\chapter{The Fourier transform}
\label{fouri}
\index{transform}
\index{Fourier transform}
\index{Fourier, Jean Baptiste Joseph\\(1768--1830)} % bad break
% diagn: this chapter has been pretty heavily revised; it's long, so
% one might not review it line by line, but it should probably be read
% end to end.

The Fourier series of chapter~\ref{fours} is useful.
Applications of the series are extensive.  However, the series applies solely
to waveforms that repeat---or, at most, to waveforms that can be framed
as though they repeated.

\index{nonrepeating waveform}
\index{waveform!nonrepeating}
An effort to extend the Fourier series to the broader domain of
nonrepeating waveforms leads to the \emph{Fourier transform,} this
chapter's chief subject.

% ----------------------------------------------------------------------

\section{The Fourier transform}
\label{fouri:100}
\index{Fourier transform}
\index{transform!Fourier}

This section extends the Fourier series to derive the Fourier transform.

\subsection{Fourier's equation}
\label{fouri:100.10}
\index{Fourier's equation}

\index{pulse}
\index{pulse train}
\index{nonrepeating waveform}
\index{waveform!nonrepeating}
\index{integral!as the continuous limit of a sum}
\index{sum!continuous limit of}
Consider the nonrepeating waveform or \emph{pulse} of
Fig.~\ref{fouri:100:fig}.
\begin{figure}
  \caption{A pulse.}
  \label{fouri:100:fig}
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        \uput[u](0,\xxyb){$f(t)$}%
      }%
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         \xxaxes
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           /e 2.71828182845905 def
           /scale 0.7 def
           /pulse {
             /A     exch def
             /sigma exch def
             /to    exch def
             e x scale div to sub sigma div dup mul 2.0 div neg exp A mul scale mul
           } def
           -0.20 1.00  2.20 pulse
            0.50 0.80 -1.80 pulse
           -2.00 0.60  1.10 pulse
           add add
         }
      }
    \end{pspicture}
  \ec
\end{figure}%
Because the pulse does not repeat it has no Fourier series, yet
one can however give it something very like a Fourier series in the
following way.  First, convert the pulse $f(t)$ into the pulse train
\[
  g(t) \equiv \sum_{n=-\infty}^\infty f(t-nT_1),
\]
which naturally does repeat,%
\footnote{%
  One could divert rigorously from this point to consider formal
  requirements against $f(t)$ but for applications it probably suffices that
  $f(t)$ be limited enough in extent that $g(t)$ exist for all $\Re(T_1)
  > 0$, $\Im(T_1) = 0$.  Formally, such a condition would forbid a
  function like $f(t) = A \cos \omega_o t$, but one can evade the
  formality, among other ways, by defining the function as $f(t) =
  \lim_{T_2\ra\infty} \Pi(t/T_2) A \cos \omega_o t$, the $\Pi(\cdot)$
  being the rectangular pulse of~(\ref{fours:095:10}).
  %---even if (as would probably be the case) $T_2 \gg T_1$.
  Other pulses of \S~\ref{fours:095} might suit, as well.%
}
where $T_1 > 0$ is an arbitrary period of repetition whose value you can choose.
Second, by~(\ref{fours:100:15}), calculate the Fourier coefficients of
this pulse train $g(t)$.  Third, use these coefficients in the Fourier
series~(\ref{fours:100:10}) to reconstruct
\[
  g(t) = \sum_{j=-\infty}^{\infty}
  \left\{
    \left[
      \frac{1}{T_1} \int_{-T_1/2}^{T_1/2}
        e^{-ij\,\Delta\omega\,\tau}
          g(\tau)
        \,d\tau
    \right]
  e^{ij \,\Delta\omega\, t}
  \right\}.
\]
Fourth, observing that $\lim_{T_1\ra\infty}g(t) = f(t)$, recover from
the train the original pulse
\[
  f(t) = \lim_{T_1\ra\infty}\sum_{j=-\infty}^{\infty}
  \left\{
    \left[
      \frac{1}{T_1} \int_{-T_1/2}^{T_1/2}
        e^{-ij\,\Delta\omega\,\tau}
          f(\tau)
        \,d\tau
    \right]
  e^{ij \,\Delta\omega\, t}
  \right\}
\]
[in which we have replaced $g(\tau)$ by $f(\tau)$, supposing that~$T_1$
has grown great enough to separate the several instances $f(\tau-nT_1)$
of which, according to definition, $g(\tau)$ is composed];
or, observing per~(\ref{fours:080:08}) that $\Delta\omega\,T_1 = 2\pi$
and reordering factors,
\[
  f(t) = \lim_{\Delta\omega\ra 0^{+}}
  \frac{1}{\sqrt{2\pi}} \sum_{j=-\infty}^{\infty}
    e^{ij \,\Delta\omega\, t}
    \left[
      \frac{1}{\sqrt{2\pi}}
      \int_{-2\pi/2\,\Delta\omega}^{2\pi/2\,\Delta\omega}
        e^{-ij\,\Delta\omega\,\tau}
            f(\tau)
        \,d\tau
    \right] \Delta\omega.
\]
Fifth, defining the symbol $\omega \equiv j\,\Delta\omega$, observe
that the summation is really an integration in the limit, such that
\bq{fouri:eqn}
  f(t) =
  \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}
    e^{i\omega t}
    \left[
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty}
        e^{-i\omega\tau}
            f(\tau)
        \,d\tau
    \right] d\omega.
\eq
This is \emph{Fourier's equation,} a remarkable, highly significant
result.

\subsection{The transform and inverse transform}
\label{fouri:100.20}
\index{inverse Fourier transform}
\index{Fourier transform!inverse}

\index{\fourf}
The reader may agree that Fourier's equation~(\ref{fouri:eqn}) is
curious, but in what way is it remarkable?
To answer, let us observe
that the quantity within the square brackets of~(\ref{fouri:eqn}),
\bq{fouri:xform}
  F(\omega) \equiv \fourf\left\{f(t)\right\} \equiv
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty}
        e^{-i\omega\tau}
            f(\tau)
        \,d\tau,
\eq
is a function not of~$t$ but rather of~$\omega$.  We conventionally give
this function the capitalized symbol $F(\omega)$ and name it the
\emph{Fourier transform} of $f(t)$, introducing also the
notation $\fourf\{\cdot\}$ (where the script letter~$\fourf$,
which stands for ``Fourier,'' is only accidentally the same letter
as~$f$ and~$F$) as a short form to represent the
transformation~(\ref{fouri:xform}) serves to define.
Substituting~(\ref{fouri:xform}) into~(\ref{fouri:eqn}) and changing
$\eta \la \omega$ as the dummy variable of integration, we have that
\bq{fouri:invxform}
  f(t) = \fourf^{-1}\left\{F(\omega)\right\} =
  \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}
    e^{i\eta t}
    F(\eta) \,d\eta.
\eq
This last is the \emph{inverse Fourier transform} of the function
$F(\omega)$.

\index{frequency content}
\index{frequency!infinitesimally graded}
\index{complex exponential!infinitesimally graded}
\index{grading, infinitesimal}
\index{infinitesimal grading}
The Fourier transform~(\ref{fouri:xform}) serves as a continuous measure
of a function's frequency content.  To understand why this should be so,
consider that~(\ref{fouri:invxform}) constructs a
function $f(t)$ of an infinity of infinitesimally graded complex
exponentials and that~(\ref{fouri:xform}) provides the
weights $F(\omega)$ for the construction.  Indeed, the Fourier
transform's complementary equations~(\ref{fouri:invxform})
and~(\ref{fouri:xform}) are but continuous versions of the earlier
complementary equations~(\ref{fours:100:10}) and~(\ref{fours:100:15}) of
the discrete Fourier series.  The transform finds even wider application
than the series does.%
\footnote{%
  Regrettably, several alternate definitions and usages of the Fourier
  series are current.  Alternate
  definitions~\cite{Phillips/Parr}\cite{Couch} handle the factors of
  $1/\sqrt{2\pi}$ differently, as for instance in \S~\ref{lapl:360}.
  Alternate usages~\cite{Feynman} change $-i\la i$ in certain
  circumstances.  The essential Fourier mathematics however remains the
  same in any case.  The reader can adapt the book's presentation at
  need to the Fourier definition and usage his colleagues prefer.%
}

Figure~\ref{fouri:100:figt} plots the Fourier transform of the pulse of
Fig.~\ref{fouri:100:fig}.
\begin{figure}
  \caption{The Fourier transform of the pulse of Fig.~\ref{fouri:100:fig}.}
  \label{fouri:100:figt}
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      {%
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        \psline(-\xxxab,0)(\xxxab,0)%
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        \uput[r](\xxxab,0){$\omega$}%
        \psline(-0.39,0.80)(-0.51,0.92)(-0.71,0.92)
        \uput[l](-0.60,0.92){$\Re[F(\omega)]$}
        \psline(-1.03,-0.94)(-1.15,-1.06)(-1.35,-1.06)
        \uput[l](-1.24,-1.06){$\Im[F(\omega)]$}
      }%
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           /e 2.71828182845905 def
           /rad 57.2957795130823 def
           /scale 0.7 def
           /pulse {
             /A     exch def
             /sigma exch def
             /to    exch def
             e x scale div sigma mul dup mul 2.0 div neg exp A mul
             x scale div to mul neg rad mul cos mul scale mul
           } def
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            0.50 0.80 -1.80 pulse
           -2.00 0.60  1.10 pulse
           add add
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           /e 2.71828182845905 def
           /rad 57.2957795130823 def
           /scale 0.7 def
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             /A     exch def
             /sigma exch def
             /to    exch def
             e x scale div sigma mul dup mul 2.0 div neg exp A mul
             x scale div to mul neg rad mul sin mul scale mul
           } def
           -0.20 1.00  2.20 pulse
            0.50 0.80 -1.80 pulse
           -2.00 0.60  1.10 pulse
           add add
         }
      }
    \end{pspicture}
  \ec
\end{figure}%

\subsection{The complementary variables of transformation}
\label{fouri:100.30}
\index{transformation!variable of}
\index{variable of transformation}

\index{complementary variables of transformation, the}
\index{independent variable!Fourier transform and}
\index{Fourier transform!independent variable and}
\index{domain!time and frequency}
\index{frequency domain}
\index{time domain}
\index{transform domain}
\index{domain!transform}
If~$t$ represents time then~$\omega$ represents angular frequency
as \S~\ref{fours:085} has explained.  In this case the function $f(t)$
is said to operate in the \emph{time domain} and the corresponding
transformed function $F(\omega)$, in the \emph{frequency domain.}  The
mutually independent variables~$\omega$ and~$t$ are then the
\emph{complementary variables of transformation.}

\index{Fourier transform pair}
\index{transform pair}
\index{dummy variable}
Formally, one can use any two letters in place of the~$\omega$
and~$t$; and indeed one need not even use two different letters,
for it is sometimes easier just to write,
\bq{fouri:byu}
  \begin{split}
  F(v) = \fourf\left\{f(v)\right\} &=
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty}
        e^{-iv\theta}
            f(\theta)
        \,d\theta, \\
  f(v) = \fourf^{-1}\left\{F(v)\right\} &=
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty}
        e^{iv\theta}
            F(\theta)
        \,d\theta, \\
  \fourf &\equiv \fourf_{vv},
  \end{split}
\eq
in which the~$\theta$ is in itself no variable of transformation but
only a dummy variable.  To emphasize the distinction between the
untransformed and transformed (respectively typically time and
frequency) domains, however, scientists and engineers tend to
style~(\ref{fouri:byu}) as
\bq{fouri:byu2}
  \begin{split}
  F(\omega) = \fourf\left\{f(t)\right\} &=
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty}
        e^{-i\omega t}
            f(t)
        \,dt, \\
  f(t) = \fourf^{-1}\left\{F(\omega)\right\} &=
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty}
        e^{i\omega t}
            F(\omega)
        \,d\omega, \\
  \fourf &\equiv \fourf_{\omega t},
  \end{split}
\eq
which are just~(\ref{fouri:xform})
and~(\ref{fouri:invxform}) together with their dummy variables
\linebreak % bad break
changed.
For precision of specification, one can affix subscripts as shown:
$\fourf_{vv}$; $\fourf_{\omega t}$.  However, the
unadorned~$\fourf$ is normally clear enough in context.)

Whichever letter or letters might be used for the independent variable,
the functions
\bq{fouri:100:30}
  f(v) \fouripair F(v)
\eq
constitute a \emph{Fourier transform pair.}

\subsection{Transforms of the basic nonanalytic pulses}
\label{fouri:100.40}
\index{Fourier transform!example of}
% The writer has checked this section's two rolloff
% pulses by FFT. At this writing, the check was still kept
% in ../debian/helper/deprecated/check_nonanalytic_pulses.m.
% The three nonrolloff pulses were not directly checked but, if the
% rolloffs are good, the nonrolloffs probably are, too, since they are
% limiting cases of the rolloffs.

\index{triangular pulse!Fourier transform of}
\index{Fourier transform!of a triangular pulse}
As a Fourier example, consider the triangular pulse $\Lambda(v)$
of~(\ref{fours:095:10}).  Its Fourier transform according
to~(\ref{fouri:byu}) is
\bqb
  \fourf\left\{\Lambda(v)\right\} &=&
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty}
        e^{-iv\theta}
           \Lambda(\theta)
        \,d\theta
  \\&=&
    \frac{1}{\sqrt{2\pi}} \left\{
      \int_{-1}^{0} e^{-iv\theta} (1+\theta) \,d\theta
      +
      \int_{0}^{1}  e^{-iv\theta} (1-\theta) \,d\theta
    \right\}.
\eqb
Evaluating the integrals according to Table~\ref{inttx:470:tbl}'s
antiderivative that $\theta e^{-iv\theta} = [d/d\theta] [
e^{-iv\theta}(1+iv\theta)/v^2 ]$,
\bqb
  \fourf\left\{\Lambda(v)\right\} &=&
    \frac{1}{v^2\sqrt{2\pi}} \bigg\{
      \left[ e^{-iv\theta} [1+(iv)(1+\theta)] \right]_{\theta=-1}^{0}
  \\&&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mbox{}
      +
      \left[ e^{-iv\theta} [-1+(iv)(1-\theta)] \right]_{\theta=0}^{1}
    \bigg\}
  \\&=&
    \frac{\sinarg^2(v/2)}{\sqrt{2\pi}},
\eqb
where $\sinarg(\cdot)$ is the sine-argument function
of~(\ref{fours:160:10}).  Thus we find the Fourier transform pair
\bq{fouri:100:41}
  \Lambda(v) \fouripair \frac{\sinarg^2(v/2)}{\sqrt{2\pi}}.
\eq

\index{square pulse!Fourier transform of}
\index{Fourier transform!of a square pulse}
\index{raised-cosine pulse!Fourier transform of}
\index{Fourier transform!of a raised-cosine pulse}
One can compute other Fourier transform pairs in
like manner, such as\footnote{%
  To verify~(\ref{fouri:100:44}) and~(\ref{fouri:100:45}) is left as
  an exercise.  Hint toward~(\ref{fouri:100:45}): $\sin(v\pm\pi) =
  -\sin v$.%
  %See also \S~\ref{lapl:123.95}.
}%
\begin{eqnarray}
  \Pi(v) &\fouripair& \frac{\sinarg(v/2)}{\sqrt{2\pi}},\label{fouri:100:44}\\
  \Psi(v) &\fouripair& \frac{\sinarg v}{\sqrt{2\pi}\left[1-(v/\pi)^2\right]}.\label{fouri:100:45}
\end{eqnarray}
One can compute yet further transform pairs by the duality rule and
other properties of \S~\ref{fouri:110}.

