!}}{{5!}\over{2^22!}}{{2^4}\over{z^4}}
-{1\over{5!}}{{7!}\over{2^33!}}{{2^5}\over{z^5}}
-\ldots\cr
&=1-{2\over z}
-{1\over{2!}}{{1!}\over{0!}}{{4}\over{z^2}}
-{1\over{3!}}{{3!}\over{1!}}{{4}\over{z^3}}
-{1\over{4!}}{{5!}\over{2!}}{{4}\over{z^4}}
-{1\over{5!}}{{7!}\over{3!}}{{4}\over{z^5}}
-\ldots\cr
&=1-{2\over z}
-\sum_{n=2}^{\infty}{{4(2n-3)!}\over{n!(n-2)!z^n}}\cr}$$
\noindent and
$$\eqalign{{z\over 2}(1-(1-{4\over z})^{1\over 2})
&=1+\sum_{n=2}^{\infty}{{2(2n-3)!}\over{n!(n-2)!z^{n-1}}}\cr
&=1+\sum_{n=1}^{\infty}{{2(2n-1)!}\over{(n+1)!(n-1)!z^n}}\cr
&=1+\sum_{n=1}^{\infty}{{(2n)!}\over{(n+1)!n!z^n}}\cr
&=\sum_{n=0}^{\infty}{{(2n)!}\over{(n+1)!n!z^n}}.\cr}$$
\noindent This means that
$$a_n={{(2n)!}\over{(n+1)!n!}}$$
\noindent for every $n\ge 0$.
(To check this, run the argument backwards. Observe that
$\sequenceon{{\Bover{(2n)!}{(n+1)!n!}}}$
does have a Z-transform $A(z)$ defined for $|z|>4$ (use the ratio test),
that $A(z)={\Bover z2}(1-(1-{\Bover4z})^{1/2})$ so that $z(A-1)=A^2$, and that therefore $a_{n+1}=\sum_{j=0}^na_ja_{n-j}$ for every $n$.)
\medskip
\def\He{\mathop{\text{He}}\nolimits}
{\bf 11N Hermite polynomials (a)} Consider the real polynomials $\He_n$
defined by starting with
$$\He_0(x)= 1,\quad \He_1(x)=x,$$
\noindent and proceeding with the recurrence relation
$$\He_{n+1}(x)=x\He_n(x)-n\He_{n-1}(x),$$
\noindent so that $\He_2(x)=x^2-1$, $\He_3(x)=x^3-3x$,
$\He_4(x)=x^4-6x^2+3$, and so on.
\medskip
{\bf (b)} Fix $x$ for a while, and set $a_n={\Bover1{n!}}\He_n(x)$, so
that
$$(n+1)!a_{n+1}=xn!a_n-n(n-1)!a_{n-1},$$
\noindent and
$$(n+1)a_{n+1}=xa_n-a_{n-1},\quad (n+2)a_{n+2}=xa_{n+1}-a_n.$$
\noindent Recall our table:
$$\eqalign{a_{n}\quad&:\quad A\cr
na_n\quad&:\quad -zA'\cr
a_{n+1}\quad&:\quad z(A-a_0)\cr
(n+1)a_{n+1}\quad&:\quad z(-zA'-0)=-z^2A'\cr
(n+2)a_{n+2}\quad&:\quad z(-z^2A'-a_1)=-z^3A'-za_1.\cr}$$
\noindent In the present case we have $a_0=1$, $a_1=x$, so if $A$ is the
Z-transform of $\sequenceon{a_n}$ (assuming, as usual, that there is one
defined for sufficiently large $|z|$), we get
$$-z^3A'-zx=xz(A-1)-A,$$
\noindent that is,
$$z^3A'+A(xz-1)=0.$$
\noindent Solving this differential equation by separation of variables,
we have
$${{A'}\over A}+{x\over{z^2}}-{1\over{z^3}}=0,$$
\noindent so that
$$\Ln A-{x\over z}+{1\over{2z^2}}=\text{constant},\quad A=c\exp({x\over
z}-{1\over{2z^2}});$$
\noindent without looking too deeply into what $\Ln A$ means in this
context, we can check that this solves the differential equation
$z^3A'+A(xz-1)=0$. Now, letting $|z|\to\infty$, we find (if this is
going to work at all) that $c=\lim_{|z|\to\infty}A(z)=a_0=1$, so that
$A=\exp({\Bover xz}-{\Bover1{2z^2}})$, and
$$\sum_{n=0}^{\infty}{{\He_n(x)}\over{n!z^n}}
=\exp({x\over z}-{1\over{z^2}}).$$
\medskip
{\bf (c)} Given this formula for $A$, we can now justify all the
preceding steps, because $A(z)$ surely has a Laurent series
$\sum_{n=0}^{\infty}{\Bover{a_n}{z^n}}$ valid for all non-zero $z$, and
the differential equation which $A$ satisfies translates readily into
the correct recurrence relation for the $a_n$.
\medskip
{\bf (d)} Now let $x$ vary, and write
$$A(z,x)=\exp({x\over z}-{1\over{2z^2}})
=\sum_{n=0}^{\infty}{{\He_n(x)}\over{n!z^n}}.$$
\noindent We can compute
$$\eqalign{\int_{-\infty}^{\infty}A(z,x)A(w,x)e^{-x^2/2}dx
&=\int_{-\infty}^{\infty}\exp({x\over z}-{1\over{2z^2}})
\exp({x\over w}-{1\over{2w^2}})e^{-x^2/2}dx\cr
&=\int_{-\infty}^{\infty}\exp({x\over z}-{1\over{2z^2}}
+{x\over w}-{1\over{2w^2}}-{{x^2}\over 2})dx\cr
&=\int_{-\infty}^{\infty}\exp(-{1\over2}(x-{1\over z}-{1\over
w})^2+{1\over{wz}})dx\cr
&=\exp({1\over{wz}})\int_{\infty}^{\infty}\exp(-{1\over 2}y^2)dy,\cr}$$
\noindent at least if $w$, $z$ are real and non-zero, using the
substitution $y=x-{\Bover 1w}-{\Bover 1z}$; that is,
$$\eqalign{\int_{-\infty}^{\infty}A(z,x)A(w,x)e^{-x^2/2}dx
&=\sqrt{2\pi}\exp({1\over{wz}})\cr
&=\sqrt{2\pi}\sum_{n=0}^{\infty}{1\over{n!w^nz^n}}.\cr}$$
\noindent On the other hand, we also have
$$\eqalign{\int_{-\infty}^{\infty}A(z,x)A(w,x)e^{-x^2/2}dx
&=\int_{\infty}^{\infty}\sum_{m=0}^{\infty}{{\He_m(x)}\over{m!w^m}}
\sum_{n=0}^{\infty}{{\He_n(x)}\over{n!z^n}}e^{-x^2/2}dx\cr
&=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}
\int_{\infty}^{\infty}{{\He_m(x)}\over{m!w^m}}
{{\He_n(x)}\over{n!z^n}}e^{-x^2/2}dx\cr
&=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}{1\over{m!n!w^mz^n}}
\int_{\infty}^{\infty}\He_m(x)\He_n(x)e^{-x^2/2}dx.\cr}$$
\noindent Equating coefficients of ${\Bover1{w^mz^n}}$, we get
$$\eqalign{\int_{\infty}^{\infty}\He_m(x)\He_n(x)
e^{-x^2/2}dx
&=n!\sqrt{2\pi}\text{ if }m=n,\cr
&=0\text{ if }m\ne n.\cr}$$
\discrpage
\filename{cv2n12.tex}
\versiondate{19.2.93}
\noindent{\bf 12. Laplace Transforms}
\medskip
{\bf 12A Definition} Let $y$ be a complex-valued function.
Its {\bf Laplace transform} is the function $Y$ defined by setting
$$Y(s)=\int_0^{\infty}e^{-sx}y(x)dx$$
\noindent for all those complex numbers $s$ for which the integral is
defined.
\medskip
{\bf 12B Remarks (a)} The question immediately arises -- I mean rather:
you ought to ask at once -- which functions $y$ have Laplace transforms?
The answer depends on which theory of integration you use. For our
purposes here, Riemann integration will be adequate, taking
$\int_0^{\infty}f(x)dx$ to be
$\int_0^{\infty}\Real(f(x))dx+i\int_0^{\infty}\Imag(f(x))dx$ if both of
these are
defined as integrals of the real functions $\Real f$, $\Imag
f$, and $\int_0^{\infty}f(x)dx$ to be
$\lim_{a\to\infty}\int_0^af(x)dx$ if the limit is defined.
It will sometimes happen that we shall want a Laplace transform of a
function which is not defined on the whole of $\coint{0,\infty}$. In
particular, if for every $a\ge 0$ there are only finitely many points in
$[0,a]$ at which $y$ is undefined, we can list them in increasing order
as $a_1$, $a_2,\ldots,a_n$, and then say that
$$\int_0^ae^{-sx}y(x)dx=\int_0^{a_1}e^{-sx}y(x)dx
+\int_{a_1}^{a_2}e^{-sx}y(x)dx+\ldots+\int_{a_n}^ae^{-sx}y(x)dx,$$
\noindent ignoring the points $a_1,\ldots,a_n$ altogether.
\medskip
{\bf (b)} Under these rules, we find that there is always an
$R\in[0,\infty]$
such that $Y(s)$ is defined if $\Real s>R$ and not if $\Real sR\}$, moreover, $Y$ will be an
analytic function; its derivative will be
$$Y'(s)=-\int_0^{\infty}xe^{-sx}y(x)dx.$$
\medskip
{\bf (c)} In fact it may help if we restrict our attention to certain
functions which are adequate for our needs and are relatively easy to
visualize. I will say that a complex-valued function $y$ on a subset
of $\coint{0,\infty}$ is {\bf locally piecewise continuous} if
for every $a\in\coint{0,\infty}$ there are finitely many points
$a_0,\ldots,a_n$ such that $0=a_0R_1\}$, $\{s:\Real s>R_2\}$ then $y_1+y_2$ has Laplace
transform $Y_1+Y_2$ defined (at least) on $\{s:\Real s>\max(R_1,R_2)\}$.
\medskip
{\bf (b)} If $y$ has Laplace transform $Y$ defined on
$\{s:\Real(s)>R\}$, and $c\in\Bbb C$, then $cy$ has Laplace transform
$cY$ defined (at least) on $\{s:\Real s>R\}$.
\medskip
{\bf (c)} If $y(x)$ has Laplace transform $Y(s)$ defined on $\{s:\Real
s>R\}$, and $a\in\Bbb C$, then $e^{ax}y(x)$ has Laplace transform
$Y(s-a)$ defined on $\{s:\Real s>R+\Real a\}$. \Prf\
$\int_0^{\infty}e^{-sx}e^{ax}y(x)dx=\int_0^{\infty}e^{-(s-a)x}y(x)dx$.
\Qed
\medskip
{\bf (d)} If a function $y$ is such that
\quad (i) $y$ is defined and continuous on $\coint{0,\infty}$
\quad (ii) $y$ has Laplace transform $Y(s)$ defined on $\{s:\Real
s>R\}$
\quad (iii) its derivative $y'$ is locally piecewise continuous on
$\coint{0,\infty}$
\quad (iv) $\alpha\in\Bbb R$ is such that
$\lim_{x\to\infty}e^{-\alpha x}y(x)=0$,
\noindent then the Laplace transform of
$y'(x)$ is
$sY(s)-y(0)$, defined (at least) on $\{s:\Real s>\max(R,\alpha)\}$.
\medskip
\noindent{\bf idea of proof} Integrating by parts,
$$\eqalign{\int_0^{\infty}e^{-sx}y'(x)dx
&=\left[e^{-sx}y(x)\right]_0^{\infty}
-\int_0^{\infty}(-se^{-sx})y(x)dx\cr
&=\lim_{x\to\infty}e^{-sx}y(x)-y(0)+sY(s)\cr
&=sY(s)-y(0),\cr}$$
\noindent because $\lim_{x\to\infty}|e^{-sx}y(x)|\le\lim_{x\to\infty}
e^{-\alpha x}|y(x)|=0$ if $\Real s>\alpha$.
\medskip
{\bf (e)} If $y(x)$ has Laplace transform $Y(s)$ defined for $\Real
s>R$, then $xy(x)$ has Laplace transform $-Y'(s)$ defined for $\Real
s>R$. (See 12Bb.)
\medskip
{\bf 12D Basic transforms: three to memorise (a)} The Laplace transform
of the constant function $1$ is
$$Y(s)=\int_0^{\infty}e^{-sx}dx
=\left[-{1\over s}e^{-sx}\right]_0^{\infty}
={1\over s}(1-\lim_{x\to\infty}e^{-sx})={1\over s}$$
\noindent whenever $\Real s>0$ (so that
$\lim_{x\to\infty}|e^{-sx}|=\lim_{x\to\infty}e^{-x\Real s}=0$).
\medskip
{\bf (b)} For real $\omega$, the Laplace transform of $\cos\omega x$ is
${\Bover s{s^2+\omega^2}}$; the Laplace transform of $\sin\omega x$ is
${\Bover{\omega}{s^2+\omega^2}}$, both defined for $\Real s>0$.
\vthsp\Prf\ Consider the Laplace transform of $e^{i\omega x}$. By
12Cc, with
$a=i\omega$ and $y(x)=1$, $Y(s)={\Bover1s}$ this is
$Y(s-a)={\Bover1{s-i\omega}}$ for $\Real s>0+\Real(i\omega)=0$.
Similarly, the Laplace transform of $e^{-i\omega x}$ is
${\Bover1{s+i\omega}}$. By 12Ca-b we get the Laplace transform of
$\cos\omega x={\Bover12}(e^{i\omega x}+e^{-i\omega x})$ to be
$${1\over 2}\bigl({1\over{s-i\omega}}+{1\over{s+i\omega}}\bigr)
={s\over{s^2+\omega^2}},$$
\noindent while the Laplace transform of $\sin\omega x={\Bover1{2i}}
(e^{i\omega x}-e^{-i\omega x})$ is
$${1\over {2i}}\bigl({1\over{s-i\omega}}-{1\over{s+i\omega}}\bigr)
={{\omega}\over{s^2+\omega^2}}. \text{ \Qed}$$
\medskip
{\bf 12E Derived transforms (a)} Putting 12Da and 12Cc together, the
Laplace
transform of $e^{ax}$ is ${\Bover1{s-a}}$, defined for $\Real s>\Real a$.
\medskip
{\bf (b)} Combining this with 12De, the Laplace transform of $xe^{ax}$
is
$$-{d\over{ds}}{1\over{s-a}}={1\over{(s-a)^2}},$$
\noindent defined for $\Real s>\Real a$; now the Laplace transform of
$x^2e^{ax}$ is
$$-{d\over{ds}}{1\over{(s-a)^2}}={2\over{(s-a)^3}}.$$
\noindent In particular, the Laplace transforms of $x$ and $x^2$ are
${\Bover1{s^2}}$ and ${\Bover2{s^3}}$.
Generally, the Laplace transform of $x^k$ is
${\Bover{k!}{s^{k+1}}}$, defined for $\Real s>0$.
\medskip
{\bf (c)} Putting 12Db and 12Cc together, the Laplace transforms of
$e^{ax}\cos\omega x$ and $e^{ax}\sin\omega x$ are
${\Bover{s-a}{(s-a)^2+\omega^2}}$, ${\Bover{\omega}{(s-a)^2+\omega^2}}$.