\index{raised cosine-rolloff pulse!Fourier transform of}
\index{Fourier transform!of a raised cosine-rolloff pulse}
\index{convergence!related to the raised cosine-rolloff pulse}
\index{divergence!related to the square pulse}
\index{raised cosine-rolloff pulse!convergence related to}
\index{square pulse!divergence related to}
\index{integral!absolute}
\index{absolute integral}
The algebra can be intricate, though.  As a second Fourier example,
consider the raised cosine-rolloff pulse $\Psi_r(t)$
of~(\ref{fours:095:50}):
\bqb
  \lefteqn{\left(\!\sqrt{2\pi}\right)\fourf\left\{\Psi_r(v)\right\}}
  %\\&=&
  %  \frac 1 2 \int_{-(1+r)/2}^{-(1-r)/2} e^{-iv\theta}
  %  \left\{
  %    1 - \sin \left[\frac{\pi \left(-\theta-\frac 1 2\right)}{r}\right]
  %  \right\}d\theta
  %\\&&\quad\mbox{} +
  %  \int_{-(1-r)/2}^{(1-r)/2} e^{-iv\theta}\,d\theta
  %\\&&\quad\mbox{} +
  %  \frac 1 2 \int_{(1-r)/2}^{(1+r)/2} e^{-iv\theta}
  %  \left\{
  %    1 - \sin \left[\frac{\pi \left(\theta-\frac 1 2\right)}{r}\right]
  %  \right\}d\theta
  \\&=&
    \int_{-(1+r)/2}^{-(1-r)/2}
    \left\{
      \frac{e^{-iv\theta}}{2}
      + \frac{e^{i\left[(-v+\pi/r)\theta+2\pi/4r\right]}}{i4}
      - \frac{e^{i\left[(-v-\pi/r)\theta-2\pi/4r\right]}}{i4}
    \right\}d\theta
  \\&&\quad\mbox{} +
    \int_{-(1-r)/2}^{(1-r)/2} e^{-iv\theta}\,d\theta
  \\&&\quad\mbox{} +
    \int_{(1-r)/2}^{(1+r)/2}
    \left\{
      \frac{e^{-iv\theta}}{2}
      + \frac{e^{i\left[(-v-\pi/r)\theta+2\pi/4r\right]}}{i4}
      - \frac{e^{i\left[(-v+\pi/r)\theta-2\pi/4r\right]}}{i4}
    \right\}d\theta
  \\&=&
    \left\{
      -\frac{e^{-iv\theta}}{i2v}
      - \frac{e^{i\left[(-v+\pi/r)\theta+2\pi/4r\right]}}{4(-v+\pi/r)}
      + \frac{e^{i\left[(-v-\pi/r)\theta-2\pi/4r\right]}}{4(-v-\pi/r)}
    \right\}_{-(1+r)/2}^{-(1-r)/2}
  \\&&\quad\mbox{} +
    \left\{-\frac{e^{-iv\theta}}{iv}\right\}_{-(1-r)/2}^{(1-r)/2}
  \\&&\quad\mbox{} +
    \left\{
      -\frac{e^{-iv\theta}}{i2v}
      - \frac{e^{i\left[(-v-\pi/r)\theta+2\pi/4r\right]}}{4(-v-\pi/r)}
      + \frac{e^{i\left[(-v+\pi/r)\theta-2\pi/4r\right]}}{4(-v+\pi/r)}
    \right\}_{(1-r)/2}^{(1+r)/2},
\eqb
and so on.  After two more pages or so of algebra (left as a tedious
though not especially difficult exercise to try the accuracy of the interested
reader's pencil),
% And here is the tedious algebra, with checks, if you wish to review it,
% but it's too boring for the reader to compile it into the book.
%\bqb
%  \lefteqn{\left(\!\sqrt{2\pi}\right)\fourf\left\{\Psi_r(v)\right\}}
%  \\&=&
%    \left\{
%      -\frac{e^{iv(1-r)/2}}{i2v}
%      + \frac{e^{i\left[(v+\pi/r)(1-r)/2-2\pi/4r\right]}}{4(-v-\pi/r)}
%      - \frac{e^{i\left[(v-\pi/r)(1-r)/2+2\pi/4r\right]}}{4(-v+\pi/r)}
%    \right\}
%  \\&&\quad\mbox{} +
%    \left\{
%      \frac{e^{iv(1+r)/2}}{i2v}
%      - \frac{e^{i\left[(v+\pi/r)(1+r)/2-2\pi/4r\right]}}{4(-v-\pi/r)}
%      + \frac{e^{i\left[(v-\pi/r)(1+r)/2+2\pi/4r\right]}}{4(-v+\pi/r)}
%    \right\}
%  \\&&\quad\mbox{} +
%    \left\{-\frac{e^{-iv(1-r)/2}}{iv}+\frac{e^{iv(1-r)/2}}{iv}\right\}
%  \\&&\quad\mbox{} +
%    \left\{
%      -\frac{e^{-iv(1+r)/2}}{i2v}
%      + \frac{e^{i\left[(-v+\pi/r)(1+r)/2-2\pi/4r\right]}}{4(-v+\pi/r)}
%      - \frac{e^{i\left[(-v-\pi/r)(1+r)/2+2\pi/4r\right]}}{4(-v-\pi/r)}
%    \right\}
%  \\&&\quad\mbox{} +
%    \left\{
%      \frac{e^{-iv(1-r)/2}}{i2v}
%      - \frac{e^{i\left[(-v+\pi/r)(1-r)/2-2\pi/4r\right]}}{4(-v+\pi/r)}
%      + \frac{e^{i\left[(-v-\pi/r)(1-r)/2+2\pi/4r\right]}}{4(-v-\pi/r)}
%    \right\}
%  \\&=&
%    e^{iv(1-r)/2}\left\{
%      \frac{1}{i2v}
%      + \frac{e^{i\left[(\pi/r)(1-r)/2-2\pi/4r\right]}}{4(-v-\pi/r)}
%      - \frac{e^{i\left[(-\pi/r)(1-r)/2+2\pi/4r\right]}}{4(-v+\pi/r)}
%    \right\}
%  \\&&\quad\mbox{} +
%    e^{iv(1+r)/2}\left\{
%      \frac{1}{i2v}
%      - \frac{e^{i\left[(\pi/r)(1+r)/2-2\pi/4r\right]}}{4(-v-\pi/r)}
%      + \frac{e^{i\left[(-\pi/r)(1+r)/2+2\pi/4r\right]}}{4(-v+\pi/r)}
%    \right\}
%  \\&&\quad\mbox{} +
%    e^{iv(-1-r)/2}\left\{
%      -\frac{1}{i2v}
%      + \frac{e^{i\left[(\pi/r)(1+r)/2-2\pi/4r\right]}}{4(-v+\pi/r)}
%      - \frac{e^{i\left[(-\pi/r)(1+r)/2+2\pi/4r\right]}}{4(-v-\pi/r)}
%    \right\}
%  \\&&\quad\mbox{} +
%    e^{iv(-1+r)/2}\left\{
%      -\frac{1}{i2v}
%      - \frac{e^{i\left[(\pi/r)(1-r)/2-2\pi/4r\right]}}{4(-v+\pi/r)}
%      + \frac{e^{i\left[(-\pi/r)(1-r)/2+2\pi/4r\right]}}{4(-v-\pi/r)}
%    \right\}
%  \\&=&
%    e^{iv(1-r)/2}\left\{
%      \frac{1}{i2v}
%      + \frac{e^{-i2\pi/4}}{4(-v-\pi/r)}
%      - \frac{e^{i2\pi/4}}{4(-v+\pi/r)}
%    \right\}
%  \\&&\quad\mbox{} +
%    e^{iv(1+r)/2}\left\{
%      \frac{1}{i2v}
%      - \frac{e^{i2\pi/4}}{4(-v-\pi/r)}
%      + \frac{e^{-i2\pi/4}}{4(-v+\pi/r)}
%    \right\}
%  \\&&\quad\mbox{} +
%    e^{iv(-1-r)/2}\left\{
%      -\frac{1}{i2v}
%      + \frac{e^{i2\pi/4}}{4(-v+\pi/r)}
%      - \frac{e^{-i2\pi/4}}{4(-v-\pi/r)}
%    \right\}
%  \\&&\quad\mbox{} +
%    e^{iv(-1+r)/2}\left\{
%      -\frac{1}{i2v}
%      - \frac{e^{-i2\pi/4}}{4(-v+\pi/r)}
%      + \frac{e^{i2\pi/4}}{4(-v-\pi/r)}
%    \right\}
%\eqb
%\bqb
%  \lefteqn{\fourf\left\{\Psi_r(v)\right\}}
%  \\&=&
%    e^{iv(1-r)/2}\left\{
%      \frac{1}{i2v}
%      - \frac{1}{i4(v+\pi/r)}
%      - \frac{1}{i4(v-\pi/r)}
%    \right\}
%  \\&&\quad\mbox{} +
%    e^{iv(1+r)/2}\left\{
%      \frac{1}{i2v}
%      - \frac{1}{i4(v+\pi/r)}
%      - \frac{1}{i4(v-\pi/r)}
%    \right\}
%  \\&&\quad\mbox{} +
%    e^{iv(-1-r)/2}\left\{
%      -\frac{1}{i2v}
%      + \frac{1}{i4(v-\pi/r)}
%      + \frac{1}{i4(v+\pi/r)}
%    \right\}
%  \\&&\quad\mbox{} +
%    e^{iv(-1+r)/2}\left\{
%      -\frac{1}{i2v}
%      + \frac{1}{i4(v-\pi/r)}
%      + \frac{1}{i4(v+\pi/r)}
%    \right\}
%  \\&=&
%    e^{iv/2}\cos\left(\frac{rv} 2 \right)
%    \left\{
%      \frac{1}{iv}
%      - \frac{1}{i2(v+\pi/r)}
%      - \frac{1}{i2(v-\pi/r)}
%    \right\}
%  \\&&\quad\mbox{} +
%    e^{-iv/2}\cos\left(\frac{rv} 2 \right)
%    \left\{
%      -\frac{1}{iv}
%      + \frac{1}{i2(v-\pi/r)}
%      + \frac{1}{i2(v+\pi/r)}
%    \right\}
%  \\&=&
%    e^{iv/2}\cos\left(\frac{rv} 2 \right)
%    \left\{
%      \frac{1}{iv}
%      + \frac{iv}{v^2-(\pi/r)^2}
%    \right\}
%  \\&&\quad\mbox{} +
%    e^{-iv/2}\cos\left(\frac{rv} 2 \right)
%    \left\{
%      -\frac{1}{iv}
%      - \frac{iv}{v^2-(\pi/r)^2}
%    \right\}
%  \\&=&
%    ive^{iv/2}\cos\left(\frac{rv} 2 \right)
%    \left\{
%      -\frac{1}{v^2}
%      + \frac{1}{v^2-(\pi/r)^2}
%    \right\}
%  \\&&\quad\mbox{} +
%    ive^{-iv/2}\cos\left(\frac{rv} 2 \right)
%    \left\{
%      \frac{1}{v^2}
%      - \frac{1}{v^2-(\pi/r)^2}
%    \right\}
%  \\&=&
%    \left[iv\cos\left(\frac{rv} 2 \right)\right]
%    \left[e^{iv/2}-e^{-iv/2}\right]
%    \left[ -\frac{1}{v^2} + \frac{1}{v^2-(\pi/r)^2} \right]
%  \\&=&
%    \left[iv\cos\frac{rv} 2 \right]
%    \left[i2\sin\frac v 2\right]
%    \left[ -\frac{1}{v^2} + \frac{1}{v^2-(\pi/r)^2} \right]
%  \\&=&
%    2v\cos\left(\frac{rv} 2\right) \sin\left(\frac v 2\right)
%    \left[ \frac{1}{v^2} - \frac{1}{v^2-(\pi/r)^2} \right]
%  \\&=&
%    \cos\left(\frac{rv} 2\right) \sinarg\left(\frac v 2\right)
%    \left[ 1 - \frac{1}{1-(\pi/rv)^2} \right]
%  \\&=&
%    \cos\left(\frac{rv} 2\right) \sinarg\left(\frac v 2\right)
%    \left[ \frac{-(\pi/rv)^2}{1-(\pi/rv)^2} \right]
%  \\&=&
%    \cos\left(\frac{rv} 2\right) \sinarg\left(\frac v 2\right)
%    \left[ \frac{1}{1-(rv/\pi)^2} \right]
%  \\&=&
%    \frac
%    {\cos(rv/2) \sinarg(v/2)}
%    {\left[1-(rv/\pi)^2\right]}
%\eqb
%\[
%  \begin{split}
%    \Psi_0(v) &= \frac 1 {\sqrt{2\pi}}
%    2v \sin\frac v 2
%    \left[ \frac{1}{v^2} \right]
%    =
%      \frac 1 {\sqrt{2\pi}}\sinarg\frac v 2
%    \\
%    \Psi_1(v) &= \frac 1 {\sqrt{2\pi}}
%    2v\cos\frac{v} 2 \sin\frac v 2
%    \left[ \frac{1}{v^2} - \frac{1}{v^2-\pi^2} \right]
%    \\&= \frac 1 {\sqrt{2\pi}}
%    v \sin v
%    \left[ \frac{1}{v^2} - \frac{1}{v^2-\pi^2} \right]
%    = \frac 1 {\sqrt{2\pi}}
%    \sinarg v
%    \left[ 1 - \frac{v^2}{v^2-\pi^2} \right]
%    \\&= \frac 1 {\sqrt{2\pi}}
%    \sinarg v
%    \left[ \frac{-\pi^2}{v^2-\pi^2} \right]
%    = \frac 1 {\sqrt{2\pi}}
%    \sinarg v
%    \left[ \frac{\pi^2}{\pi^2-v^2} \right]
%    = \frac 1 {\sqrt{2\pi}}
%    \sinarg v
%    \left[ \frac{1}{1-(v/\pi)^2} \right]
%  \end{split}
%\]
the result is that\footnote{%
    The conscientious reader might check~(\ref{fouri:100:46})
    against~(\ref{fouri:100:44}) and~(\ref{fouri:100:45}), the former with
    $r=0$, the latter with $r=1$.  The former looks all right but the latter
    looks wrong at first glance, until one recalls from
    Table~\ref{trig:275:table} the trigonometric identity that $\sin2\alpha =
    2\sin\alpha\cos\alpha$.
}
\bq{fouri:100:46}
  \Psi_r(v) \fouripair \frac{\cos(rv/2)\sinarg(v/2)}{\sqrt{2\pi}\left[1-(rv/\pi)^2\right]}.
\eq
Unless $r=0$, the~$v^2$ in the transform's denominator lends $\fourf\{\Psi_r(v)\}$
the possibly significant property that
$\int_{-\infty}^\infty|\fourf\{\Psi_r(v)\}|\,dv$
converges as $\int dv/v^3$ does, whereas
$\int_{-\infty}^\infty|\fourf\{\Pi(v)\}|\,dv$ does not converge but
diverges as $\int dv/v$.  Using a small but nonzero~$r$, applications
(\S~\ref{fouri:100.90}) sometimes replace $\Pi(v)$ by $\Psi_r(v)$
% diagn: the new parenthetical (actually, square-bracketed) note that
% follows wants review; the rest of the sentence following the note has been
% modified and also wants review.
[or, alternately, if convergence\footnote{%
  \index{band confinement}%
  Rather than ``convergence,'' an electrical engineer might instead say
  ``band confinement,'' where by ``band'' the engineer means ``range of
  frequencies.''  The engineer would change $t \la v$, transform
  via~$\fourf_{\omega t}$ rather than~$\fourf_{vv}$, and, in the
  result, observe that $\Psi_r(t)$ is even less active at high values
  of the frequency~$\omega$ than $\Lambda_r(t)$ is.%
}
less aggressive than that of $\Psi_r(v)$ is tolerable,
maybe instead by the simpler $\Lambda_r(v)$] for this or related reasons.