Adding 12Ce, the Laplace transform of
$xe^{ax}\cos\omega x$ is
$$-{d\over{ds}}{{(s-a)}\over{(s-a)^2+\omega^2}}
={{\omega^2-(s-a)^2}\over{((s-a)^2+\omega^2)^2}};$$
\noindent while the Laplace transform of $xe^{ax}\sin\omega x$ is
$$-{d\over{ds}}{{\omega}\over{(s-a)^2+\omega^2}}
={{2\omega (s-a)}\over{((s-a)^2+\omega^2)^2}}.$$
\medskip
{\bf (d)} Using 12Cd twice, we see that if $y$ is a function such that
\quad (i) $y$ and $y'$ are defined and continuous on $\coint{0,\infty}$
\quad (ii) $y$ has Laplace transform $Y(s)$ defined on $\{s:\Real
s>R\}$
\quad (iii) the second derivative $y''$ is locally piecewise continuous
on $\coint{0,\infty}$
\quad (iv) $\alpha\in\Bbb R$ is such that
$\lim_{x\to\infty}e^{-\alpha x}y(x)=\lim_{x\to\infty}e^{\alpha
x}y'(x)=0$,
\noindent then the Laplace transform of
$y'(x)$ is
$s^2Y(s)-sy(0)-y'(0)$, defined (at least) on $\{s:\Real
s>\max(R,\alpha)\}$.
\medskip
{\bf 12F Uniqueness Theorem} Suppose $y_1$, $y_2$ are two functions with
Laplace transforms $Y_1$ and $Y_2$, and suppose that $Y_1(s)=Y_2(s)$ at
least for all large real $s$. Then in fact $Y_1(s)=Y_2(s)$ wherever
either is defined, and $y_1(x)=y_2(x)$ wherever both are continuous.
\medskip
\noindent {\bf Remark} Note that of course we cannot say that $y_1$,
$y_2$ are exactly the same function, because it might be (for instance)
that they differ at only finitely many points, and then they will still
have the same Laplace transform.
\medskip
{\bf 12G Solution of differential equations} One of the uses of the
Laplace transform is in the solution of differential equations.
Consider, for instance, the equation
$$y''+4y=1,\,y(0)=0,\,y'(0)=1.$$
\noindent We proceed by {\it assuming} that there is a solution in which
$y$, $y'$ are continuous, $y''$ is piecewise continuous, and all have
Laplace transforms defined for $\Real s>R$ for some finite $R$. In
this case, if the Laplace transform of $y$ is $Y$, the Laplace transform
of $y''$ will be $s^2Y-sy(0)-y'(0)=s^2Y-1$; on the other side, the
Laplace transform of the constant function $1$ is ${\Bover1s}$, so
(using 12Ca-b to find the Laplace transform of $y''+4y$) we get
$$s^2Y - 1 + 4Y = {1\over s}.$$
\noindent Now this is easy to solve for $Y$; we have
$$Y={{s+1}\over{s(s^2+4)}}.$$
\noindent We seek a function $y$ with this Laplace transform. There
are various approaches to this: the one I favour is to say that we know
some relevant ones, being
$$\eqalign{1\quad&:\quad{1\over s}\cr
\cos 2x\quad&:\quad{s\over{s^2+4}}\cr
\sin 2x\quad&:\quad{2\over{s^2+4}}.\cr}$$
\noindent An appropriate combination of these has a good chance of
working. Specifically, the Laplace transform of $\alpha +\beta \cos
2x+\gamma \sin 2x$
will have Laplace transform
${\Bover{\alpha}s} + {\Bover{\beta s}{s^2+4}} + {\Bover{2\gamma}{s^2+4}}.$
\noindent So we seek constants $\alpha $, $\beta $, $\gamma $ such that
$${\alpha \over s} + {{\beta s}\over{s^2+4}} + {{2\gamma }\over{s^2+4}}
={{s+1}\over{s(s^2+4)}}.$$
\noindent This is almost a problem in partial fractions, and indeed can
be done that way if you wish; but I think it is quicker and safer just
to multiply up and say
$$\alpha (s^2+4)+\beta s^2+2\gamma s=s+1.$$
\noindent Equating coefficients, this becomes
$$\vbox{
\halign{#1\hfil&\quad\hfil#2&=#3\hfil\cr
(coefficient of $s^2$)&$\alpha+\beta$&$0$,\cr
(coefficient of $s$)&$2\gamma$&$1$,\cr
(coefficient of $1$)&$4\alpha$&$1$,\cr}}$$
%$$\eqalign{(\text{coefficient of }s^2)\quad \alpha +\beta &=0,\cr
%(\text{coefficient of }s)\quad 2\gamma &=1,\cr
%(\text{coefficient of }1)\quad 4\alpha &=1,\cr}$$
\noindent giving $\alpha ={\Bover14}$, $\beta =-{\Bover14}$ and $\gamma
={\Bover12}$, so
that
$$y={1\over 4}-{1\over 4}\cos 2x + {1\over 2}\sin 2x.$$
\medskip
{\bf 12H Notes (a)} We can use this method on a wide variety of linear
ordinary differential equations with constant coefficients. In
principle, it will work on equations of all orders, because we can find
the Laplace transform of all the repeated derivatives $y'$, $y''$,
$y''',\ldots$ in terms of the Laplace transform $Y$ of $y$ -- for
instance, the Laplace transform of $y'''$ is $s^3Y -s^2y(0) - sy'(0) -
y''(0)$. We do need to be able to find the Laplace transform of the
function on the right-hand side of the equation, and we do need to be
able to calculate, or guess, the function whose Laplace transform is the
$Y$ we finally extract from the equation.
\medskip
{\bf (b)} An equation with constant coefficients, of the form
$$a_ky^{(k)}+\ldots+a_1y'+a_0y=f,$$
\noindent will always turn into an equation of the form
$$(a_ks^k+\ldots+a_1s+a_0)Y(s) + p(s) = F(s),$$
\noindent where $p$ is a polynomial depending on the initial conditions
$y(0)=c_0$, $y'(0)=c_1,\ldots,y^{(n-1)}(0)=c_{n-1}$ imposed, and $F$ is
the Laplace transform of $f$. You will I hope recognise the polynomial
$a_ks^k+\ldots+a_1s+a_0$ as the characteristic polynomial of the
equation, and the method depends on factorising it, at least into
quadratic factors. It may well happen, of course, that the transform
$F(s)$ introduces further factors into the denominator of the final
expression for $Y(s)$.
Given this factorisation, a suitable combination of transforms of the
form
$${1\over{s+a}},{1\over{(s+a)^2}},{2\over{(s+a)^3}},\ldots,$$
$${{(s+a)}\over
{(s+a)^2+\omega^2}},{{\omega}\over{(s+a)^2+\omega^2}},
{{\omega^2-(s+a)^2}\over{((s+a)^2+\omega^2)^2}},
{{2\omega(s+a)}\over{(s+a)^2+\omega^2)^2}},\ldots$$
\noindent corresponding to the functions
$$e^{-ax},xe^{-ax},x^2e^{-ax},\ldots,$$
$$e^{-ax}\cos\omega x,
e^{-ax}\sin\omega x,
xe^{-ax}\cos\omega x,xe^{-ax}\sin\omega x,\ldots$$
\noindent ought to work, at least when $F(s)$ is a rational function of
$s$.
\medskip
{\bf (c)} In the example above, we can identify a \lq particular
solution' as the constant function ${\Bover14}$, while the terms in
$\cos 2x$ and $\sin 2x$ are solutions of the homogeneous equation
$y''+4y=0$.
\medskip
{\bf (d)} Note the use of the Uniqueness Theorem here. We started by
assuming that there was a solution, and that it would have a Laplace
transform. We worked out what that Laplace transform would have to be,
and found a function which had that transform. I then cheerfully
asserted that that function did indeed solve the equation. The point
is that I -- if not you -- could have proved, from the general theory of
ordinary differential equations, that there was a continuous solution,
and that it would have a Laplace transform defined at least for $s$ with
large real part. So the first part of the argument was in fact sound,
even if it relied on deeper results than I wish to go into here. Now,
having found a function with the right Laplace transform, the uniqueness
theorem tells me that it must be the right function, because they agree
at least at all points where they are both continuous, and my function
is continuous everywhere, like the true solution.
Of course any prudent person does not rely entirely on such abstract
considerations, but substitutes the supposed solution into the original
equation (and especially the given initial values) as a check.
\medskip
{\bf (e)} Of course in this form the method cannot solve any equation
not soluble by the basically more elementary method of trying suitable
combinations of the functions on the list just given, choosing $a$ and
$\omega$ among those parameters occuring in the function $f$, or from
roots of the characteristic equation in the form $a\pm i\omega$.
For the real power of the method, we need some more theorems.
\medskip
{\bf 12I Delay switch} Suppose a function $y$ has Laplace transform $Y$
defined for $\Real s>R$. Suppose that $\alpha\ge 0$ and that we define
a new function $y_1$ by setting
\centerline{$y_1(x)=0$ for $0\le x<\alpha$, $y_1(x)=y(x-\alpha)$ if
$x\ge\alpha$ (and $x-\alpha\in\dom y$).}
\noindent Then the Laplace transform of $y_1$ is
$e^{-as}Y(s)$, defined for $\Real s>R$.
\medskip
\noindent{\bf proof}
$$\eqalign{\int_0^{\infty}e^{-sx}y_1(x)dx
&=\int_{\alpha}^{\infty}e^{-sx}y(x-\alpha)dx\cr
&=\int_0^{\infty}e^{-s(x+\alpha)}y(x)dx\cr
&=e^{-\alpha s}Y(s)\cr}$$
\noindent whenever $Y(s)$ is defined.
\medskip
{\bf 12J Example} Consider the equation $y'+y=f$, where $f(x)=e^x$ for
$0\le x<2$, $0$ for $x\ge 2$, with $y(0) = 0$.
Express $f(x)$ as $e^x-g(x)$, where $g(x)=0$ for $0\le x<2$, $e^x$ for
$x\ge 2$. The Laplace transform of $e^x$ is ${1\over{s-1}}$. Now we
can express $g$ as $g(x)=0$ for $x<2$, $e^2e^{x-2}$ for $x\ge 2$; so
that the Laplace transform of $g$ is $e^2e^{-2s}{\Bover1{s-1}}$, by 12I,
and the Laplace transform of $f$ is ${\Bover{1-e^2e^{-2s}}{s-1}}$.
If the Laplace transform of $y$ is $Y$, then the Laplace transform of
$y'+y$ is $sY+y(0)+Y=sY+Y$, so we have
$$sY+Y={{1-e^2e^{-2s}}\over{s-1}}$$
\noindent and
$$Y={{1-e^2e^{-2s}}\over{(s+1)(s-1)}}.$$
\noindent As before, we have to assemble this out of a table of
appropriate Laplace transforms:
$$\eqalign{e^x\quad&:\quad{1\over{s-1}}\cr
e^{-x}\quad&:\quad{1\over{s+1}}\cr
e^x\text{ delayed by }2\quad&:\quad {e^{-2s}\over{s-1}}\cr
e^{-x}\text{ delayed by }2\quad&:\quad {e^{-2s}\over{s+1}}.\cr}$$
\noindent Thus we need to find constants $\alpha $, $\beta $, $\gamma $,
$\delta $ such that
$${\alpha \over{s-1}}+{\beta \over{s+1}}+{{\gamma e^{-2s}}\over{s-1}}
+{{\delta e^{-2s}}\over{s+1}}={{1-e^2e^{-2s}}\over{(s+1)(s-1)}}.$$
\noindent Multiplying up, this becomes
$$\alpha (s+1)+\beta (s-1)+\gamma e^{-2s}(s+1)+\delta e^{-2s}(s-1)=1-
e^2e^{-2s}.$$
\noindent Equating coefficients of $s$, $1$, $e^{-2s}s$ and $e^{-2s}$ we
get
$$\eqalign{\alpha +\beta &=0\cr
\alpha -\beta &=1\cr
\gamma +\delta &=0\cr
\gamma -\delta &=-e^2,\cr}$$
\noindent so $\alpha ={\Bover12}$, $\beta =-{\Bover12}$, $\gamma =-
{\Bover12}e^2$,
$\delta ={\Bover12}e^2$. Putting these together, our solution is
$$y(x)={1\over 2}e^x-{1\over 2}e^{-x}-{{e^2}\over 2}h_1(x)+{{e^2}\over
2}h_2(x),$$
\noindent where $h_1$, $h_2$ are the functions $e^x$, $e^{-x}$ \lq
delayed to 2', that is,
\centerline{$h_1(x) = 0$ if $x<2$, $e^{x-2}=e^{-2}e^x$ if $x\ge 2$,}
\centerline{$h_2(x)=0$ if $x<2$, $e^{-(x-2)}=e^2e^{-x}$ if $x\ge 2$.}
\noindent Accordingly
$$\eqalign{y(x)&={1\over 2}e^x-{1\over 2}e^{-x}\text{ if }x<2,\cr
&={1\over 2}e^x-{1\over 2}e^{-x}-{{e^2}\over 2}e^{-2}e^x+{{e^2}\over
2}e^2e^{-x}
={1\over 2}(e^4-1)e^{-x}\text{ if }x\ge 2.\cr}$$
\medskip
{\bf 12K Notes (a)} Observe that on each section the answer is in the
expected form. For $02$ we have
just a multiple of $e^{-x}$, because in that region we have $y'+y=0$.
The constant ${\Bover12}(e^4-1)$ is just that required to make the value
of $y(2)={\Bover12}(e^2-e^{-2})=\sinh 2$ match up from both sides.
\medskip
{\bf (b)} In this particular case it would I think be easier to solve
the equations in the two halves independently, doing $[0,2]$ first,
calculating $y(2)$ from the solution there, and then doing
$\coint{2,\infty}$. In other cases (and in particular with
higher-order equations) the method here is quicker and more reliable.
\medskip
{\bf (c)} Note the way in which the Laplace transform method deals with
the jump in the function $f$. If we are to have $y'+y=f$, at least one
of $y$, $y'$ must jump at $2$. The Laplace transform method
automatically puts the jump into $y'$. This is because the formula
$sY(s)-y(0)$ for the Laplace transform of $y'$ assumes that $y$ is
continuous, but allows $y'$ to jump. In the same way, if we have a
higher-order equation of the same type -- such as $y''+y'+y=f$ -- the
jump will be put into $y''$, and $y$ and $y'$ will be kept continuous.
Happily, this is the appropriate thing to do in most applications.
\medskip
{\bf 12L Convolutions} The next theorem which gives further scope to the
Laplace transform method is the {\bf convolution theorem}, as follows:
\medskip
\noindent{\bf Theorem} Let $f$, $g$ be two functions with Laplace
transforms $F$, $G$, defined respectively for $\Real s>R$ and $\Real
s>S$. Set
$$h(x)=\int_0^xf(t)g(x-t)dt$$
\noindent for $x\ge 0$. Then $h$ has Laplace transform
$H(s)=F(s)G(s)$, at
least for $\Real s>\max(R,S)$.
\medskip
\noindent{\bf proof} We have
$$\eqalignno{H(s)
&=\int_0^{\infty}e^{-sx}\bigl(\int_0^xf(t)g(x-t)dt\bigr)dx\cr
&=\int_0^{\infty}f(t)\bigl(\int_t^{\infty}e^{-sx}g(x-t)dx\bigr)dt\cr
\noalign{\noindent reversing the order of integration over the domain
$\{(x,t):0\le t\le x\}$}
&=\int_0^{\infty}f(t)\bigl(\int_0^{\infty}e^{-s(u+t)}g(u)du\bigr)dt\cr
\noalign{\noindent substituting $u=x-t$}
&=\int_0^{\infty}f(t)e^{-st}
\bigl(\int_0^{\infty}e^{-su}g(u)du\bigr)dt\cr
&=\int_0^{\infty}f(t)e^{-st}
G(s)dt\cr
&=F(s)G(s),\cr}$$
\noindent as claimed.
\medskip
{\bf 12M Example} Let us return to the equation $y'+y=f$, $y(0)=0$, but
this time without specifying the function $f$, except to declare that it
does have a Laplace transform $F(s)$ defined at least when $\Real s$ is
large enough.