Similarly,\footnote{%
  If you work these out for yourself and get
  \[
    \Lambda_r(v) \fouripair
    \frac{2}{\left(\!\sqrt{2\pi}\right)rv^2}\left[
      \cos\frac{(1-r)v}{2} - \cos\frac{(1+r)v}{2}
    \right]
  \]
  then your answer is right.  Apply a sum-of-angles
  identity of Table~\ref{trig:275:table} to reach the formula the
  narrative gives.%
}
\bq{fouri:100:48}
    \Lambda_r(v) \fouripair
    \frac{\sinarg(rv/2)\sinarg(v/2)}{\sqrt{2\pi}}.
\eq

\index{l'H\^opital's rule}
\index{raised cosine-rolloff pulse!vanishing denominator in the Fourier transform of}
\index{vanishing denominator!in the Fourier transform of a raised cosine-rolloff pulse}
\index{denominator!vanishing, in the Fourier transform of a raised cosine-rolloff pulse}
One thing left to consider is the values of the raised-cosine transforms when the
denominators of~(\ref{fouri:100:45}) and~(\ref{fouri:100:46}) vanish.
For the latter,
\[
  \begin{split}
    \lim_{rv\ra\pm\pi}\frac{\cos(rv/2)\sinarg(v/2)}{\sqrt{2\pi}\left[1-(rv/\pi)^2\right]}
    &=
      \lim_{rv\mp\pi\ra 0}\frac{\sin[(\pi\mp rv)/2]
      \sinarg(v/2)}{\sqrt{2\pi}(1\mp rv/\pi)(1\pm rv/\pi)}
    \\&=
      \lim_{rv\mp\pi\ra 0}\frac{\pi^2\sin[(\pi\mp rv)/2]
      \sinarg(v/2)}{\sqrt{2\pi}(\pi\mp rv)(\pi\pm rv)}
    \\&=
      \lim_{rv\mp\pi\ra 0}\frac{(2\pi)^{3/2}\sinarg[(\pi\mp rv)/2]
      \sinarg(v/2)}{8(\pi\pm rv)}
    \\&=
      \frac{\left(\!\sqrt{2\pi}\right)\sinarg(v/2)}{8}
    =
      \frac{\left(\!\sqrt{2\pi}\right)\sinarg(2\pi/4r)}{8}.
  \end{split}
\]
For the former, using l'H\^opital's rule~(\ref{drvtv:260:lhopital}) and the
derivative~(\ref{fours:160:30}),
\[
  \begin{split}
    \lim_{v\ra\pm\pi} \frac{\sinarg v}{\sqrt{2\pi}\left[1-(v/\pi)^2\right]}
    = \left.\frac{\sinarg v-\cos v}{2\sqrt{2\pi}(v/\pi)^2}\right|_{v=\pm\pi}
    = \frac{1}{2\sqrt{2\pi}},
  \end{split}
\]
which, reassuringly, agrees with the former at $r=1$.
Summarizing,
\bq{fouri:100:47}
  \begin{split}
    \fourf\{\Psi(v)\}_{v=\pm\pi} &= \frac{1}{2\sqrt{2\pi}},
    \\
    \fourf\{\Psi_r(v)\}_{rv=\pm\pi}
    &= \frac{\left(\!\sqrt{2\pi}\right)\sinarg(v/2)}{8}
    = \frac{\left(\!\sqrt{2\pi}\right)\sinarg(2\pi/4r)}{8}.
  \end{split}
\eq

% diagn: this recently added subsection wants one more review
\subsection{Approximation to an arbitrary pulse}
\label{fouri:100.90}
\index{pulse!arbitrary}
\index{arbitrary pulse}
\index{approximation to an arbitrary pulse}
\index{construction of an arbitrary pulse}

\index{integral!absolute}
\index{absolute integral}
\index{sampling}
\index{interval, sampling}
\index{abruptness}
\index{slope!infinite}
\index{infinite slope}
Before closing the section, we should briefly notice an important
application of the several nonanalytic pulses of \S~\ref{fours:095},
whose Fourier transforms \S~\ref{fouri:100.40} has computed and
Table~\ref{fouri:120:tbl20} will list.
% diagn: revised sentence wants review.
Sampling a pulse $f(t)$ at equal, finite intervals of~$\Delta t$, one
can approximate the pulse as
\bq{fouri:100:90}
  \tilde f(t) \equiv \sum_{k=-\infty}^{\infty}
  f(k\,\Delta t)h\left(\frac{t-k\,\Delta t}{\Delta t}\right),
\eq
where $h(t)$ is any of the basic nonanalytic pulses
of \S~\ref{fours:095.10} and Fig.~\ref{fours:095:fig1},
whether $\Pi(t)$, $\Lambda(t)$ or $\Psi(t)$;
or is either of the rolloff variants
of \S~\ref{fours:095.50} and Fig.~\ref{fours:095:fig5},
whether $\Lambda_r(t)$ or $\Psi_r(t)$.  Naturally,
there are constraints for this to work.  For instance, $f(t)$ must not
change so quickly or abruptly that sufficiently dense samples fail
to approximate the function.  Fortunately, whether $f(t)$
submits to such constraints tends to be clearer in an actual
application with a specific, concrete $f(t)$ than it is in the
abstract; so, other than to warn the reader regarding any $f(t)$
that suffers an infinite slope---like
% The parenthesis on the right side of the next line's equation are not
% strictly necessary but, because of the / divisions that operate at
% different levels on the right side, make the equation visually easier
% to read.
$f(t)=(te^{-|t|^2/2})/\sqrt{|t|}$,
for example---we shall not
further study the constraints here (but see \S~\ref{lapl:123}, which
develops constructive functions that impose fewer constraints).

\index{superposition!of pulses}
Having approximated the pulse by~(\ref{fouri:100:90}), having thus
reduced the pulse to a superposition $\tilde f(t)$ of pulses whose Fourier
transforms we already know, we can approximate that $\fourf\{f(t)\}
\approx \fourf\{\tilde f(t)\}$, where
\bq{fouri:100:91}
  \fourf\left\{\tilde f(t)\right\} = \sum_{k=-\infty}^{\infty}
  f(k\,\Delta t)\fourf\left\{h\left(\frac{t-k\,\Delta t}{\Delta t}\right)\right\}.
\eq
Qualifications regarding convergence and such could be multiplied against
% bad break
(\ref{fouri:100:91})
but, again, that is not our purpose here. Other than to note that the
square pulse $\Pi(t)$ might, in some applications,
be a poor choice for $h(t)$ because the
absolute integral $\int_{-\infty}^\infty|\fourf\{\Pi(t)\}|\,d\omega$ of
its transform fails to converge, we will leave the question in that
form.

We said that we already knew the transforms of the various basic
nonanalytic pulses.  However,~(\ref{fouri:100:90})
and~(\ref{fouri:100:91}) have used $h[(t-k\,\Delta t)/\Delta t]$ rather
than a plain $h(t)$.  Fortunately, the conversion is not difficult.
It is that
\bq{fouri:100:95}
  \fourf\left\{h\left(\frac{t-k\,\Delta t}{\Delta t}\right)\right\}
  = e^{-ik\,\Delta t\,\omega}\,\Delta t\,H(\omega\,\Delta t),
\eq
as we shall see in Table~\ref{fouri:110:tbl20} of \S~\ref{fouri:110}, next.

% ----------------------------------------------------------------------

\section{Properties of the Fourier transform}
\label{fouri:110}
\index{Fourier transform!properties of}

The Fourier transform obeys an algebra of its own, exhibiting properties
the mathematician can exploit to extend the transform's reach.  This
section derives and lists several such properties.

\subsection{Symmetries of the real and imaginary parts}
\label{fouri:110.15}
\index{symmetry!in the Fourier transform}
\index{Fourier transform!symmetries of}

In Fig.~\ref{fouri:100:figt}, page~\pageref{fouri:100:figt}, each of the
real and imaginary parts of the Fourier transform is
symmetrical (or at least each \emph{looks} symmetrical), though the
imaginary symmetry differs from the real.
% diagn: the next sentence, revised, wants a brief review
The present subsection analytically develops the symmetries.

\index{conjugate!Fourier transform of}
\index{Fourier transform!of a complex conjugate}
\index{Fourier transform!real and imaginary parts of}
\index{real part!of the Fourier transform}
\index{imaginary part!of the Fourier transform}
The Fourier transform of a function's conjugate according
to~(\ref{fouri:byu}) is
\[
  \fourf\{f^{*}(v)\} =
    \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-iv\theta}
    f^{*}(\theta)\, d\theta
  = \left[ \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{iv^{*}\theta} f(\theta)\, d\theta\right]^{*},
\]
in which we have taken advantage of the fact that the integrand's dummy
variable $\theta = \theta^{*}$ happens to be real.
On the other hand, within the paragraph's first equation just above,
\[
    \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{iv^{*}\theta}
    f(\theta)\, d\theta =
    \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-i(-v^{*})\theta}
    f(\theta)\, d\theta =
    F(-v^{*}),
\]
implying that%
\footnote{
  From past experience with complex conjugation, an applied
  mathematician might naturally have expected of~(\ref{fouri:110:18a})
  that $f^{*}(v) \fouripair F^{*}(v)$, but the expectation though
  natural would have been incorrect.
  Unlike most of the book's
  mathematics before chapter~\ref{fours}, eqns.~(\ref{fours:100:10})
  and~(\ref{fours:100:15})---and thus ultimately also the Fourier
  transform's definition~(\ref{fouri:xform})
  or~(\ref{fouri:byu})---have arbitrarily chosen a
  particular sign for the~$i$ in the phasing factor
  $e^{-ij\,\Delta\omega\,\tau}$ or~$e^{-iv\theta}$, which phasing factor
  the Fourier integration bakes into the transformed function $F(v)$, so
  to speak.  The Fourier transform as such therefore does not meet
  \S~\ref{alggeo:225.2}'s condition for~(\ref{alggeo:225:conj2}) to
  hold.  Fortunately,~(\ref{fouri:110:18a}) does hold.

  Viewed from another angle, it must be so, because Fourier transforms
  real functions into complex ones.  See Figs.~\ref{fouri:100:fig}
  and~\ref{fouri:100:figt}.
}
\bq{fouri:110:18a}
  \begin{split}
    f^{*}(v) &\fouripair F^{*}(-v^{*});\\
    f^{*}(t) &\fouripair F^{*}(-\omega^{*}).
  \end{split}
\eq
%
If we express the real and imaginary parts of $f(v)$ in the style
of~(\ref{alggeo:225:25}) as
\[
  \begin{split}
    \Re[f(v)] &= \frac{f(v) + f^{*}(v)}{2}, \\
    \Im[f(v)] &= \frac{f(v) - f^{*}(v)}{i2},
  \end{split}
\]
then the Fourier transforms of these parts according
to~(\ref{fouri:110:18a}) are%
\footnote{
  The precisely orderly reader might note that a forward reference
  to Table~\ref{fouri:110:tbl20} is here implied; but the property
  referred to, Fourier superposition $A_1f_1(v) + A_2f_2(v) \fouripair
  A_1F_1(v) + A_2F_2(v)$, which does not depend on this subsection's
  results anyway, is so trivial to prove that we will not bother about
  the precise ordering in this instance.
}
\bq{fouri:110:19}
  \begin{split}
    \Re[f(v)] &\fouripair \frac{F(v) + F^{*}(-v^{*})}{2}, \\
    \Im[f(v)] &\fouripair \frac{F(v) - F^{*}(-v^{*})}{i2}.
  \end{split}
\eq
For real~$v$ and an $f(v)$ which is itself real for all real~$v$, the
latter line becomes
\[
  0 \fouripair \frac{F(v) - F^{*}(-v)}{i2}
  \ \ \mbox{if $\Im(v) = 0$ and, for all such~$v$, $\Im[f(v)] = 0$,}
\]
whereby\footnote{%
  A professional mathematician might object that we had
  never established a one-to-one correspondence between a function and
  its transform.  On the other hand, recognizing the applied spirit of
  the present work, the professional might waive the objection---if not
  with pleasure, then at least with a practical degree of
  indulgence---but see also \S~\ref{fours:100.80}.%
}
\bq{fouri:110:19b}
  F(v) = F^{*}(-v)
  \ \ \mbox{if $\Im(v) = 0$ and, for all such~$v$, $\Im[f(v)] = 0$.}
\eq
Interpreted,~(\ref{fouri:110:19b}) says for real~$v$ and $f(v)$ that the
plot of $\Re[F(v)]$ must be symmetric about the vertical axis whereas
the plot of $\Im[F(v)]$ must be symmetric about the origin, as
Fig.~\ref{fouri:100:figt} has illustrated.

\subsection{Duality}
\label{fouri:110.10}
\index{duality}
\index{Fourier transform!dual of}

\index{Fourier transform!reversing the independent variable of}
Changing $-v \la v$ makes (\ref{fouri:byu})'s second line to read
\[
  f(-v) =
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty}
        e^{-iv\theta}
            F(\theta)
        \,d\theta.
\]
However, according to (\ref{fouri:byu})'s
% or (\ref{fouri:byu2})'s?
first line, this says neither more nor less than that
\bq{fouri:110:10}
  \begin{split}
    F(v) &\fouripair f(-v),\\
    F(t) &\fouripair f(-\omega),
  \end{split}
\eq
which is that the transform of the transform is the original function
with the independent variable reversed, an interesting and useful
property.  It is entertaining,
and moreover enlightening, to combine~(\ref{fouri:100:30})
and~(\ref{fouri:110:10}) to form the endless transform progression
\bq{fouri:110:12}
  \cdots
  \fouripair f(v)
  \fouripair F(v)
  \fouripair f(-v)
  \fouripair F(-v)
  \fouripair f(v)
  \fouripair \cdots
\eq
Equation~(\ref{fouri:110:10}) or alternately~(\ref{fouri:110:12})
expresses the Fourier transform's property of \emph{duality.}

\index{pulse!basic nonanalytic, dual transform of}
\index{nonanalytic pulse!dual transform of}
\index{square pulse!dual transform of}
\index{triangular pulse!dual transform of}
\index{raised-cosine pulse!dual transform of}
For an example of duality, recall that \S~\ref{fouri:100.40} has
computed the transform pairs
\[
  \begin{split}
    \Pi(v) &\fouripair \frac{\sinarg(v/2)}{\sqrt{2\pi}}, \\
    \Lambda(v) &\fouripair \frac{\sinarg^2(v/2)}{\sqrt{2\pi}}, \\
    \Psi(v) &\fouripair \frac{\sinarg v}{\sqrt{2\pi}\left[1-(v/\pi)^2\right]}.
  \end{split}
\]
Application of~(\ref{fouri:110:10}) yields the additional, dual
transform pairs\footnote{The last of these three duals though logical is
admittedly, probably not very useful---except maybe in a few special
cases in which the applied mathematician, already expecting it, will
hardly need to look it up.  Table~\ref{fouri:120:tbl20} omits it.}
\bq{fouri:110:15}
  \begin{split}
    \frac{\sinarg(v/2)}{\sqrt{2\pi}} &\fouripair \Pi(v), \\
    \frac{\sinarg^2(v/2)}{\sqrt{2\pi}} &\fouripair \Lambda(v), \\
    \frac{\sinarg v}{\sqrt{2\pi}\left[1-(v/\pi)^2\right]} &\fouripair \Psi(v),
  \end{split}
\eq
in which that $\Pi(-v)=\Pi(v)$, $\Lambda(-v)=\Lambda(v)$ and
$\Psi(-v)=\Psi(v)$ are
observed (but eqn.~\ref{fouri:110:10} works as well on pulses that lack
such symmetry).  Without duality, to use~(\ref{fouri:byu}) to compute
the transform pairs of~(\ref{fouri:110:15}) might have been hard, but
with duality it's pretty easy, as you see.
(Section~\ref{fouri:120} incidentally will further improve
eqn.~\ref{fouri:110:15}.)

\subsection{The Fourier transform of the Dirac delta}
\label{fouri:110.20}
\index{delta function, Dirac!Fourier transform of}
\index{Dirac delta function!Fourier transform of}
\index{Fourier transform!of the Dirac Delta}

\index{constant!Fourier transform of}
\index{Fourier transform!of a constant}
\index{$1$ (one)!Fourier transform of}
\index{Dirac delta function!sifting property of}
\index{delta function, Dirac!sifting property of}
Section~\ref{fouri:120}
will compute several Fourier transform pairs but one particular pair
is so significant, so illustrative, so curious, and so easy to compute
that we will pause to compute it and its dual now.
Applying~(\ref{fouri:byu}) to the Dirac delta~(\ref{integ:670:20}) and
invoking its sifting property~(\ref{integ:670:sift}), we find that
\bq{fouri:110:20}
  \delta(v) \fouripair \frac{1}{\sqrt{2\pi}},
\eq
the dual of which according to~(\ref{fouri:110:12}) is
\bq{fouri:110:22}
  1 \fouripair
  \left(\!\sqrt{2\pi}\right) \delta(v)
\eq
inasmuch as $\delta(-v) = \delta(v)$.

The duality rule again incidentally proves its worth in
eqn.~\ref{fouri:110:22}.  Had we tried to calculate the Fourier
transform of~$1$---that is, of $f(v)\equiv 1$---directly according to
eqn.~\ref{fouri:byu} we would have found the pair
$
  1 \fouripair
\linebreak % bad break
  (1/\sqrt{2\pi})
    \int_{-\infty}^{\infty} e^{-iv\theta} \,d\theta,
$
the right side of which features an integral impossible to evaluate.  A
limit of some kind might perhaps have been enforced to circumvent the
impossibility, but as you see again, duality is easier.

\subsection{Delay and shift}
\label{fouri:110.25}
\index{Fourier transform!of a delayed function}
\index{Fourier transform!frequency-shifted}
\index{delay}
\index{frequency shift}
\index{shift!in frequency}

Applying~(\ref{fouri:byu}) to $f(v-a)$ and changing $\xi \la \theta - a$,
we have the transform property of \emph{delay:}
\bq{fouri:110:25}
  f(v-a) \fouripair e^{-iav}F(v).
\eq
Applying~(\ref{fouri:byu}) to $e^{iav}f(v)$, we have the transform
property of \emph{frequency shift:}
\bq{fouri:110:26}
  e^{iav}f(v) \fouripair F(v-a).
\eq

\subsection{Metaduality}
\label{fouri:110.27}
\index{duality}
\index{metaduality}
\index{Fourier transform!dual of}
\index{Fourier transform!metadual of}

Section~\ref{fouri:110.10} has shown how to compose the dual of a
Fourier transform pair.  One can likewise compose the
\emph{metadual}\footnote{%
  % diagn: new footnote wants review
  Other books the writer has read just call it a ``dual.'' However, the
  book you are reading finds it expedient to disambiguate by modifying
  the term where the term is used of properties rather than of mere
  pairs.%
}
of a Fourier transform \emph{property,} but to do so correctly wants
careful handling.

Defining\footnote{%
  The Roman~$w$ of this subsection is not the Greek~$\omega$ of
  \S~\ref{fouri:100.10}.%
}%
\[
  \begin{split}
    F(w) &\equiv \fourf \{f(u)\}, \\
    \Phi(u) &\equiv \fourf \{\phi(w)\}, \\
    \phi(w) &\equiv F(w),
  \end{split}
\]
one can write~(\ref{fouri:110:12}) as
\bq{fouri:110:27f}
  \br{ccccccccc}
    \cdots\ \fouripair &
    \Phi(-u) & \fouripair &
    \phi(w) & \fouripair &
    \Phi(u) & \fouripair &
    \phi(-w) & \fouripair \ \cdots
    \\
    \cdots\ \fouripair &
    f(u) & \fouripair &
    F(w) & \fouripair &
    f(-u) & \fouripair &
    F(-w) & \fouripair \ \cdots
  \er
\eq
in which $\phi(w)=F(w)$ are vertically aligned, $\Phi(u)=f(-u)$ are
vertically aligned, and so on.
Similarly,
\bq{fouri:110:27g}
  \br{ccccccccc}
    \cdots\ \fouripair &
    \Gamma(-v) & \fouripair &
    \gamma(v) & \fouripair &
    \Gamma(v) & \fouripair &
    \gamma(-v) & \fouripair \ \cdots
    \\
    \cdots\ \fouripair &
    g(v) & \fouripair &
    G(v) & \fouripair &
    g(-v) & \fouripair &
    G(-v) & \fouripair \ \cdots
  \er
\eq
These are just to repeat~(\ref{fouri:110:12}) in various symbols $\phi
\la F$ and $\gamma \la G$, so they say nothing new, but the variant
symbology helps for example as follows.

% diagn: this paragraph, revised, wants review, especially of its
% prose style.
Let the delay property~(\ref{fouri:110:25}) be styled as
\[
  \gamma(v) \fouripair \Gamma(v),
\]
where
\[
  \begin{split}
    \gamma(v) &\equiv \phi(v-a), \\
    \Gamma(v) &= e^{-iav}\Phi(v), \\
    \gamma(-v) &= \phi(-v-a), \\
    \Gamma(-v) &= e^{iav}\Phi(-v), \\
    w &\equiv v-a, \\
    u &\equiv v.
  \end{split}
\]
Observe the technique: $w$ and~$u$ are defined such that the
$\gamma(v)$ and $\Gamma(v)$ lines just above mention
$\phi(v-a)=\phi(w)$ and $\Phi(v)=\Phi(u)$ on their respective right
sides, whereas the $\gamma(-v)$ and $\Gamma(-v)$ lines do not purposely
mention $\phi(-w)$ or $\Phi(-u)$ (although they might mention the one
or the other by accident); thus, the latter two lines are written
merely by substituting $-v\la v$ in the former two.
%[One might be tempted to write $\gamma(-v) = \phi(-v+a)$ just above
%because $-w=-v+a$, but this would be a mistake; for though indeed
%$\gamma(v) = \phi(w)$, nothing has said that $\gamma(-v) = \phi(-w)$.
%No, one must evaluate $\gamma(-v)$ consistently with the
%definition that $\gamma(v) \equiv \phi(v-a)$, substituting $-v \la v$.]
Having written the four lines (for $\gamma[v]$, $\Gamma[v]$,
$\gamma[-v]$ and $\Gamma[-v]$),
we can change symbols downward a row with respect to each
of~(\ref{fouri:110:27f}) and~(\ref{fouri:110:27g}), revealing the pair
\[
  G(v) \fouripair g(-v),
\]
where
\[
  \begin{split}
    G(v) &= F(v-a), \\
    g(-v) &= e^{-iav}f(-v), \\
    G(-v) &= F(-v-a), \\
    g(v) &= e^{iav}f(v),
  \end{split}
\]
among which, again, the latter two lines are written by substituting
$-v\la v$.
Since $g(v) \fouripair G(v)$, the four lines mean that
\[
  e^{iav}f(v) \fouripair F(v-a),
\]
which is~(\ref{fouri:110:26}).  Apparently, the frequency-shifting
property is the metadual of the delay property.
\index{exercises}
(Exercise: pursuing this paragraph's pattern, show that the delay
property is the metadual of the frequency-shifting property.)

\index{letter}
Besides~(\ref{fouri:110:27f}) and~(\ref{fouri:110:27g}), the reverse
skew is just as possible:
\[%{fouri:110:28f}
  \br{ccccccccc}
    \cdots\ \fouripair &
    \phi(w) & \fouripair &
    \Phi(u) & \fouripair &
    \phi(-w) & \fouripair &
    \Phi(-u) & \fouripair \ \cdots
    \\
    \cdots\ \fouripair &
    F(-w) & \fouripair &
    f(u) & \fouripair &
    F(w) & \fouripair &
    f(-u) & \fouripair \ \cdots
  \er
\]
and
\[%{fouri:110:28g}
  \br{ccccccccc}
    \cdots\ \fouripair &
    \gamma(v) & \fouripair &
    \Gamma(v) & \fouripair &
    \gamma(-v) & \fouripair &
    \Gamma(-v) & \fouripair \ \cdots
    \\
    \cdots\ \fouripair &
    G(-v) & \fouripair &
    g(v) & \fouripair &
    G(v) & \fouripair &
    g(-v) & \fouripair \ \cdots
  \er
\]
though this writer seems (for whatever reason) to have
found the forward
skew of~(\ref{fouri:110:27f}) and~(\ref{fouri:110:27g}) to be the
more convenient.  Either way, a lot of letters are used---$\phi\Phi f
F(wu)$ and $\gamma \Gamma g G(v)$ [and you can use yet more
letters like $\chi \mathrm X h H(w_\chi u_{\mathrm X})$ if you have an extra
function to transform as, for instance, while deriving
eqn.~\ref{fouri:110:43}]---but the letters serve to keep the various
relations straight and, anyway, you don't need so many letters to
compute the dual~(\ref{fouri:110:10}) of an ordinary transform pair but
only to compute the metadual of a transform property.

\subsection{Summary of properties}
\label{fouri:110.30}
\index{Fourier transform!shifting of}
\index{Fourier transform!of a shifted function}
\index{Fourier transform!differentiation of}
\index{Fourier transform!of a derivative}
\index{Fourier transform!linearity of}
\index{derivative!of a Fourier transform}
\index{derivative!Fourier transform of}

Table~\ref{fouri:110:tbl20} summarizes properties of the Fourier
transform.
\begin{table}
  \caption{Properties of the Fourier transform.}
  \label{fouri:110:tbl20}
  \settowidth\tla{$\ds\int_{-\infty}^{\infty} H^{*}(v) F(v) \,dv$}
  \bqb
    F(v) &=& F^{*}(-v)
    \ \ \mbox{if $\Im(v) = 0$ and,} \\&&\quad
      \mbox{for all such~$v$, $\Im[f(v)] = 0$.}\\
    f(v-a) &\fouripair& e^{-iav}F(v) \\
    e^{iav}f(v) &\fouripair& F(v-a) \\
    Af(\alpha v) &\fouripair&
      \frac{A}{\left|\alpha\right|}F\left(\frac{v}{\alpha}\right) \\
    \left(\!\sqrt{|\alpha|}\right)Af(\alpha v) &\fouripair&
      \frac{A}{\sqrt{\left|\alpha\right|}}F\left(\frac{v}{\alpha}\right) \\
    Af[(\alpha)(v-a)] &\fouripair&
      \frac{Ae^{-iav}}{\left|\alpha\right|}F\left(\frac{v}{\alpha}\right) \\
    &&\quad \Im(\alpha)=0,\ \Re(\alpha)\neq 0\\
    A_1f_1(v) + A_2f_2(v) &\fouripair& A_1F_1(v) + A_2F_2(v) \\
    \frac{d}{dv}f(v) &\fouripair& ivF(v) \\
    -ivf(v) &\fouripair& \frac{d}{dv}F(v) \\
    \frac{d^n}{dv^n}f(v) &\fouripair& (iv)^nF(v) \\
    (-iv)^nf(v) &\fouripair& \frac{d^n}{dv^n}F(v) \\
    &&\quad n\in\mathbb Z,\ n\ge 0\\
    \int_{-\infty}^v f(\tau) \,d\tau &\fouripair&
      \frac{F(v)}{iv} + \pi F(0) \delta(v) \\
    \int_{-\infty}^{\infty} h^{*}(v) f(v) \,dv &=&
      \int_{-\infty}^{\infty} H^{*}(v) F(v) \,dv
      % The next qualification should be kept, but probably commented out.
      %\ \ \mbox{if $\Im(v)=0$}
      \\
    \int_{-\infty}^{\infty} \left|f(v)\right|^2 \,dv &=&
      \makebox[\tla][l]{$\ds\int_{-\infty}^{\infty} \left|F(v)\right|^2 \,dv$}
      %\ \ \mbox{if $\Im(v)=0$}
      \\
    a_j &=& \frac{\Delta\omega}{\sqrt{2\pi}} F(j\,\Delta\omega)
  \eqb
\end{table}

\index{scaling property of the Fourier transform}
\index{Fourier transform!scaling of}
\index{superpositional property of the Fourier transform}
\index{Fourier transform!superpositional property of}
The table's first three properties have been proved earlier in this
section.  The table's fourth property, that
\bq{fouri:110:33}
    Af(\alpha v) \fouripair \frac{A}{\left|\alpha\right|}F\left(\frac{v}{\alpha}\right)
      \ \ \mbox{if $\Im(\alpha)=0$, $\Re(\alpha)\neq 0$},
\eq
which is the \emph{scaling property} of the Fourier transform, is
proved by applying~(\ref{fouri:byu}) to $Af(\alpha v)$ and then changing
$\xi\la\alpha\theta$ (the magnitude sign~$|\cdot|$ coming
because~$\alpha$, if negative, reverses Fourier's infinite limits of
integration in eqn.~\ref{fouri:byu}; see \S~\ref{inttx:reversal} and
its eqn.~\ref{inttx:reversal:70}).
The table's fifth property is merely the fourth with an alternate scale.
The table's sixth property applies the fourth and then the
second, in that sequence.  The table's seventh property is proved
trivially.

\index{Fourier transform!of differentiation}
\index{differentiation!Fourier transform of}
The table's eighth through eleventh properties begin from the derivative of
the inverse Fourier transform; that is, of (\ref{fouri:byu})'s second
line.  This derivative is
\[
  \frac{d}{dv}f(v) =
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty}
        e^{iv\theta}
        [i\theta F(\theta)]
        \,d\theta
  = \fourf^{-1}\{ivF(v)\},
\]
which implies that
\[
  \fourf\left\{\frac{d}{dv}f(v) \right\} = ivF(v),
\]
the table's eighth property.  The metadual (\S~\ref{fouri:110.27}) of the
eighth property is the table's ninth property, during the
computation of which\footnote{%
  % diagn: closely check this new footnote.
  If working out this metadual with your own pencil according to the
  pattern of \S~\ref{fouri:110.27}, if you reach
  \[
    \begin{split}
      G(v) &= \frac{d}{dv}F(v), \\
      g(-v) &= ivf(-v),
    \end{split}
  \]
  then you are probably on the right track.  Note that, during the
  calculation of this particular metadual, it happens that $w=v$ and
  $u=v$, so this particular metadual is a little easier than some
  others to work out.