In this case, taking the Laplace transform of $y$ to be $Y$ as usual, we
get $sY+Y=F$ and $Y(s)=F(s){\Bover1{s+1}}$. Now this is a product of
Laplace transforms, so will be the Laplace transform of a convolution:
$F(s)$ is the transform of $f(x)$ and ${\Bover1{s+1}}$ is the transform
of $e^{-x}$, so ${\Bover{F(s)}{s+1}}$ will be the transform of $y(x)$,
where
$$y(x)=\int_0^xf(t)e^{-(x-t)}dt=e^{-x}\int_0^xe^tf(t)dt.$$
\noindent We can run this through the previous analysis, with $f(t)=e^t$
for $t<2$, $0$ thereafter: if $x\le 2$, then
$$y(x)=e^{-x}\int_0^xe^{2t}dt={1\over 2}e^{-x}(e^{2x}-1)
={1\over 2}(e^{x}-e^{-x}),$$
\noindent while if $x\ge 2$ then
$$y(x)=e^{-x}\int_0^2e^{2t}dt={1\over 2}e^{-x}(e^4-1),$$
\noindent just as before.
\medskip
{\bf 12N Notes} Of course the real value of this approach is when we
don't know the function $f$, or can't calculate its Laplace transform.
Of course, if we can't find a formula for the Laplace transform of $f$,
it's unlikely that we shall be able to get a formula for such an
expression as $\int_0^xe^tf(t)dt$. But it may be practicable to
estimate values of this integral by numerical methods, and thereby get a
table of values for $y$, even if we don't have it in \lq closed form'.
\medskip
{\bf 12O Example} Consider the equation
$$y(x)={1\over{x^2}}\int_0^xy(t)y(x-t)dt.$$
\noindent Let us, optimistically, assume that this equation has a
solution $y$ defined on $\coint{0,\infty}$ which has a Laplace transform
$Y(s)$ defined for at least some $s$.
Then we have
$$x^2y(x)=\int_0^xy(t)y(x-t)dt,$$
\noindent so that
$$Y''=Y^2.$$
\noindent There is a standard trick for solving differential equations
of the form $Y''=f(Y,Y')$; set $V=Y'$, so that
$$Y''={{d^2Y}\over{d^2s}}={{dv}\over{ds}}={{dV}\over{dY}}{{dY}\over{ds}}
=V{{dV}\over{dY}},$$
\noindent and
$$V{{dV}\over{dY}}=Y^2,\quad V\,dV=Y^2\,dY,
\quad {1\over 2}V^2={1\over 3}Y^3+\text{constant}.$$
\noindent To determine the value of the constant here, note that (if
this is going to work at all) $\lim_{\Real s\to\infty}V(s)=\lim_{\Real
s\to\infty}Y(s)=0$ (see 12Be), so we must have
$$\bigl({{dY}\over{ds}}\bigr)^2=V^2={2\over 3}Y^3.$$
\noindent Without fretting too much about whether $Y(s)^{3/2}$ is really
definable on a strict interpretation, write
$${{dY}\over{ds}}=\sqrt{2\over 3}Y^{3/2},\quad
{{dY}\over{Y^{3/2}}}=\sqrt{2\over 3}ds,
\quad -{2\over{Y^{1/2}}}=\sqrt{2\over 3}(s+c),\quad
Y={6\over{(s+c)^2}}.$$
\noindent This gives us
$$y=6e^{-cx}x.$$
All the arguments above are shaky enough so that we must immediately
check that this works: we have
$$\eqalign{\int_0^xy(t)y(x-t)dt
&=\int_0^x{6}e^{-ct}t.{6}e^{-c(x-t)}(x-t)dt\cr
&={{36}}e^{-cx}\int_0^xt(x-t)dt\cr
&={{36}}e^{-cx}\left[{1\over 2}xt^2-{1\over 3}t^3\right]_0^x\cr
&={6}x^3e^{-cx}\cr
&=x^2y(x),\cr}$$
\noindent as required. Of course this check is no guarantee that there
are not other, equally good solutions; for instance, it misses the
solution $y=0$ (corresponding to $c\to\infty$). But at least the
method picks up one non-trivial family of solutions.
\discrpage
\filename{cv2n13.tex}
\versiondate{14.1.99}
\def\varcheck#1{{\setover
{\lower 0.4ex\hbox{$\scriptscriptstyle\vee$}}{#1}}}
\def\varcheckf{{\setover{\,\lower 0.4ex
\hbox{$\scriptscriptstyle\vee$}}{f}}}
\def\varspcheck{^{\scriptscriptstyle\vee}}
\def\varhat#1{{\setover{\lower 0.4ex
\hbox{$\scriptscriptstyle\wedge$}}{#1}}}
\def\varhatf{{\setover{\,\lower 0.4ex
\hbox{$\scriptscriptstyle\wedge$}}{f}}}
\def\varsphat{^{\scriptscriptstyle\wedge}}
\noindent{\bf 13. Fourier Transforms}
For the last five lectures of the course, I seek to give a taste of
Fourier analysis. This will necessarily be unnaturally curtailed, but
I hope I can indicate both something of what it is and something of what
it's for.
\medskip
{\bf 13A The formulae} If $f$ is a {\it complex-valued} function of a
{\it real} variable, write
\Centerline{$\varhatf(y) = \Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{-iyx}f(x)dx$,}
\Centerline{$\varcheckf(y) = \Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{iyx}f(x)dx=\varhatf(-y)$.}
\noindent $\varhatf$ is the {\bf Fourier transform} of $f$, and
$\varcheckf$ is the {\bf inverse Fourier transform}.
\medskip
{\bf 13B Notes (a)} There are two particular things to watch in these
formulae; the constant $\Bover1{\sqrt{2\pi}}$ and the sign $-$ in
$e^{-ixy}$ in the first. Several choices are being made here, and
different authors do different things.
The six commonest variations are, I think,
\Centerline{$\varhatf(y)=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{\mp iyx}f(x)dx$,}
\Centerline{$\varhatf(y)=\int_{-\infty}^{\infty}
e^{\mp iyx}f(x)dx$,}
\Centerline{$\varhatf(y)=\int_{-\infty}^{\infty}
e^{\mp 2\pi iyx}f(x)dx$,}
\noindent corresponding to inverse transforms
\Centerline{$\varcheckf(y)=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{\pm iyx}f(x)dx$,}
\Centerline{$\varcheckf(y)=\Bover1{{2\pi}}\int_{-\infty}^{\infty}
e^{\pm iyx}f(x)dx$,}
\Centerline{$\varcheckf(y)=\int_{-\infty}^{\infty}
e^{\pm 2\pi iyx}f(x)dx$.}
\noindent The mathematics of the theory is identical whichever of these
you use, but the formulae are different, in sometimes surprising ways.
{\it I therefore strongly recommend} that you pick one and stick to it
(as I shall do). If, at any point in your future career, you find
yourself using one of the other systems, your first task will be to
rewrite all your notes, from beginning to end, in the new system, before
locking the original versions away where they can't confuse you.
In particular, I see that Kreyszig uses $\pm$ where I use $\mp$. Since
he hardly touches on the theory of (complex) Fourier transforms I do not
think this will affect you.
\medskip
{\bf (b)} I am deliberately not saying what I mean by
$\int_{-\infty}^{\infty}$ here. The `real' theory of Fourier analysis
is astonishingly robust under changing interpretations of its formulae.
Any pedantic insistence on saying exactly what one means is liable to
exclude some important idea. Of course when one comes to {\it proofs},
it is necessary to be precise; and there are many traps into which
blind faith can lead. This is not a course about proofs anyway, and I
think that the theory as a whole is best introduced not as a logical
structure but as a pattern of ideas; ideas which for a century and
three-quarters have been among the most fertile in mathematics.
Fourier's thesis itself was criticized, and sent back for revision, on
the grounds that the proofs were inadequate. And this was not just
that the thesis committee was being fussy. If you take a serious pure
mathematician's approach to Fourier analysis (e.g., T.W.K\"orner, {\it
Fourier Analysis,} Cambridge U.P., 1988), and look at the theorems
there, you will find that hardly one was proved by, or even known to,
Fourier himself. The very language in which they are stated
(`continuous function', `differentiable function', `function of bounded
variation', `square-integrable function') did not exist in his time.
It was sixty years after Fourier started the subject before anybody had
any solid idea of when the formulae worked and when they didn't. But
Fourier's {\it idea}, that one could study functions (whatever they
might be -- even that was not clear then!) in terms of their Fourier
series and transforms, has created as much mathematics as any since the
invention of calculus.
\lectureend{98/15}
\medskip
{\bf 13C Manipulating Fourier transforms} I think it may help if I write
down a list of the basic formulae of Fourier analysis, leaving
explanations to one side for the moment. The fundamental ideas are
encapsulated in the principles
\medskip
{\it Linearity} $\widehat{f+g}=\varhatf+\varhat{g}$,
\quad$\widehat{cf}=c\varhatf$.
\medskip
{\it Shift} If $g(x)=f(x+a)$ then $\varhat{g}(y)=e^{iya}\varhatf(y)$.
\medskip
{\it Scale} If $g(x)=f(ax)$, where $a>0$, then
$\varhat{g}(y)=\Bover1a\varhatf(\Bover{y}a)$.
\medskip
{\it Symmetry} If $g(x)=f(-x)$ then $\varhat{g}(y)=\varhatf(-y)$.
\medskip
{\it Complex Conjugate} If $g(x)=\overline{f(x)}$ then
$\varhat{g}(y)=\overline{\varhatf(-y)}$.
\medskip
{\it Differentiation} If $g(x)=f'(x)$ then
$\varhat{g}(y)=iy\varhatf(y)$.
\medskip
{\it Duality} $\int_{-\infty}^{\infty}f(x)\varhat{g}(x)dx
=\int_{-\infty}^{\infty}\varhatf(y)g(y)dy$.
\medskip
{\it Convolution} If we write $(f*g)(x)
=\int_{-\infty}^{\infty}f(t)g(x-t)dt$, then
\Centerline{$(f*g)\varsphat(y)=\sqrt{2\pi}\varhatf(y)\varhat{g}(y)$,
\quad$(f\times g)\varsphat(y)
=\Bover1{\sqrt{2\pi}}(\varhatf*\varhat{g})(y)$.}
\medskip
{\it Inversion} $\varhatf\varcheck{\phantom{h}}
=\varcheckf\varhat{\phantom{h}}=f$.
\medskip
{\it Reversal} $\varcheckf(y)=\varhatf(-y)$.
\medskip
{\it Parseval} $\int_{-\infty}^{\infty}f(x)\overline{g(x)}dx
=\int_{-\infty}^{\infty}\varhatf(y)\overline{\varhat{g}(y)}dy$.
\medskip
{\bf 13D Where the rules come from (I)} I say again, we are not here in
the business of `proving' anything. But I think that they will be
easier to remember, and make better sense, if we take a moment to look
at ways in which some, at least, of the rules above grow out of the
defining formulae.
\medskip
{\bf (a) Linearity} All this says is that
\Centerline{$\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{-ixy}(f+g)(x)dx
=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ixy}f(x)dx
+\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ixy}g(x)dx$,}
\Centerline{$\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{-ixy}(cf)(x)dx
=c\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ixy}f(x)dx$;}
\noindent that is, that $\int$ is a linear operator.
\medskip
{\bf (b) Shift} This is a simple substitution $\xi=x+c$:
$$\eqalign{\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{-ixy}f(x+c)dx
&=\Bover1{\sqrt{2\pi}}
\int_{-\infty}^{\infty}e^{-i(\xi-c)y}f(\xi)d\xi\cr
&=e^{icy}\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{-i\xi y}f(\xi)d\xi.\cr}$$
\medskip
{\bf (c) Scale} This is the substitution $\xi=cx$:
$$\eqalign{\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{-ixy}f(cx)dx
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{-i\xi y/c}f(\xi)\Bover{d\xi}c\cr
&=\Bover1c\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{-i\xi y/c}f(\xi)d\xi.\cr}$$
\medskip
{\bf (d) Symmetry} This is the substitution $\xi=-x$:
$$\eqalign{\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{-ixy}f(-x)dx
&=\Bover1{\sqrt{2\pi}}\int_{\infty}^{-\infty}
e^{i\xi y}f(\xi)(-d\xi)\cr
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{-i\xi(-y)}f(\xi)d\xi.\cr}$$
\medskip
{\bf (e) Complex Conjugate} This is because
$\overline{e^{-ixy}}=e^{ixy}$ and $\int\overline{f}=\overline{\int f}$:
$$\eqalign{\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{-ixy}\overline{f(x)}dx
&=\overline{\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{ixy}f(x)dx}.\cr}$$
\medskip
{\bf 13E Where the rules come from (II)} The formulae in 13D rely only
on the most fundamental properties of integration on the real line, and
are accordingly universal and robust and can be trusted in almost any
context in which they make sense at all. The next group use more
sophisticated ideas and are correspondingly both less reliable and more
important.
\medskip
{\bf (a) Differentiation} This is integration by parts:
$$\eqalign{\Bover1{\sqrt{2\pi}}&\int_{-\infty}^{\infty}
e^{-ixy}f'(x)dx\cr
&=\Bover1{\sqrt{2\pi}}\biggl[e^{-ixy}f(x)\biggr]_{-\infty}^{\infty}
-\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
(-iy)e^{-ixy}f(x)dx\cr
&=\Bover1{\sqrt{2\pi}}\bigl(\lim_{x\to\infty}e^{-ixy}f(x)
-\lim_{x\to-\infty}e^{-ixy}f(x)\bigr)
+iy\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{-ixy}f(x)dx\cr
&=iy\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{-ixy}f(x)dx\cr}$$
\noindent at least when $\lim_{x\to\infty}f(x)
=\lim_{x\to-\infty}f(x)=0$.
\medskip
{\bf (b) Duality} This involves a change in the order of integration:
$$\eqalign{\Bover1{\sqrt{2\pi}}&\int_{-\infty}^{\infty}f(x)
\biggl(\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{-ixy}g(y)dy\biggr)dx\cr
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(y)
\biggl(\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{-ixy}f(x)dx\biggr)dy\cr}$$
\noindent(we hope).
\medskip
{\bf (c) Convolution} Another change in the order of integration:
$$\eqalignno{\Bover1{\sqrt{2\pi}}&\int_{-\infty}^{\infty}e^{-ixy}
\biggl(\int_{-\infty}^{\infty}f(t)g(x-t)dt\biggr)dx\cr
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)
\biggl(\int_{-\infty}^{\infty}e^{-ixy}g(x-t)dx\biggr)dt\cr
\noalign{\noindent (we hope)}
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)
\biggl(\int_{-\infty}^{\infty}e^{-i(\xi+t)y}g(\xi)d\xi\biggr)dt\cr
\noalign{\noindent (substituting $\xi=x-t$)}
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{-ity}f(t)\sqrt{2\pi}\varhat{g}(y)dt\cr
&=\sqrt{2\pi}\varhatf(y)\varhat{g}(y).\cr}$$
\noindent The other `convolution' formula comes from putting this
together with the `inversion' formula.
\medskip
{\bf 13F The hard one: inversion} Here we do not really have the
apparatus to make sense of the general phenomenon. What I will do
is to give a handful of examples in which we can do the calculations and
see at least (i) a sort of explanation of the factor $\sqrt{2\pi}$ we
use (ii) an indication of why the inverse Fourier transform is not
identical to the Fourier transform.