  For a harder metadual, see \S~\ref{fouri:110.40}.
}
one observes that,
\bq{fouri:110:35}
  \mbox{if}\ \gamma(v) = \frac{d}{dv}\phi(v),\ \mbox{then}\
  \gamma(-v) = -\frac{d}{dv}\phi(-v),
\eq
a fact whose truth can be demonstrated via eqn.~\ref{drvtv:defz}'s
definition of the derivative or, easier, can be seen by sketching on a
sheet of paper some arbitrary, asymmetric function (like, say, $\phi[v]
\equiv e^{v/3}$) and a visual approximation to its derivative.
The table's tenth and eleventh properties come by repeated application of
the eighth and ninth.

The table's twelfth property is~(\ref{fouri:125:10}).
Section~\ref{fouri:125} will derive it.

The table's final three properties are~(\ref{fouri:110:parseval}),
(\ref{fouri:110:parseval2}) and~(\ref{lapl:140:22}).
Sections~\ref{fouri:110.55} and~\ref{lapl:140.20} will derive them.

\subsection{Convolution and correlation}
\label{fouri:110.40}
\index{convolution}
\index{correlation}
\index{convolution!Fourier transform of}
\index{correlation!Fourier transform of}
\index{product!Fourier transform of}
\index{Fourier transform!of a convolution}
\index{Fourier transform!of a correlation}
\index{Fourier transform!of a product}

\index{transfer function}
\index{Dirac delta function!response to}
\index{delta function, Dirac!response to}
\index{convolution!balanced expression of}
\index{convolution!unbalanced expression of}
\index{balanced expression of convolution}
\index{unbalanced expression of convolution}
\index{engineer}
\index{mechanical engineer}
\index{electrical engineer}
In mechanical and electrical engineering, the concept of
\emph{convolution} emerges during the analysis of a linear system whose
response to a Dirac impulse $\delta(t)$ is
some characteristic \emph{transfer function} $h(t)$.
% diagn: the next two lines, until the semicolon, want one last review.
To explore the mechanical origin and engineering application of the
transfer function would exceed the book's remit;
but, inasmuch as
a system is linear and its response to a Dirac impulse $\delta(t)$ is
indeed $h(t)$, its response to an arbitrary input $f(t)$ cannot but be
\bq{fouri:110:40a}
  g_1(t) \equiv \int_{-\infty}^{\infty} h(t-\tau)f(\tau) \,d\tau;
\eq
or, changing $t/2+\tau \la \tau$ to improve the equation's symmetry,
\bq{fouri:110:40}
  g_1(t) \equiv \int_{-\infty}^{\infty} h\left(\frac{t}{2}-\tau\right)
    f\left(\frac{t}{2}+\tau\right) d\tau.
\eq
This integral defines%
\footnote{\cite[\S~2.2]{Hsu:sig}}
convolution of the two functions $f(t)$ and $h(t)$.

Changing $v\la t$ and $\psi\la\tau$ in~(\ref{fouri:110:40}) to comport
with the notation found elsewhere in this section and then
applying~(\ref{fouri:byu}) yields that%
\footnote{%
  See \S~\ref{fours:100.10}.%
}
\bqb
  \lefteqn{
    \fourf\left\{
      \int_{-\infty}^{\infty}
        h\left(\frac{v}{2}-\psi\right)
        f\left(\frac{v}{2}+\psi\right)
      d\psi
    \right\}
  }&&\\&=&
    \frac{1}{\sqrt{2\pi}}
    \int_{-\infty}^{\infty}
      e^{-iv\theta}
    \int_{-\infty}^{\infty}
        h\left(\frac{\theta}{2}-\psi\right)
        f\left(\frac{\theta}{2}+\psi\right)
      d\psi
      \,d\theta
  \\&=&
    \frac{1}{\sqrt{2\pi}}
    \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}
      e^{-iv\theta}
        h\left(\frac{\theta}{2}-\psi\right)
        f\left(\frac{\theta}{2}+\psi\right)
      d\theta
      \,d\psi.
\eqb
Now changing $\phi \la \theta/2 + \psi$ within the inner integral,
\bqb
  \lefteqn{
    \fourf\left\{
      \int_{-\infty}^{\infty}
        h\left(\frac{v}{2}-\psi\right)
        f\left(\frac{v}{2}+\psi\right)
      d\psi
    \right\}
  }&&\\&=&
    \frac{2}{\sqrt{2\pi}}
    \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}
      e^{-iv(2\phi-2\psi)}
        h(\phi-2\psi)
        f(\phi)
      \,d\phi
      \,d\psi
  \\&=&
    \frac{2}{\sqrt{2\pi}}
    \int_{-\infty}^{\infty}
      e^{-iv\phi} f(\phi)
      \int_{-\infty}^{\infty}
        e^{-iv(\phi-2\psi)} h(\phi-2\psi)
      \,d\psi
      \,d\phi.
\eqb
Again changing $\mu \la \phi - 2\psi$ within the inner integral,
\bqb
  \lefteqn{
    \fourf\left\{
      \int_{-\infty}^{\infty}
        h\left(\frac{v}{2}-\psi\right)
        f\left(\frac{v}{2}+\psi\right)
      d\psi
    \right\}
  }&&\\&=&
    \frac{1}{\sqrt{2\pi}}
    \int_{-\infty}^{\infty}
      e^{-iv\phi} f(\phi)
      \int_{-\infty}^{\infty}
        e^{-iv\mu} h(\mu)
      \,d\mu
      \,d\phi
  \\&=&
    \left[\!\sqrt{2\pi}\right]
    \left[
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty} e^{-iv\mu} h(\mu) \,d\mu
    \right]
    \left[
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty} e^{-iv\phi} f(\phi) \,d\phi
    \right]
  \\&=&
    \left(\!\sqrt{2\pi}\right) H(v) F(v).
\eqb
That is,
\bq{fouri:110:42}
  \int_{-\infty}^{\infty}
    h\left(\frac{v}{2}-\psi\right)
    f\left(\frac{v}{2}+\psi\right)
  d\psi
  \fouripair \left(\!\sqrt{2\pi}\right) H(v) F(v).
\eq

Symbolizing~(\ref{fouri:110:42}) in the manner of \S~\ref{fouri:110.27}
as
\[
  \gamma(v) \fouripair \Gamma(v),
\]
where
\[
  \begin{split}
    \gamma(v) &\equiv \int_{-\infty}^{\infty}
    \chi\left(\frac{v}{2}-\psi\right)
    \phi\left(\frac{v}{2}+\psi\right)d\psi,\\
    \Gamma(v) &= \left(\!\sqrt{2\pi}\right) \mathrm X(v) \Phi(v),\\
    \gamma(-v) &= \int_{-\infty}^{\infty}
    \chi\left(-\frac{v}{2}-\psi\right)
    \phi\left(-\frac{v}{2}+\psi\right)d\psi,\\
    \Gamma(-v) &= \left(\!\sqrt{2\pi}\right) \mathrm X(-v) \Phi(-v),\\
     w_\chi &\equiv \frac{v}{2} - \psi,\\
     w_\phi &\equiv \frac{v}{2} + \psi,\\
     u &\equiv v,
  \end{split}
\]
or, after changing symbols downward a row with respect to each
of~(\ref{fouri:110:27f}) and~(\ref{fouri:110:27g}),
\[
  G(v) \fouripair g(-v),
\]
where
\[
  \begin{split}
    G(v) &= \int_{-\infty}^{\infty}
    H\left(\frac{v}{2}-\psi\right)
    F\left(\frac{v}{2}+\psi\right)d\psi,\\
    g(-v) &= \left(\!\sqrt{2\pi}\right) h(-v) f(-v),\\
    G(-v) &= \int_{-\infty}^{\infty}
    H\left(-\frac{v}{2}-\psi\right)
    F\left(-\frac{v}{2}+\psi\right)d\psi,\\
    g(v) &= \left(\!\sqrt{2\pi}\right) h(v) f(v),
  \end{split}
\]
one finds the metadual $g(v)\fouripair G(v)$ of~(\ref{fouri:110:42})
to be
\bq{fouri:110:43}
  h(v) f(v)
  \fouripair
  \frac{1}{\sqrt{2\pi}}
  \int_{-\infty}^{\infty}
    H\left(\frac{v}{2}-\psi\right)
    F\left(\frac{v}{2}+\psi\right)
  d\psi.
\eq

Whether by~(\ref{fouri:110:42}) or by~(\ref{fouri:110:43}), convolution
in the one domain evidently transforms to multiplication in the other.

\newboolean{showFouriCorrelation}
\setboolean{showFouriCorrelation}{false}
\index{argument!real}
\index{real argument}
\index{Fourier transform!of a function whose argument is complex}
Closely related to the convolutional integral~(\ref{fouri:110:40}) is
the integral
\bq{fouri:110:45}
  g_2(t) \equiv \int_{-\infty}^{\infty} h\left(\tau-\frac{t}{2}\right)
    f\left(\tau+\frac{t}{2}\right) d\tau,
\eq
whose transform and dual transform are computed as above,
with one extra step using~(\ref{fouri:110:12}), to be
\bq{fouri:110:47}
  \begin{split}
    \int_{-\infty}^{\infty}
      h\left(\psi-\frac{v}{2}\right)
      f\left(\psi+\frac{v}{2}\right)
    d\psi
    &\fouripair \left(\!\sqrt{2\pi}\right) H(-v) F(v),
  \\
    h(-v) f(v)
    &\fouripair
    \frac{1}{\sqrt{2\pi}}
    \int_{-\infty}^{\infty}
      H\left(\psi-\frac{v}{2}\right)
      F\left(\psi+\frac{v}{2}\right)
    d\psi.
  \end{split}
\eq
\ifthenelse{\boolean{showFouriCorrelation}}{%
  Details:
  \[
    g(t) \equiv \int_{-\infty}^{\infty} h\left(\tau-\frac{t}{2}\right)
      f\left(\tau+\frac{t}{2}\right) d\tau.
  \]
  Changing $v\la t$ and $\psi\la\tau$ and applying~(\ref{fouri:byu})
  yields that
  \bqb
    \lefteqn{
      \fourf\left\{
        \int_{-\infty}^{\infty}
          h\left(\psi-\frac{v}{2}\right)
          f\left(\psi+\frac{v}{2}\right)
        d\psi
      \right\}
    }&&\\&=&
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty}
        e^{-iv\theta}
      \int_{-\infty}^{\infty}
          h\left(\psi-\frac{\theta}{2}\right)
          f\left(\psi+\frac{\theta}{2}\right)
        d\psi
        \,d\theta
    \\&=&
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}
        e^{-iv\theta}
          h\left(\psi-\frac{\theta}{2}\right)
          f\left(\psi+\frac{\theta}{2}\right)
        d\theta
        \,d\psi.
  \eqb
  Now changing $\phi \la \theta/2 + \psi$ within the inner integral,
  \bqb
    \lefteqn{
      \fourf\left\{
        \int_{-\infty}^{\infty}
          h\left(\psi-\frac{v}{2}\right)
          f\left(\psi+\frac{v}{2}\right)
        d\psi
      \right\}
    }&&\\&=&
      \frac{2}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}
        e^{-iv(2\phi-2\psi)}
          h(2\psi-\phi)
          f(\phi)
        \,d\phi
        \,d\psi
    \\&=&
      \frac{2}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty}
        e^{-iv\phi} f(\phi)
        \int_{-\infty}^{\infty}
          e^{-iv(\phi-2\psi)} h(2\psi-\phi)
        \,d\psi
        \,d\phi.
  \eqb
  Again changing $\mu \la \phi - 2\psi$ within the inner integral,
  \bqb
    \lefteqn{
      \fourf\left\{
        \int_{-\infty}^{\infty}
          h\left(\psi-\frac{v}{2}\right)
          f\left(\psi+\frac{v}{2}\right)
        d\psi
      \right\}
    }&&\\&=&
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty}
        e^{-iv\phi} f(\phi)
        \int_{-\infty}^{\infty}
          e^{-iv\mu} h(-\mu)
        \,d\mu
        \,d\phi
    \\&=&
      \left[\!\sqrt{2\pi}\right]
      \left[
        \frac{1}{\sqrt{2\pi}}
        \int_{-\infty}^{\infty} e^{-iv\mu} h(-\mu) \,d\mu
      \right]
      \left[
        \frac{1}{\sqrt{2\pi}}
        \int_{-\infty}^{\infty} e^{-iv\phi} f(\phi) \,d\phi
      \right]
    \\&=&
      \left[\!\sqrt{2\pi}\right]
      \left[
        -\frac{1}{\sqrt{2\pi}}
        \int_{-\mu=\infty}^{-\infty} e^{-i(-v)(-\mu)} h(-\mu) \,d(-\mu)
      \right]\\&&\qquad\qquad\times
      \left[
        \frac{1}{\sqrt{2\pi}}
        \int_{-\infty}^{\infty} e^{-iv\phi} f(\phi) \,d\phi
      \right]
    \\&=&
      \left[\!\sqrt{2\pi}\right]
      \left[
        \frac{1}{\sqrt{2\pi}}
        \int_{-\mu=-\infty}^{\infty} e^{-i(-v)(-\mu)} h(-\mu) \,d(-\mu)
      \right]\\&&\qquad\qquad\times
      \left[
        \frac{1}{\sqrt{2\pi}}
        \int_{-\infty}^{\infty} e^{-iv\phi} f(\phi) \,d\phi
      \right]
    \\&=&
      \left(\!\sqrt{2\pi}\right) H(-v) F(v).
  \eqb
  That is,
  \[
    \int_{-\infty}^{\infty}
      h\left(\psi-\frac{v}{2}\right)
      f\left(\psi+\frac{v}{2}\right)
    d\psi
    \fouripair \left(\!\sqrt{2\pi}\right) H(-v) F(v).
  \]
  Symbolizing in the manner of \S~\ref{fouri:110.27} as
  \[
    \gamma(v) \fouripair \Gamma(v),
  \]
  where
  \[
    \begin{split}
      \gamma(v) &\equiv \int_{-\infty}^{\infty}
      \chi\left(\psi-\frac{v}{2}\right)
      \phi\left(\psi+\frac{v}{2}\right)d\psi,\\
      \Gamma(v) &= \left(\!\sqrt{2\pi}\right) \mathrm X(-v) \Phi(v),\\
      \gamma(-v) &= \int_{-\infty}^{\infty}
      \chi\left(\psi+\frac{v}{2}\right)
      \phi\left(\psi-\frac{v}{2}\right)d\psi,\\
      \Gamma(-v) &= \left(\!\sqrt{2\pi}\right) \mathrm X(v) \Phi(-v),\\
       w_\chi &\equiv \psi - \frac{v}{2},\\
       w_\phi &\equiv \psi + \frac{v}{2},\\
       u &\equiv v,
    \end{split}
  \]
  or, after changing symbols downward a row with respect to each
  of~(\ref{fouri:110:27f}) and~(\ref{fouri:110:27g}),
  \[
    G(v) \fouripair g(-v),
  \]
  where
  \[
    \begin{split}
      G(v) &= \int_{-\infty}^{\infty}
      H\left(\psi-\frac{v}{2}\right)
      F\left(\psi+\frac{v}{2}\right)d\psi,\\
      g(-v) &= \left(\!\sqrt{2\pi}\right) h(v) f(-v),\\
      G(-v) &= \int_{-\infty}^{\infty}
      H\left(\psi+\frac{v}{2}\right)
      F\left(\psi-\frac{v}{2}\right)d\psi,\\
      g(v) &= \left(\!\sqrt{2\pi}\right) h(-v) f(v),
    \end{split}
  \]
  one finds the metadual $g(v)\fouripair G(v)$ to be
  \[
    h(-v) f(v)
    \fouripair
    \frac{1}{\sqrt{2\pi}}
    \int_{-\infty}^{\infty}
      H\left(\psi-\frac{v}{2}\right)
      F\left(\psi+\frac{v}{2}\right)
    d\psi.
  \]
  (End of details.)
}{}
Furthermore, according to~(\ref{fouri:110:18a}), $h^{*}(t) \fouripair
H^{*}(-\omega^{*})$, so
\bq{fouri:110:48}
  \begin{split}
    \int_{-\infty}^{\infty}
      h^{*}\left(\psi-\frac{v}{2}\right)
      f\left(\psi+\frac{v}{2}\right)
    d\psi
    &\fouripair \left(\!\sqrt{2\pi}\right) H^{*}(v^{*}) F(v),
  \\
    h^{*}(-v) f(v)
    &\fouripair
    \frac{1}{\sqrt{2\pi}}
    \int_{-\infty}^{\infty}
      H^{*}\bigg(\frac{v^{*}}{2}-\psi\bigg)
      F\bigg(\frac{v}{2}+\psi\bigg)
    d\psi;
  \end{split}
\eq
and indeed one can do the same to the transforms~(\ref{fouri:110:42})
and~(\ref{fouri:110:43}) of the convolutional integral, obtaining
\bq{fouri:110:49}
  \begin{split}
    \int_{-\infty}^{\infty}
      h^{*}\left(\frac{v}{2}-\psi\right)
      f\left(\frac{v}{2}+\psi\right)
    d\psi
    &\fouripair \left(\!\sqrt{2\pi}\right) H^{*}(-v^{*}) F(v),
  \\
    h^{*}(v) f(v)
    &\fouripair
    \frac{1}{\sqrt{2\pi}}
    \int_{-\infty}^{\infty}
      H^{*}\bigg(\psi-\frac{v^{*}}{2}\bigg)
      F\bigg(\psi+\frac{v}{2}\bigg)
    d\psi.
  \end{split}
\eq
(The~$v^{*}$ of eqns.~\ref{fouri:110:48} and~\ref{fouri:110:49} seems
to imply that the argument~$v$ might be complex.  Though this writer has
encountered applications with complex~$h$ and~$f$, and though
complex~$H$ and~$F$ are the norm, the writer has never yet met
an application with complex~$v$.  How to interpret the case of
complex~$v$, or whether such a case is even valid in Fourier work, are
questions left open to the reader's consideration.  It is perhaps
interesting that H.~F.\ Davis, author of a book on Fourier mathematics,
does not in his book seem to consider the transform of a function whose
argument is complex at all.\footnote{\cite[\S~6.7]{HFDavis}}
Still, it appears that one can consider complex $v\neq v^{*}$ at
least in a formal sense, as in \S~\ref{fouri:110.15}; yet in
applications at any rate, normally and maybe always, $v=v^{*}$ will
be real.)