\medskip
{\bf (a) Example} Set $f(x)=1$ for $-1\le x\le 1$, $0$ otherwise. Then
$$\eqalign{\varhatf(y)
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ixy}f(x)dx
=\Bover1{\sqrt{2\pi}}\int_{-1}^1e^{-ixy}dx\cr
&=\Bover1{\sqrt{2\pi}}\biggl[e^{-ixy}\biggr]_{-1}^1
=\Bover1{\sqrt{2\pi}}\Bover1{-iy}(e^{-iy}-e^{iy})\cr
&=\Bover1{\sqrt{2\pi}}\Bover1{iy}(e^{iy}-e^{-iy})
=\Bover1{\sqrt{2\pi}}\Bover{2\sin y}{y}.\cr}$$
\noindent So
$$\eqalign{\varhatf\varcheck{\phantom{h}}(0)
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{i0y}\varhatf(y)dy\cr
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
\Bover1{\sqrt{2\pi}}\Bover{2\sin y}{y}dy\cr
&=\Bover1{\pi}\int_{-\infty}^{\infty}
\Bover{\sin y}{y}dy\cr
&=\Bover2{\pi}\int_{0}^{\infty}
\Bover{\sin y}{y}dy
=1
=f(0)\cr}$$
\noindent by the calculation in 7E above.
Of course this is extraordinarily specialized; we cannot even find
$\varhatf\varcheck{\phantom{h}}(x)$ for any non-zero $x$ by this method,
let alone for other functions $f$.
\medskip
{\bf (b)} Consider the function $f(x)=e^{-x}$ for $x\ge 0$, $0$ for
$x<0$. Then
$$\eqalign{\varhatf(y)
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ixy}f(x)dx
=\Bover1{\sqrt{2\pi}}\int_{0}^{\infty}e^{-ixy}e^{-x}dx\cr
&=\Bover1{\sqrt{2\pi}}
\biggl[\Bover1{-1-iy}e^{-ixy}e^{-x}\biggr]_0^{\infty}
=\Bover1{\sqrt{2\pi}}
\Bover1{1+iy}.\cr}$$
\noindent So
$$\eqalign{\varhatf\varcheck{\phantom{h}}(x)
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{ixy}\varhatf(y)dy\cr
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
\Bover1{\sqrt{2\pi}}\Bover{e^{ixy}}{1+iy}dy\cr
&=\Bover1{2\pi}\int_{-\infty}^{\infty}
\Bover{e^{ixy}}{1+iy}dy.\cr}$$
\noindent Now, at least for $x>0$, this is made for one of our standard
techniques from Chapter 7. Integrate $\Bover{e^{ixz}}{1+iz}$ around a
large semicircle based on $[-R,R]$. By Jordan's Lemma, the integral
around the curved section will converge to $0$ (because $x>0$ and
$\lim_{|z|\to\infty}|\bover1{1+iz}|=0$), while the integral around the
whole contour will be
\Centerline{$2\pi i\Res_{i}=2\pi i\Bover{e^{ixi}}{i}=2\pi e^{-x}$.}
\noindent So
$$\eqalign{\Bover1{2\pi}\int_{-\infty}^{\infty}
\Bover{e^{ixy}}{1+iy}dy
=\Bover1{2\pi}\lim_{R\to\infty}\int_{-R}^R
\Bover{e^{ixy}}{1+iy}dy
=\Bover1{2\pi}2\pi e^{-x}
=e^{-x}
=f(x).\cr}$$
\noindent For $x<0$ we can use the same method, provided we first change
the integral to
\Centerline{$\Bover1{2\pi}\int_{-\infty}^{\infty}
\Bover{e^{i(-x)(-y)}}{1+iy}dy
=\Bover1{2\pi}\int_{-\infty}^{\infty}
\Bover{e^{i(-x)y}}{1-iy}dy$.}
\noindent Again using the big-semicircle method, we find that (because
$-x>0$) we just have to integrate $\Bover{e^{-ixz}}{1-iz}$ around large
semicircles above the real axis; but this is always zero, because the
singularity is now at $-i$, outside the curve. So
$\varhatf\varcheck{\phantom{h}}(x)=0$ for $x<0$.
Finally, {\it provided} we interpret it as
\Centerline{$\lim_{R\to\infty}\Bover1{2\pi}\int_{-R}^{R}
\Bover1{1+iy}dy$,}
\noindent we can compute $\varhatf\varcheck{\phantom{h}}(0)$, as follows.
$\Bover1{1+iy}=\Bover1i\Bover{d}{dy}\Ln(1+iy)$ for all real $y$
($1+iy\in\dom\Ln$ for all real $y$, and $\Ln$ is analytic). So the
integral is
$$\eqalign{\lim_{R\to\infty}\Bover1{2\pi}\int_{-R}^{R}
\Bover1{1+iy}dy
&=\Bover1{2\pi}\lim_{R\to\infty}
\biggl[\Bover1i\Ln(1+iy)\biggr]_{-R}^R\cr
&=\Bover1{2\pi i}\lim_{R\to\infty}\Ln(1+iR)-\Ln(1-iR)\cr
&=\Bover1{2\pi i}\lim_{R\to\infty}
\ln(|1+iR|)+i\arg(1+iR)\cr
&\qquad\qquad\qquad\qquad -\ln(|1-iR|)-i\arg(1-iR)\cr
&=\Bover1{2\pi}\lim_{R\to\infty}
\arg(1+iR)-\arg(1-iR)\cr
&=\Bover1{2\pi}\bigl(\Bover{\pi}2-(-\Bover{\pi}2)\bigr)
=\Bover12.\cr}$$
\noindent I include this last step because it is characteristic of the
theory of Fourier transforms, as in Fourier series, that the inverse
Fourier transform, when interpreted as
$\Bover1{\sqrt{2\pi}}\lim_{R\to\infty}\int_{-R}^Re^{ixy}\varhatf(y)dy$,
converges at jumps to $\bover12(\lim_{t\downarrow x}f(t)+\lim_{t\uparrow
x}f(t))$; here we have
\Centerline{$\Bover12(\lim_{t\downarrow 0}f(t)+\lim_{t\uparrow 0}f(t))
=\Bover12(\lim_{t\downarrow 0}e^{-t})+\lim_{t\uparrow 0}0)
=\Bover12$.}
\medskip
{\bf (c)} For a third example, I calculate the Fourier transform of the
function $f(x)=e^{-x^2/2}$. Here we have
\Centerline{$\varhatf(y)
=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ixy}e^{-x^2/2}dx$.}
\noindent Once again, we have a contour-integration trick for this.
Integrate $e^{-z^2/2}$ around a large rectangle with vertices at $\pm
R$, $\pm R-iy$. $e^{-z^2/2}$ is an entire function, defined everywhere
in $\Bbb C$, so the whole integral will be zero, by Cauchy's theorem.
The four sides of the rectangle give contributions
(along base) parametrize as $z=x$ for $-R\le x\le R$, $dz=dx$,
\Centerline{$I_1=\int_{-R}^Re^{-x^2/2}dx
\to\int_{-\infty}^{\infty}e^{-x^2/2}dx=\sqrt{2\pi}$}
\noindent (I hope you knew this from MA102 or MA257).
(along right edge) Length of edge is $|y|$, modulus of function is at
most
$$\eqalign{\sup_{|t|\le|y|}|\exp(-\Bover12(R+it)^2)|
&=\sup_{|t|\le|y|}|\exp\Bover12(-R^2+t^2-2iRt)|\cr
&=\sup_{|t|\le|y|}|\exp\Bover12(-R^2+t^2)\exp(-iRt)|\cr
&=\sup_{|t|\le|y|}\exp\Bover12(-R^2+t^2)
=e^{-R^2/2}e^{y^2/2}.\cr}$$
\noindent So
\Centerline{$|I_2|\le|y|e^{-R^2/2}e^{y^2/2}\to 0$ as $R\to\infty$.}
(other horizontal edge) parametrize as $z=-x-iy$ for $-R\le x\le R$, $dz=-dx$, to get
$$\eqalign{I_3
&=\int_{-R}^R\exp(-(-x-iy)^2/2)(-dx)\cr
&=-\int_{-R}^R\exp(\Bover12(-x^2+y^2-2ixy))dx\cr
&=-e^{y^2/2}\int_{-R}^Re^{-ixy}e^{-x^2/2}dx\cr
&\to-e^{y^2/2}\int_{-\infty}^{\infty}e^{-ixy}e^{-x^2/2}dx\cr}$$
\noindent as $R\to\infty$.