Unlike the operation the integral~(\ref{fouri:110:40}) expresses,
known as convolution, the operation the integral~(\ref{fouri:110:45})
expresses has no special name as far as the writer is aware.  However,
the operation its variant
\bq{fouri:110:45a}
  g_3(t) \equiv \int_{-\infty}^{\infty} h^{*}\left(\tau-\frac{t}{2}\right)
    f\left(\tau+\frac{t}{2}\right) d\tau
\eq
%of~(\ref{fouri:110:48})
expresses does have a name.  It is called
\emph{correlation,}\footnote{%
  % diagn: briefly review this new footnote (is the comma necessary?).
  Also called \emph{cross-correlation,} as in~\cite{EWW-web}.%
}
being a
measure of the degree to which one function tracks another with an
offset in the independent variable.  Reviewing this subsection, we see
that in~(\ref{fouri:110:48}) we have already determined the transform
of the correlational integral~(\ref{fouri:110:45a}).  Moreover, assuming
that $\Im(t) = 0$, we see in (\ref{fouri:110:49})'s second line that
we have already determined the dual of this transform, as well.
Convolution and correlation arise often enough in applications to enjoy
their own, peculiar notations%
\footnote{\cite[\S~19.4]{JJH}}
\bq{fouri:110:50}
  h(t) \ast f(t) \equiv
    \int_{-\infty}^{\infty} h\left(\frac{t}{2}-\tau\right)
    f\left(\frac{t}{2}+\tau\right) d\tau
\eq
for convolution and
\bq{fouri:110:53}
  R_{fh}(t) \equiv \int_{-\infty}^{\infty} h^{*}\left(\tau-\frac{t}{2}\right)
    f\left(\tau+\frac{t}{2}\right) d\tau
\eq
for correlation
% diagn: review the rest of the sentence, which is new.
(in the latter of which one can read the symbol~$R_{fh}$ as ``the
correlation of~$f$ against~$h$'').

\index{autocorrelation}
\index{commutivity!of convolution}
\index{associativity!of convolution}
\index{convolution!commutivity of}
\index{convolution!associativity of}
Nothing prevents one from correlating a function against
itself, incidentally.  The \emph{autocorrelation}
\bq{fouri:110:54}
  R_{f\!f}(t) = \int_{-\infty}^{\infty} f^{*}\left(\tau-\frac{t}{2}\right)
    f\left(\tau+\frac{t}{2}\right) d\tau
\eq
proves useful at times.\footnote{\cite[\S~1.6A]{Hsu:comm}}
For convolution, commutative and associative properties that
\bq{fouri:110:51}
  \begin{split}
    f(t) \ast h(t) &= h(t) \ast f(t), \\
    f(t) \ast [g(t) \ast h(t)] &= [f(t) \ast g(t)] \ast h(t),
  \end{split}
\eq
may be demonstrated, the former by changing $-\tau \la
\tau$ in~(\ref{fouri:110:50}) and the latter
by Fourier transformation
as $f(v) \ast [g(v) \ast h(v)]
\fouripair
(\!\sqrt{2\pi})F(v)
\linebreak\times % bad break: this whole line is deletable if the break is resolved
[(\!\sqrt{2\pi})G(v)H(v)] =
(\!\sqrt{2\pi})[(\!\sqrt{2\pi})F(v)G(v)]H(v)
\invfouripair [f(v) \ast g(v)] \ast h(v)$.

\index{energy spectral density}
\index{spectral density}
\index{density!spectral}
\index{noise}
\index{signaling}
\index{electronic signaling}
\index{electrical engineer}
In most cases of practical interest in applications,~$v$ is probably
real even when $H(v)$
and $F(v)$ are not, so one can use~(\ref{fouri:110:54}) to write a
transform pair like the first line of~(\ref{fouri:110:48}) in the style of
\bq{fouri:110:57}
  R_{f\!f}(t) \fouripair
  \left(\!\sqrt{2\pi}\right) \left|F(\omega)\right|^{2},\ \ \Im(t) = 0.
\eq
Electrical engineers call the quantity $\left|F(\omega)\right|^2$
on (\ref{fouri:110:57})'s right the \emph{energy spectral density} of
$f(t)$.\footnote{\cite[\S~1.6B]{Hsu:comm}}
Equation~(\ref{fouri:110:57}) is significant among other reasons because,
in electronic signaling---especially where the inevitable imposition of
random environmental \emph{noise} has degraded the signal---it may
happen that an adequate estimate of $R_{f\!f}(t)$ is immediately available
while sufficient information regarding $F(\omega)$ is unavailable.  When
this is the case,~(\ref{fouri:110:57}) affords an elegant, indirect way
to calculate an energy spectral density even if more direct methods
cannot be invoked.

Tables~\ref{fouri:110:tbl40} and~\ref{fouri:110:tbl41} summarize.
\begin{table}
  \caption%
  [Convolution and correlation, and their Fourier properties.]%
  {%
    Convolution and correlation, and their Fourier properties.
    (Note that, though the table provides for complex~$v$, $v$ is
    typically real.)%
  }
  \label{fouri:110:tbl40}
  \bqb
      \int_{-\infty}^{\infty}
        h\left(\frac{v}{2}-\psi\right)
        f\left(\frac{v}{2}+\psi\right)
      d\psi
    &\fouripair&
      \left(\!\sqrt{2\pi}\right) H(v) F(v)
    \\
      h(v) f(v)
    &\fouripair&
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty}
        H\left(\frac{v}{2}-\psi\right)
    \\&&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mbox{}\times
        F\left(\frac{v}{2}+\psi\right)
      d\psi
    \\
      \int_{-\infty}^{\infty}
        h\left(\psi-\frac{v}{2}\right)
        f\left(\psi+\frac{v}{2}\right)
      d\psi
    &\fouripair&
      \left(\!\sqrt{2\pi}\right) H(-v) F(v)
    \\
      h(-v) f(v)
    &\fouripair&
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty}
        H\left(\psi-\frac{v}{2}\right)
    \\&&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mbox{}\times
        F\left(\psi+\frac{v}{2}\right)
      d\psi
    \\
      \int_{-\infty}^{\infty}
        h^{*}\left(\frac{v}{2}-\psi\right)
        f\left(\frac{v}{2}+\psi\right)
      d\psi
    &\fouripair&
      \left(\!\sqrt{2\pi}\right) H^{*}(-v^{*}) F(v)
    \\
      h^{*}(v) f(v)
    &\fouripair&
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty}
        H^{*}\left(\psi-\frac{v^{*}}{2}\right)
    \\&&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mbox{}\times
        F\left(\psi+\frac{v}{2}\right)
      d\psi
    \\
      \int_{-\infty}^{\infty}
        h^{*}\left(\psi-\frac{v}{2}\right)
        f\left(\psi+\frac{v}{2}\right)
      d\psi
    &\fouripair&
      \left(\!\sqrt{2\pi}\right) H^{*}(v^{*}) F(v)
    \\
      h^{*}(-v) f(v)
    &\fouripair&
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty}
        H^{*}\left(\frac{v^{*}}{2}-\psi\right)
      \\&&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mbox{}\times
        F\left(\frac{v}{2}+\psi\right)
      d\psi
  \eqb
\end{table}
\begin{table}
  \caption[Convolution and correlation in their peculiar notation.]{%
    Convolution and correlation in their peculiar notation.
    (Note that the~$\ast$ which appears in the table as $h[t] \ast f[t]$
    differs in meaning from the~$\mbox{}^{*}$ in $h^{*}[v]$.)%
  }
  \label{fouri:110:tbl41}
  \bqb
    \Im(t) &=& 0\\
      f(t) \ast h(t) = h(t) \ast f(t)
    &\equiv&
      \int_{-\infty}^{\infty} h\left(\frac{t}{2}-\tau\right)
      f\left(\frac{t}{2}+\tau\right) d\tau
    \\
      R_{fh}(t)
    &\equiv&
      \int_{-\infty}^{\infty} h^{*}\left(\tau-\frac{t}{2}\right)
      f\left(\tau+\frac{t}{2}\right) d\tau
    \\
      h(t) \ast f(t)
    &\fouripair&
      \left(\!\sqrt{2\pi}\right) H(\omega) F(\omega)
    \\
      h(t) f(t)
    &\fouripair&
      \frac{1}{\sqrt{2\pi}}[H(\omega) \ast F(\omega)]
    \\
      R_{fh}(t)
    &\fouripair&
      \left(\!\sqrt{2\pi}\right) H^{*}(\omega) F(\omega)
    \\
      h^{*}(t) f(t)
    &\fouripair&
      \frac{1}{\sqrt{2\pi}}R_{FH}(\omega)
    \\
      R_{f\!f}(t)
    &\fouripair&
      \left(\!\sqrt{2\pi}\right) \left|F(\omega)\right|^{2}
    \\
    f(t) \ast [g(t) \ast h(t)] &=& [f(t) \ast g(t)] \ast h(t)
  \eqb
\end{table}

See also \S~\ref{lapl:140}.

\subsection{Parseval's theorem}
\label{fouri:110.55}
\index{Parseval's theorem}
\index{Parseval, Marc-Antoine (1755--1836)}

Provided that
\[
  \Im(v) = 0,
\]
one finds that
\bqb
  \int_{-\infty}^{\infty} h^{*}(v) f(v) \,dv
  &=&
    \int_{-\infty}^{\infty} h^{*}(v) \left[
      \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{iv\theta}
      F(\theta) \,d\theta
    \right] dv
  \\&=&
    \int_{-\infty}^{\infty} \left[
      \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{iv\theta} h^{*}(v) \,dv
    \right] F(\theta) \,d\theta
  \\&=&
    \int_{-\infty}^{\infty} \left[
      \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-iv\theta} h(v) \,dv
    \right]^{*} F(\theta) \,d\theta
  \\&=&
    \int_{-\infty}^{\infty} H^{*}(\theta) F(\theta) \,d\theta,
\eqb
in which an interchange of integrations between two
applications of~(\ref{fouri:byu}) has been interposed.
Changing $v \la \theta$ on the right,
\bq{fouri:110:parseval}
  \int_{-\infty}^{\infty} h^{*}(v) f(v) \,dv =
    \int_{-\infty}^{\infty} H^{*}(v) F(v) \,dv.
\eq
This is \emph{Parseval's theorem.}%
\footnote{
  \cite[\S~2-2]{Couch}\cite[\S~1.6B]{Hsu:comm}
}
It is related to Parseval's principle of \S~\ref{fours:080}
and Parseval's equality of \S~\ref{fours:140.10}.

\index{time}
\index{energy}
\index{frequency}
\index{frequency!angular}
\index{angular frequency}
\index{frequency!spatial}
\index{frequency content}
\index{space}
\index{spatial frequency}
\index{dimension}
\index{quadrature}
\index{channel}
\index{$I$ channel}
\index{$Q$ channel}
Especially interesting to Parseval's theorem is the case of
$h(v) = f(v)$, in which
\bq{fouri:110:parseval2}
  \int_{-\infty}^{\infty} \left|f(v)\right|^2 \,dv =
    \int_{-\infty}^{\infty} \left|F(v)\right|^2 \,dv.
\eq
When this is written as
\[
  \int_{-\infty}^{\infty} \left|f(t)\right|^2 \,dt =
    \int_{-\infty}^{\infty} \left|F(\omega)\right|^2 \,d\omega,
\]
and~$t$, $\left|f(t)\right|^2$, $\omega$ and $\left|F(\omega)\right|^2$
respectively have physical dimensions of time, energy per unit time,
angular frequency and energy per unit angular frequency, then the
theorem conveys the important physical insight that energy transferred
at various times can equally well be regarded as energy
transferred \emph{at various frequencies.}  This works for space and
spatial frequencies, too: see \S~\ref{fours:085}.  For real $f(v)$, one
can write~(\ref{fouri:110:parseval2}) as
\bq{fouri:110:parseval2a}
  \int_{-\infty}^{\infty} f^2(v) \,dv =
    \int_{-\infty}^{\infty} \Re^2[F(v)] \,dv
    + \int_{-\infty}^{\infty} \Im^2[F(v)] \,dv,\ \ \Im[f(v)] = 0,
\eq
which expresses the principle of \emph{quadrature,} conveying the
additional physical insight that a single frequency can carry energy
in not only one but each of two distinct, independent channels; namely,
a \emph{real-phased, in-phase} or~$I$ channel and an
\emph{imaginary-phased, quadrature-phase} or~$Q$ channel.%
\footnote{\cite[\S~5-1]{Couch}}
Practical digital electronic communications systems, wired or wireless,
often do precisely this---effectively transmitting two, independent
streams of information at once, without conflict, in the selfsame band.

\subsection{Oddness and evenness}
\label{fouri:110.65}
\index{function!odd or even}
\index{odd function!Fourier transform of}
\index{even function!Fourier transform of}
\index{Fourier transform!of an odd or even function}

Odd functions have odd transforms.  Even functions have even transforms.
Symbolically,
\bi
  \item \mbox{[\textsc{odd}]} if $f(-v) = -f(v)$ for all~$v$, then $F(-v) = -F(v)$;
  \item \mbox{[\textsc{even}]} if $f(-v) = f(v)$ for all~$v$, then $F(-v) = F(v)$.
\ei
The odd case is proved by expressing $F(-v)$ per~(\ref{fouri:byu}) as
\[
  F(-v) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-i(-v)\theta} f(\theta) \,d\theta
\]
and then changing $-\theta \la \theta$ to get that
\[
  F(-v)
    = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-i(-v)(-\theta)} f(-\theta) \,d\theta
    = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-iv\theta}
    f(-\theta) \,d\theta.
\]
That $f(v)$ should be odd means by definition that $f(-\theta) =
-f(\theta)$, so
\[
  F(-v)
    = -\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-iv\theta}
    f(\theta) \,d\theta = -F(v).
\]
The even case is proved similarly.  See \S~\ref{taylor:365}.

% ----------------------------------------------------------------------

\section[Fourier transforms of selected functions]%
  {The Fourier transforms of selected functions}
\label{fouri:120}
\index{Fourier transform!of selected functions}

\index{sine-argument function!Fourier transform of}
\index{Fourier transform!of the sine-argument function}
We have already computed the Fourier transforms of several functions
in \S\S~\ref{fouri:100.40} and~\ref{fouri:110.20}.  We have also
already computed duals of these but would still like to put the duals
into more pleasing forms.  For example, the dual
of
\[
  \Lambda(v) \fouripair \frac{\sinarg^2(v/2)}{\sqrt{2\pi}}
\]
according to~(\ref{fouri:110:15}) is
\[
  \frac{\sinarg^2(v/2)}{\sqrt{2\pi}} \fouripair \Lambda(v).
\]
The scaling property~(\ref{fouri:110:33}) of
Table~\ref{fouri:110:tbl20}, using $\alpha = 2$,
makes it
\[
  \sinarg^2 v \fouripair
  \frac{\sqrt{2\pi}}{2}\Lambda\left(\frac{v}2\right),
\]
which is probably more convenient than~(\ref{fouri:110:15}) to use when
one meets a $\sinarg^2(\cdot)$ and wishes to transform it.

Tables~\ref{fouri:120:tbl20} and~\ref{fouri:120:tbl25} list the last
transform pair and others similarly derived.\footnote{%
  In electronic signaling systems, including radio, the table's
  transform pair
  $\sinarg(v) \fouripair \frac{\sqrt{2\pi}}{2}\Pi\left(\frac v2\right)$
  implies significantly that, to spread energy evenly over an available
  ``baseband'' but to let no energy leak outside that band, one should
  transmit sine-argument-shaped pulses as in Fig.~\ref{fours:160:fig}.%
}
% diagn: the references and logic in next sentence are right but the
% sentence's prose wants one last review (it's probably good, but just
% check once more).
The tables list further transform pairs as well---some as gleaned from
earlier in the chapter; others as computed in the last paragraph's way,
as adapted by the properties of Table~\ref{fouri:110:tbl20} (especially
the properties of delay, shift and scaling), or as derived in the
subsections that follow.
\begin{table}
  \caption[Fourier transform pairs.]{%
    Fourier transform pairs.  (See also Table~\ref{fouri:120:tbl25}.)%
  }
  \label{fouri:120:tbl20}
  \index{Fourier transform pair}
  \index{transform pair}
  \setlength\tabcolsep{0.3333333333333333\arraycolsep}
  \renewcommand\arraystretch{2.25}
  \nc\psirv{\ensuremath{\ds%
    \frac{\cos(rv/2)\sinarg(v/2)}{\sqrt{2\pi}\left[1-(rv/\pi)^2\right]}}}
  \nc\fourpsirv{\ensuremath{\ds \fourf\{\Psi_r(v)\}_{rv=\pm\pi}}}
  \nc\expra{\ensuremath{\ds \frac{n!}{\left(\!\sqrt{2\pi}\right)(a+iv)^{n+1}},}}
  \settowidth\tla{$\ds\psirv$}
  \settowidth\tlb{$\ds\fourpsirv$}
  \settowidth\tlc{$\Psi_r$}
  \settowidth\tld{$\ds\fouripair$}
  \settowidth\tle{$\ds\expra$}
  \nc\wideequal{\ensuremath{\ds\makebox[\tld][c]{$\ds =$}}}
  \begin{center}
    \begin{tabular}{rclcrcl}
      $1$ &$\fouripair$& $\ds\left(\!\sqrt{2\pi}\right) \delta(v)$
      &\mbox{\ \ }&
      $\delta(v)$ &$\fouripair$& $\ds\frac{1}{\sqrt{2\pi}}$ \\
      $\Lambda(v)$ &$\fouripair$& $\ds\frac{\sinarg^2(v/2)}{\sqrt{2\pi}}$ &&
      $u(v)$ &$\fouripair$& $\ds\frac{1}{\left(\!\sqrt{2\pi}\right)iv}
        + \frac{\sqrt{2\pi}}{2} \delta(v)$ \\
      $\Lambda_r(v)$ &$\fouripair$& $\ds\frac{\sinarg(rv/2)\sinarg(v/2)}{\sqrt{2\pi}}$ &&
      $\Pi(v)$ &$\fouripair$& $\ds\frac{\sinarg(v/2)}{\sqrt{2\pi}}$ \\
      $\Psi(v)$ &$\fouripair$& $\ds\frac{\sinarg v}{\sqrt{2\pi}\left[1-(v/\pi)^2\right]}$ &&
      $\ds\fourf\{\Psi(v)\}_{v=\pm\pi}$ &$\wideequal$& $\ds\frac{1}{2\sqrt{2\pi}}$ \\
      $\Psi_r(v)$ &$\fouripair$& $\ds\psirv$ &&
      $\ds\fourf\{\Psi_r(v)\}_{rv=\pm\pi}$ &$\wideequal$&
        $\ds\frac{\left(\!\sqrt{2\pi}\right)\sinarg(v/2)}{8}$ \\
      &&&&
      &$\wideequal$& $\ds\frac{\left(\!\sqrt{2\pi}\right)\sinarg(2\pi/4r)}{8}$ \\
      $\sinarg^2(v)$ &$\fouripair$& $\ds\frac{\sqrt{2\pi}}{2}\Lambda\left(\frac v2\right)$ &&
      $\sinarg(v)$ &$\fouripair$& $\ds\frac{\sqrt{2\pi}}{2}\Pi\left(\frac v2\right)$
    \end{tabular}
  \end{center}
  \bqb
    u(v)e^{-av} &\fouripair& \makebox[\tle][l]{$\ds\frac{1}{\left(\!\sqrt{2\pi}\right)(a+iv)},$}
      \ \ \Re(a) > 0 \\
    u(v)e^{-av}v^n &\fouripair& \makebox[\tle][l]{$\expra$}\ \
      \Re(a) > 0,\ n \in \mathbb Z, \ n \ge 0 \\
    e^{iav} &\fouripair& \makebox[\tle][l]{$\ds\left(\!\sqrt{2\pi}\right) \delta(v-a),$}
      \ \ \Im(a) = 0\\
    \sin av &\fouripair& \frac{\sqrt{2\pi}}{i2}\left[ \delta(v-a) - \delta(v+a) \right],
      \ \,\Im(a) = 0\\
    \cos av &\fouripair& \frac{\sqrt{2\pi}}{2}\left[ \delta(v-a) + \delta(v+a) \right],
      \ \,\Im(a) = 0\\
    \sum_{j=-\infty}^{\infty} \delta(v-jT_1) &\fouripair&
      \frac{\sqrt{2\pi}}{T_1} \sum_{j=-\infty}^{\infty} \delta\left(v-j\frac{2\pi}{T_1}\right)
        = \frac{\Delta\omega}{\sqrt{2\pi}} \sum_{j=-\infty}^{\infty} \delta(v-j\,\Delta\omega)\\
    \makebox[0em][r]{$\sqrt{T_1}$}\sum_{j=-\infty}^{\infty} \delta(v-jT_1) &\fouripair&
        \sqrt{\Delta\omega} \sum_{j=-\infty}^{\infty}
        \delta(v-j\,\Delta\omega),\ \ \Delta\omega T_1 = 2\pi
  \eqb
\end{table}
\begin{table}
  \caption{Fourier autotransform pairs.}
  \label{fouri:120:tbl25}
  \index{Fourier autotransform pair}
  \index{autotransform pair}
  \bqb
    \sum_{j=-\infty}^{\infty} \delta\left(v-j\sqrt{2\pi}\right) &\fouripair&
    \sum_{j=-\infty}^{\infty} \delta\left(v-j\sqrt{2\pi}\right) \\
    \Omega(v) &\fouripair& \Omega(v)
  \eqb
\end{table}
% diagn: check this sentence.
The final pair of Table~\ref{fouri:120:tbl25} is derived in
\S~\ref{fouri:130}.

\subsection{Exponential decay and Heaviside's unit step}
\label{fouri:120.30}

\index{Heaviside unit step function!Fourier transform of}
\index{unit step function, Heaviside!Fourier transform of}
\index{Fourier transform!of the Heaviside unit step}
\index{exponential decay!Fourier transform of}
\index{Fourier transform!of an exponential decay}
\index{exponential, natural!Fourier transform of}
\index{natural exponential!Fourier transform of}
\index{Fourier transform!of the natural exponential}
Applying the Fourier transform's definition~(\ref{fouri:byu}) to
$u(v)e^{-av}$, where $u(v)$ is Heaviside's unit
step~(\ref{integ:670:10}), yields that
\[
  \fourf \left\{u(v)e^{-av}\right\} =
      \frac{1}{\sqrt{2\pi}}
      \int_{0}^{\infty}
        e^{-(a+iv)\theta}
        \,d\theta
  =
      \frac{1}{\sqrt{2\pi}}
      \left[
        \frac{e^{-(a+iv)\theta}}{-(a+iv)}
      \right]_{\theta=0}^{\infty},
\]
revealing the transform pair
\bq{fouri:120:21}
  u(v)e^{-av} \fouripair
    \frac{1}{\left(\!\sqrt{2\pi}\right)(a+iv)},
  \ \ \Re(a) > 0.
\eq
Interesting is the limit $a \ra 0^{+}$ in~(\ref{fouri:120:21}),
\[
  u(v) \fouripair
    \frac{1}{\left(\!\sqrt{2\pi}\right)(iv)} + C\delta(v),
\]
where the necessary term $C\delta(v)$, with scale~$C$ to be determined,
merely admits that we do not yet know how to
evaluate~(\ref{fouri:120:21}) when both~$a$ and~$v$ vanish at once.
What we do know from \S~\ref{fouri:110.65} is that odd functions have
odd transforms and that (as one can see in
Fig.~\ref{integ:670:fig-u}) one can convert $u(v)$ to an odd function by
the simple expedient of subtracting $1/2$ from it.  Since $1/2
\fouripair (\!\sqrt{2\pi}/2) \delta(v)$ according to~(\ref{fouri:110:22}),
we have then that
\[
  u(v) - \frac 12 \fouripair
    \frac{1}{\left(\!\sqrt{2\pi}\right)(iv)}
    + \left(C-\frac{\sqrt{2\pi}}{2}\right)\delta(v),
\]
which to make its right side odd demands that $C=\sqrt{2\pi}/2$.
The transform pair
\bq{fouri:120:22}
  u(v) \fouripair
  \frac{1}{\left(\!\sqrt{2\pi}\right)iv} + \frac{\sqrt{2\pi}}{2} \delta(v)
\eq
results.

Invoking~(\ref{fouri:byu}),
\[
    u(v)e^{-av}v^n
  \fouripair \frac{1}{\sqrt{2\pi}} \int_0^\infty
    e^{-(a+iv)\theta}\theta^n \,d\theta.
\]
Evaluating the antiderivative via Table~\ref{inttx:470:tbl} with
$\alpha \la -(a+iv)$,
\[
    u(v)e^{-av}v^n
  \fouripair -\frac{e^{-(a+iv)\theta}}{\sqrt{2\pi}} \sum_{k=0}^n \left.
    \frac{ (n!/k!)\theta^k }{ (a+iv)^{n-k+1} }
  \right|_{\theta=0}^\infty.
\]
Since all but the $k=0$ term vanish, the last equation implies the
transform pair\footnote{\cite[Table~5.2]{Phillips/Parr}}%
\bq{fouri:120:24}
  u(v)e^{-av}v^n \fouripair \frac{n!}{\sqrt{2\pi}(a+iv)^{n+1}},
  \ \ \Re(a) > 0, \ n \in \mathbb Z, \ n \ge 0.
\eq

\subsection{Sinusoids}
\label{fouri:120.31}

\index{Fourier transform!of a sinusoid}
\index{sine!Fourier transform of}
\index{cosine!Fourier transform of}
\index{trigonometric function!Fourier transform of}
The Fourier transforms of $\sin av$ and $\cos av$ are interesting and
important.  One can compute them from the pairs
\bq{fouri:120:26}
  \begin{split}
    e^{iav} &\fouripair \left(\!\sqrt{2\pi}\right) \delta(v-a), \\
    e^{-iav} &\fouripair \left(\!\sqrt{2\pi}\right) \delta(v+a),
  \end{split}
\eq
which result by applying to~(\ref{fouri:110:22})
Table~\ref{fouri:110:tbl20}'s property that $e^{iav}f(v) \fouripair
F(v-a)$.  Composing by Table~\ref{cexp:tbl-prop} the trigonometrics
from their complex parts, we have that
\bq{fouri:120:23}
  \begin{split}
    \sin av &\fouripair
      \frac{\sqrt{2\pi}}{i2}\left[
        \delta(v-a) - \delta(v+a)
      \right], \\
    \cos av &\fouripair
      \frac{\sqrt{2\pi}}{2}\left[
        \delta(v-a) + \delta(v+a)
      \right].
  \end{split}
\eq

\subsection{The Dirac delta pulse train}
\label{fouri:120.35}

\index{pulse train!Fourier transform of}
\index{Dirac delta pulse train!Fourier transform of}
\index{Fourier transform!of a Dirac delta pulse train}
\index{autotransform}
\index{Fourier autotransform}
Curiously, the Fourier transform of the Dirac delta pulse train of
Fig.~\ref{fours:100:fig4} turns out to be another Dirac delta pulse
train.  The reason is that the Dirac delta pulse train's Fourier series
according to~(\ref{fours:100:37}) and~(\ref{fours:100:10}) is
\[
  \sum_{j=-\infty}^{\infty} \delta(v-jT_1) = \sum_{j=-\infty}^{\infty}
  \frac{e^{ij(2\pi/T_1)v}}{T_1},
\]
the transform of which according to~(\ref{fouri:120:26}) is
\bq{fouri:120:25}
  \sum_{j=-\infty}^{\infty} \delta(v-jT_1) \fouripair
  \frac{\sqrt{2\pi}}{T_1} \sum_{j=-\infty}^{\infty}
    \delta\left(v-j\frac{2\pi}{T_1}\right).
\eq
Apparently, the farther the pulses of the original train, the nearer the
pulses of the transformed train, and vice versa; yet, even when
transformed, the train remains a train of Dirac deltas.  Letting $T_1 =
\sqrt{2\pi}$ in~(\ref{fouri:120:25}) yields the pair
\bq{fouri:120:25a}
  \sum_{j=-\infty}^{\infty} \delta\left(v-j\sqrt{2\pi}\right) \fouripair
  \sum_{j=-\infty}^{\infty} \delta\left(v-j\sqrt{2\pi}\right),
\eq
discovering a pulse train whose Fourier transform is itself.