(left-hand edge) Again, the length of the curve is $|y|$ and the
greatest value of the modulus of the integrand is
$$\eqalign{\sup_{|t|\le|y|}|\exp(-\Bover12(-R+it)^2)|
&=\sup_{|t|\le|y|}|\exp\Bover12(-R^2+t^2+2iRt)|\cr
&=\sup_{|t|\le|y|}|\exp\Bover12(-R^2+t^2)\exp iRt|\cr
&=\sup_{|t|\le|y|}\exp\Bover12(-R^2+t^2)\cr
&=e^{-R^2/2}e^{y^2/2}.\cr}$$
\noindent So once again we have
\Centerline{$|I_4|\le|y|e^{-R^2/2}e^{y^2/2}\to 0$ as $R\to\infty$.}
Putting these together, we have
$$\eqalign{0
&=I_1+I_2+I_3+I_4\cr
&\to\int_{-\infty}^{\infty}e^{-x^2/2}dx+0
-e^{y^2/2}\int_{-\infty}^{\infty}e^{-ixy}e^{-x^2/2}dx+0\cr}$$
\noindent as $R\to\infty$. So we must have
\Centerline{$\int_{-\infty}^{\infty}e^{-ixy}e^{-x^2/2}dx
=e^{-y^2/2}\int_{-\infty}^{\infty}e^{-x^2/2}dx
=\sqrt{2\pi}e^{-y^2/2}$,}
\noindent and
\Centerline{$\varhatf(y)
=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ixy}e^{-x^2}dx
=e^{-y^2/2}$.}
(Note that since I have not decided what the sign of $y$ is, I don't
know whether $I_1$ or $I_3$ is the `bottom' edge, so I don't know
whether I am going round the rectangle in a positive or negative
orientation. But as we are applying Cauchy's theorem, rather than the Residue Theorem, it doesn't matter.)
\lectureend{98/17}
\medskip
\noindent{\bf Remarks} It is one of the delights of mathematics that a
function like $e^{-x^2/2}$, of enormous importance in probability theory,
can reappear in a pivotal role in a quite different subject. The
result in (c) above is itself of great importance in probability theory,
where we need to be able to calculate the `characteristic functions' of
distributions.
Note that in each of the three examples, the factor $\Bover1{2\pi}$ or
$\Bover1{\sqrt{2\pi}}$ introduced by the Fourier transforms cancels out
nicely against formulae
($\int_{-\infty}^{\infty}\Bover{\sin x}xdx=\pi$,
$\int_{\Gamma}f(z)dz=2\pi i\sum\Res$,
$\int_{-\infty}^{\infty}\exp(-x^2/2)dx=\sqrt{2\pi}$) which had $\pi$'s
in them for apparently unrelated reasons.
\medskip
{\bf 13G Parseval} is now easy. Observe that
$$\eqalign{\overline{\varhat{g}(y)}
&=\overline{\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{-ixy}g(x)dx}\cr
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{ixy}\overline{g(x)}dx\cr
&=\varcheck{\bar g}(y).\cr}$$
\noindent So
$$\eqalign{\int_{-\infty}^{\infty}
&\varhatf(y)\overline{\varhat{g}(y)}dy\cr
&=\int_{-\infty}^{\infty}\varhatf(y)\varcheck{\bar g}(y)dy\cr
&=\int_{-\infty}^{\infty}f(y)\varcheck{\bar g}{\varsphat}(y)dy\cr
&=\int_{-\infty}^{\infty}f(y)\overline{g(y)}dy\cr}$$
\noindent because $\varcheck{\bar g}{\varsphat}=\bar g$.
(I repeat: this is just to show that the formulae hang together. The
question of just which functions $f$ and $g$ behave in such a way that
the integrals here make sense is quite another matter.)
\medskip
{\bf 13H The heat equation} One of Fourier's own applications of the
ideas here (and one of the ones people criticized as having
unsatisfactory foundations) was to a partial differential equation
\Centerline{$\Bover{\partial f}{\partial t}=\Bover12\Bover{\partial^2
f}{\partial x^2}$.}
\noindent This is one of the most important equations of applied
mathematics. The characteristic feature of partial differential
equations is that (because we have a function of two or more variables)
the specification of boundary conditions is enormously subtle. Here I
want to look at the problem in which we are given $f(x,0)=f_0(x)$ and
seek to find $f(x,t)$ for $t\ge 0$. Now, let us take the Fourier
transform of the equation. If we look at
\Centerline{$\widehat{\Bover{\partial f}{\partial t}}(y)
=\Bover1{\sqrt{2\pi}}\int_{\infty}^{-\infty}e^{-ixy}\Bover{\partial
f}{\partial t}(x,t)dx$}
\noindent then, provided the operators $\int dx$ and
$\bover{\partial}{\partial t}$ commute (which does happen, for
well-enough-behaved functions -- never mind how well-behaved they have
to be) we can hope that, setting $g(x,t)=\Bover{\partial f}{\partial
t}(x,t)$, we shall get
\Centerline{$\varhat{g}(y,t)
=\Bover1{\sqrt{2\pi}}\Bover{\partial}{\partial
t}\int_{\infty}^{\infty}e^{-ixy}f(x,t)dx
=\Bover{\partial}{\partial t}\varhatf(y,t)$.}
\noindent On the other hand, using the standard formula
\Centerline{$(f')\varsphat(y)=iy\varhatf(y)$,
\quad$(f'')\varsphat(y)=-y^2\varhatf(y)$}
\noindent we get
\Centerline{$\varhat{g}(y,t)=-\Bover12y^2\varhatf(y,t)$}
\noindent because $g=\Bover12\Bover{\partial^2 f}{\partial x^2}$. So
we have the equation
\Centerline{$\Bover{\partial}{\partial t}\varhatf
=-\Bover12y^2\varhatf$.}
\noindent Now this we can solve; the solution is
\Centerline{$\varhatf(y,t)=e^{-ty^2/2}\varhatf(y,0)
=e^{-ty^2/2}\varhatf_0(y)$.}
\noindent Now we need to know the inverse Fourier transform of
$e^{-ty^2/2}$. But we know that if $h(x)=e^{-x^2/2}$ then
$\varhat{h}(y)=e^{-y^2/2}$ (13Fc); so if we set $h_{\gamma}(x)=h(\gamma
x)=e^{-\gamma^2x^2/2}$, then
\Centerline{$\varhat{h}_{\gamma}(y)
=\Bover1{\gamma}\varhat{h}(\Bover{y}{\gamma})
=\Bover1{\gamma}e^{-y^2/2\gamma^2}$,}
\noindent at least for $\gamma>0$, by the `scaling' rule. Setting
$\gamma=1/\sqrt{t}$, therefore, we see that the Fourier transform of
$k_t(x)=\Bover1{\sqrt{t}}e^{-x^2/2t}$ is $e^{-ty^2/2}$. Accordingly
the Fourier transform of $k_t*f_0$ is
\Centerline{$(k_t*f_0)\varsphat
=\sqrt{2\pi}\varhat{k}_t\times\varhatf_0
=\sqrt{2\pi}\varhat
f$}
\noindent by the `convolution' rule. So we get back to $f$ by the
formula
\Centerline{$f_t=\Bover1{\sqrt{2\pi}}k_t*f_0$,}
\lectureend{98/18}
\noindent that is,
\Centerline{$f(x,t)
=\Bover1{\sqrt{2\pi t}}\int_{-\infty}^{\infty}
e^{-u^2/2t}f_0(x-u)du$.}
(Let me repeat, as a pure mathematician, that Fourier's critics were
absolutely right to doubt the validity of the method. Any `proof'
along the lines here must rely on some exceedingly deep ideas.
But once we have
the formula, derived by whatever method, we can then set out to check
that it makes sense and gives valid solutions to the equations starting
from whatever functions $f_0$ we are interested in. For instance, the
final formula allows such a function as $f_0(x)=|x|$, for which the
Fourier transform is a very doubtful construction.)
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$w=f(z)=1/z$ transforms
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