This completes the derivations of the Fourier transform pairs of
Tables~\ref{fouri:120:tbl20} and~\ref{fouri:120:tbl25}---except one pair.
The one pair will be the subject of \S~\ref{fouri:130}, next.

% ----------------------------------------------------------------------

\section{The Gaussian pulse}
\label{fouri:130}
\index{Gauss, Carl Friedrich (1777--1855)}
\index{Gaussian pulse}
\index{pulse!Gaussian}

\index{autoderivative}
While studying the derivative in chapters~\ref{drvtv} and~\ref{cexp}, we
asked among other questions whether any function could be its own
derivative.  We found that a sinusoid could be its own derivative after
a fashion---differentiation shifted its curve leftward without altering
its shape or scale---but that the only nontrivial function to be
exactly its own derivative was the natural exponential $f(z)=Ae^z$.  We
later found the
same natural exponential to fill several significant mathematical
roles---largely, whether directly or indirectly, because it was indeed
its own derivative.

\index{autotransform}
\index{Fourier autotransform}
\index{pulse train!contrasted against a single pulse}
\index{pulse!contrasted against a pulse train}
As we study the Fourier transform, a similar question arises: can any
function be its own \emph{transform?}  We have already found in
\S~\ref{fouri:120.35} that the Dirac delta pulse train can be its own
transform; but this train unlike the natural exponential is
nonanalytic, perhaps not the sort of function one had in mind.  One
should like an analytic function, and preferably not a train but a
single pulse.

\index{bell curve}
\index{cleverness}
\index{$\Omega$ as the Gaussian pulse}
\index{square, completing the}
\index{completing the square}
\index{contour integration!complex}
\index{contour!complex}
In chapter~\ref{prob}, during the study of the mathematics of
probability, we shall encounter a most curious function,
the \emph{Gaussian pulse,} also known as the \emph{bell curve} among
other names.  We will defer discussion of the Gaussian pulse's
provenance to the coming chapters but, for now, we can just copy here
the pulse's definition
from~(\ref{prob:normdist}) as
\bq{fouri:130:10}
  \Omega(t) \equiv \frac{\exp\left(-t^2/2\right)}{\sqrt{2\pi}},
\eq
plotted on pages~\pageref{prob:normdist-fig} below
and~\pageref{fours:095:fig8} above, respectively in
Figs.~\ref{prob:normdist-fig} and~\ref{fours:095:fig8}.
The Fourier transform of the Gaussian pulse is even trickier to compute
than were the transforms of \S~\ref{fouri:120}, but known techniques to
compute it include the following.%
\footnote{
  An alternate technique is outlined in \cite[Prob.~5.43]{Hsu:sig}.
}
From the the Fourier transform's definition~(\ref{fouri:byu}),
\[
  \fourf\{ \Omega(v) \}
  = \frac{1}{2\pi} \int_{-\infty}^{\infty}
      \exp\left( -\frac{\theta^2}{2} - iv \theta \right)
    d\theta.
\]
Completing the square (\S~\ref{alggeo:240}),%
\footnote{\cite{Davis-Westin-conversation}}
\bqb
  \fourf\{ \Omega(v) \}
  &=& \frac{\exp\left(-v^2/2\right)}{2\pi}
    \int_{-\infty}^{\infty}
      \exp\left( -\frac{\theta^2}{2} - iv \theta + \frac{v^2}{2} \right)
    d\theta
  \\&=& \frac{\exp\left(-v^2/2\right)}{2\pi}
    \int_{-\infty}^{\infty}
      \exp\left[ -\frac{\left(
        \theta + iv
      \right)^2}{2} \right]
    d\theta.
\eqb
Changing $\xi \la \theta + iv$,
\[
  \fourf\{ \Omega(v) \}
  = \frac{\exp\left(-v^2/2\right)}{2\pi}
    \int_{-\infty+iv}^{\infty+iv}
      \exp\left( -\frac{\xi^2}{2} \right)
    d\xi.
\]
Had we not studied complex contour integration in \S~\ref{inttx:250} we
should find such an integral hard to integrate in closed form.  However,
since happily we have studied it, observing that the integrand
$\exp(\xi^2)$ is an entire function (\S~\ref{taylor:330})
of~$\xi$---that is, that it is everywhere analytic---we recognize that
one can trace the path of integration from $-\infty+iv$ to
$\infty+iv$ along any contour one likes.  Let us trace it along the
real Argand axis from~$-\infty$ to~$\infty$, leaving only the two, short
complex segments at the ends which (as is easy enough to see, and the
formal proof is left as an exercise to the interested reader%
\footnote{
  The short complex segments at the ends might integrate to something
  were the real part of~$\xi^2$ negative, but the real part happens to
  be positive---indeed, most extremely positive---over the domains
  of the segments in question.
}%
) lie so far away that---for this integrand---they integrate to nothing.
So tracing leaves us with
\bq{fouri:130:20}
  \fourf\{ \Omega(v) \}
  = \frac{\exp\left(-v^2/2\right)}{2\pi}
    \int_{-\infty}^{\infty}
      \exp\left( -\frac{\xi^2}{2} \right)
    d\xi.
\eq
How to proceed from here is not immediately obvious.  None of the
techniques of chapter~\ref{inttx} seems especially suitable to evaluate
\[
  I \equiv \int_{-\infty}^{\infty}
    \exp\left( -\frac{\xi^2}{2} \right)
  d\xi,
\]
though if a search for a suitable contour of integration failed one might
fall back on the Taylor-series technique of \S~\ref{inttx:450}.
Fortunately, mathematicians have been searching hundreds of years for
clever techniques to evaluate just such integrals and, when occasionally
they should discover such a technique and reveal it to us, why, we
record it in books like this, not to forget.

\index{integral!closed complex contour}
\index{contour integration!closed complex}
\index{integration!by closed contour}
\index{function!entire}
\index{entire function}
\index{integration technique}
\index{integration!by conversion to cylindrical or polar form}
\index{cylindrical coordinates!integration in}
Here is the technique.%
\footnote{
  \cite[\S~I:40-4]{Feynman}
  %The author has never heard who first discovered the technique---it
  %might have been Gauss himself---and regrettably has misplaced the
  %citation by which he first learned of it in print long ago (see
  %appendix~\ref{hist}).  However, he has since heard the technique
  %orally explained at least twice, including
  %by~\cite{Brown-conversation} if memory serves.  The technique thus
  %would not seem especially obscure.  In any event, the technique is now
  %recorded here.
}
The equations
\[
  \begin{split}
    I &= \int_{-\infty}^{\infty}
      \exp\left( -\frac{x^2}{2} \right)
    dx, \\
    I &= \int_{-\infty}^{\infty}
      \exp\left( -\frac{y^2}{2} \right)
    dy,
  \end{split}
\]
express the same integral~$I$ in two different ways, the only difference
being in the choice of letter for the dummy variable.  What if we
multiply the two?  Then
\bqb
    I^2 &=& \int_{-\infty}^{\infty}
      \exp\left( -\frac{x^2}{2} \right)
    dx \int_{-\infty}^{\infty}
      \exp\left( -\frac{y^2}{2} \right)
    dy
    \\&=&
    \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}
      \exp\left( -\frac{x^2+y^2}{2} \right)
    dx\,dy.
\eqb
One, geometrical way to interpret this~$I^2$ is as a double integration
over a plane in which $(x,y)$ are rectangular coordinates.  If we
interpret thus, nothing then prevents us from double-integrating by the
cylindrical coordinates $(\rho; \phi)$, instead, as
\bqb
  I^2
  &=&
    \int_{-\pi}^{\pi} \int_{0}^{\infty}
      \exp\left( -\frac{\rho^2}{2} \right)
    \rho\,d\rho\,d\phi
  \\&=&
    2\pi \left[ \int_{0}^{\infty}
      \exp\left( -\frac{\rho^2}{2} \right)
    \rho\,d\rho \right].
\eqb
At a casual glance, the last integral in square brackets seems to
differ little from the integral with which we started, but see: it is
not only that the lower limit of integration and the letter of the dummy
variable have changed, but that an extra factor of the dummy variable
has appeared---that the integrand ends not with~$d\rho$ but
with~$\rho\,d\rho$.  Once we have realized this, the integral's solution
by antiderivative (\S~\ref{inttx:210}) becomes suddenly easy to guess:
\[
    I^2
    =
      2\pi \left[
        -\exp\left( -\frac{\rho^2}{2} \right)
      \right]_{0}^{\infty}
    = 2\pi.
\]
So evidently,
\[
  I = \sqrt{2\pi},
\]
which means that
\bq{fouri:130:25}
  \int_{-\infty}^{\infty}
    \exp\left( -\frac{\xi^2}{2} \right)
  d\xi = \sqrt{2\pi}
\eq
as was to be calculated.

Finally substituting~(\ref{fouri:130:25}) into~(\ref{fouri:130:20}), we
have that
\[
  \fourf\{ \Omega(v) \}
  = \frac{\exp\left(-v^2/2\right)}{\sqrt{2\pi}},
\]
which in view of~(\ref{fouri:130:10}) reveals the remarkable transform pair
\bq{fouri:130:50}
  \Omega(v) \fouripair \Omega(v).
\eq
The Gaussian pulse transforms to itself.  Old Fourier, who can twist and
knot other curves with ease, seems powerless to bend Gauss' mighty
curve.

\index{Gaussian pulse!to implement the Dirac delta by}
\index{pulse!Gaussian, to implement the Dirac delta by}
\index{Dirac delta function!as implemented by the Gaussian\\pulse} % bad break
\index{delta function, Dirac!as implemented by the Gaussian\\pulse} % bad break
\index{localization}
\index{analytic function!localized}
\index{function!localized analytic}
It is worth observing incidentally in light of~(\ref{fouri:130:10})
and~(\ref{fouri:130:25}) that
\bq{fouri:130:60}
  \int_{-\infty}^{\infty} \Omega(t) \,dt = 1,
\eq
the same as for $\Pi(t)$, $\Lambda(t)$, $\Psi(t)$ and indeed $\delta(t)$.
Section~\ref{fours:095} and its~(\ref{fours:095:33}) have recommended
the shape of the Gaussian pulse, in the tall, narrow limit, to implement
the Dirac delta $\delta(t)$.  This section lends more force to the
recommendation, for not only is the Gaussian pulse analytic (unlike the
Dirac delta) but it also behaves uncommonly well under Fourier
transformation (like the Dirac delta), thus rendering the Dirac delta
susceptible to an analytic limiting process which transforms amenably.
Too, the Gaussian pulse is about as tightly localized as a nontrivial,
uncontrived analytic function can be.%
\footnote{
  Consider that $\Omega(t) \approx \mbox{0x0.6621}, \mbox{0x0.3DF2},
  \mbox{0x0.0DD2}, \mbox{0x0.0122}, \mbox{0x0.0009}, \mbox{0x0.0000}$
  at $t=0,\pm 1,\pm 2,\pm 3,\pm 4, \pm 5$; and that $\Omega(\pm 8) <
  2^{-\mbox{\tiny 0x2F}}$.  Away from its middle region $\left|t\right|
  \lesssim 1$, the Gaussian pulse evidently vanishes rather
  convincingly.
}
The passion of one of the author's mentors in extolling the Gaussian
pulse as ``absolutely a beautiful function'' seems well supported by the
practical mathematical virtues exhibited by the function itself.

\index{function!versatile}
The Gaussian pulse resembles the natural exponential in its general
versatility.  Indeed, though the book has required several chapters
through this chapter~\ref{fouri} to develop the fairly deep mathematics
underlying the Gaussian pulse and supporting its basic application, now
that we have the Gaussian pulse in hand we shall find that it ably fills
all sorts of roles---not least the principal role of chapter~\ref{prob} to
come.

% ----------------------------------------------------------------------

\section[The Fourier transform of integration]%
  {The Fourier transform of the integration operation}
\label{fouri:125}
\index{Fourier transform!of integration}
\index{integration!Fourier transform of}

Table~\ref{fouri:110:tbl20}, page~\pageref{fouri:110:tbl20}, includes
a heretofore unproved Fourier property,
\bq{fouri:125:10}
  \int_{-\infty}^v f(\tau) \,d\tau \fouripair
  \frac{F(v)}{iv}
  + \pi F(0) \delta(v).
\eq
This property has remained unproved because, when we compiled the table
in \S~\ref{fouri:110}, we lacked the needed theory.  We have
the theory now and can proceed with the
proof.\footnote{\cite[Prob.~5.33]{Hsu:sig}}

The proof begins with the observation that
\[
  \int_{-\infty}^v f(\tau) \,d\tau =
  u(v) \ast f(v) =
  \int_{-\infty}^\infty u\left(\frac v2 -\tau\right)
    f\left(\frac v2 + \tau\right) d\tau,
\]
where the $u(t) \ast f(t)$ exercises the convolution operation of
\S~\ref{fouri:110.40}.  The convolution's correctness is probably easier
to see if the convolution is expressed according
to~(\ref{fouri:110:40a}) rather than to~(\ref{fouri:110:40}), as
\[
  \int_{-\infty}^v f(\tau) \,d\tau =
  u(v) \ast f(v) =
  \int_{-\infty}^\infty u(v -\tau) f(\tau) \,d\tau;
\]
but the two forms are equivalent and, since we are free to work with
either and the earlier is the form that appears in
Table~\ref{fouri:110:tbl40}, we will prefer the earlier form.

\index{Dirac delta function!sifting property of}
\index{delta function, Dirac!sifting property of}
\index{sifting property}
Table~\ref{fouri:110:tbl40} records the transform pair
\bqb
  \lefteqn{
    \int_{-\infty}^{\infty}
      u\left(\frac{v}{2}-\tau\right)
      f\left(\frac{v}{2}+\tau\right)
      d\tau}
  &&\\
  &\fouripair&
      \left(\!\sqrt{2\pi}\right) \fourf\{u(v)\} F(v) =
        \left[\frac{1}{iv} + \pi \delta(v)\right] F(v)
\eqb
in which $\fourf\{u(v)\}$ is evaluated according to
Table~\ref{fouri:120:tbl20}.  Substituting the observation with which we
began,
\[
    \int_{-\infty}^v f(\tau) \,d\tau =
    u(v) \ast f(v)
  \fouripair
        \left[\frac{1}{iv} + \pi \delta(v)\right] F(v).
\]
Sifting, $\delta(v)F(v) = \delta(v)F(0)$, so the last pair
is in fact~(\ref{fouri:125:10}) which was to be proved.

\index{Heaviside unit step function!convolution by}
\index{convolution!by Heaviside's unit step}
A fact that has incidentally emerged during the proof,
\bq{fouri:125:20}
  \int_{-\infty}^v f(\tau) \,d\tau = u(v) \ast f(v),
\eq
is interesting enough to merit here an equation number of its own.

