\chapter{The Fourier and Laplace transforms}
\label{fouri}
\index{transform}
\index{Fourier transform}
\index{Laplace transform}
\index{Fourier, Jean Baptiste Joseph\\(1768--1830)} % bad break
\index{Laplace, Pierre-Simon (1749--1827)}

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

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}
  \bc
    \nc\xxxab{4.3}
    \nc\xxya{-0.7}
    \nc\xxyb{1.6}
    \nc\fxa{-5.0} \nc\fxb{5.0}
    \nc\fya{-1.1} \nc\fyb{2.5}
    \nc\xxaxes{%
      {%
        \psset{linewidth=0.5pt}%
        \psline(-\xxxab,0)(\xxxab,0)%
        \psline(0,\xxya)(0,\xxyb)%
        \uput[r](\xxxab,0){$t$}%
        \uput[u](0,\xxyb){$f(t)$}%
      }%
    }
    \begin{pspicture}(\fxa,\fya)(\fxb,\fyb)
      %\psframe[linewidth=0.5pt,dimen=outer](\fxa,\fya)(\fxb,\fyb)
      %\localscalebox{1.0}{1.0}
      \small
      {
         \xxaxes
         \psplot[linewidth=2.0pt,plotpoints=300]{-4.05}{4.05}{
           /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}

The reader may agree that Fourier's equation~(\ref{fouri:eqn}) is
curious, but in what way is it remarkable? the reader might ask.
To answer, let us observe
that the quantity within the square braces of~(\ref{fouri:eqn}),
\bq{fouri:xform}
  F(\omega) \equiv \mathcal{F}\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 $\mathcal{F}\{\cdot\}$ (where the script letter~$\mathcal{F}$,
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) = \mathcal{F}^{-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}
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.  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 to the Fourier definition and usage his colleagues prefer
  at need.
}

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}
  \bc
    \nc\xxxab{4.3}
    \nc\xxya{-1.40}
    \nc\xxyb{1.40}
    \nc\fxa{-5.0} \nc\fxb{5.0}
    \nc\fya{-1.7} \nc\fyb{1.7}
    \nc\xxaxes{%
      {%
        \psset{linewidth=0.5pt}%
        \psline(-\xxxab,0)(\xxxab,0)%
        \psline(0,\xxya)(0,\xxyb)%
        \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){$\Im[F(\omega)]$}
      }%
    }
    \begin{pspicture}(\fxa,\fya)(\fxb,\fyb)
      %\psframe[linewidth=0.5pt,dimen=outer](\fxa,\fya)(\fxb,\fyb)
      %\localscalebox{1.0}{1.0}
      \small
      {
         \xxaxes
         \psplot[linewidth=2.0pt,plotpoints=300]{-4.05}{4.05}{
           /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
           -0.20 1.00  2.20 pulse
            0.50 0.80 -1.80 pulse
           -2.00 0.60  1.10 pulse
           add add
         }
         \psplot[linewidth=2.0pt,plotpoints=300,linestyle=dashed]{-4.05}{4.05}{
           /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 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}
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) = \mathcal{F}\left\{f(v)\right\} &=
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty}
        e^{-iv\theta}
            f(\theta)
        \,d\theta, \\
  f(v) = \mathcal{F}^{-1}\left\{F(v)\right\} &=
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty}
        e^{iv\theta}
            F(\theta)
        \,d\theta, \\
  \mathcal{F} &\equiv \mathcal{F}_{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) = \mathcal{F}\left\{f(t)\right\} &=
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty}
        e^{-i\omega t}
            f(t)
        \,dt, \\
  f(t) = \mathcal{F}^{-1}\left\{F(\omega)\right\} &=
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty}
        e^{i\omega t}
            F(\omega)
        \,d\omega, \\
  \mathcal{F} &\equiv \mathcal{F}_{\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:
$\mathcal F_{vv}$; $\mathcal F_{\omega t}$.  However, the
unadorned~$\mathcal{F}$ 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{An example}
\label{fouri:100.40}
\index{Fourier transform!example of}

\index{triangular pulse!Fourier transform of}
\index{Fourier transform!of a triangular pulse}
As a Fourier example, consider the triangular pulse $\Lambda(t)$
of~(\ref{fours:095:10}).  Its Fourier transform according
to~(\ref{fouri:byu}) is
\bqb
  \mathcal{F}\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
  \mathcal{F}\left\{\Lambda(v)\right\} &=&
    \frac{1}{v^2\sqrt{2\pi}} \bigg\{
      \left[ e^{-iv\theta} [1+(iv)(1+\theta)] \right]_{-1}^{0}
  \\&&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mbox{}
      +
      \left[ e^{-iv\theta} [-1+(iv)(1-\theta)] \right]_{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}
One can compute other Fourier transform pairs in
like manner, such as that\footnote{%
  To verify~(\ref{fouri:100:44}) and~(\ref{fouri:100:45}) is left as
  an exercise.  Hint toward~(\ref{fouri:100:45}): $\sin(v\pm\pi) =
  -\sin v$.%
}%
\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}.

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

\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}

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.  This subsection
analytically develops the symmetries.

\index{dummy variable}
\index{conjugate!Fourier transform of}
\index{conjugate!Fourier transform of}
The Fourier transform of a function's conjugate according
to~(\ref{fouri:byu}) is
\[
  \mathcal{F}\{f^{*}(v)\} =
    \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-iv\theta}
    f^{*}(\theta)\, d\theta
  = \left[ \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{iv^{*}\theta} f(\theta)\, d\theta\right]^{*},
\]
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,
\[
    \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{iv^{*}\theta}
    f(\theta)\, d\theta =
    \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-i(-v^{*})\theta}
    f(\theta)\, d\theta,
\]
whereas according to~(\ref{fouri:byu}),
\bqb
    \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{iv^{*}\theta}
      f(\theta)\, d\theta &=& \mathcal{F}^{-1}\{f(v^{*})\},\\
    \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-i(-v^{*})\theta}
      f(\theta)\, d\theta &=& \mathcal{F}\{f(-v^{*})\};
\eqb
so the paragraph's first equation has that
\bq{fouri:110:18}
  \mathcal{F}\{f^{*}(v)\}
    = \mathcal{F}^{-*}\{f(v^{*})\} = \mathcal{F}^{*}\{f(-v^{*})\},
\eq
in which the symbology $\mathcal{F}^{-*}\{\cdot\} \equiv
[\mathcal{F}^{-1}\{\cdot\}]^{*}$ is used.
% Upon review, I am unsure what the rest of the sentence was for.  It
% is commented out for now.
%and
%$
%  \mathcal{F}^{-1}\{g(w)\}
%  =
%  \linebreak % b-a-d b-r-e-a-k
%  \mathcal{F}\{\mathcal{F}^{-1}\{\mathcal{F}^{-1}\{g(w)\}\}\}
%  = \mathcal{F}\{g(-w)\}
%$.
In the arrow notation, since
$F(\cdot) = \mathcal{F}\{f(\cdot)\}$,
and thus $F^{*}(\cdot) = \mathcal{F}^{*}\{f(\cdot)\}$,
even for $(\cdot) = (-v^{*})$, it implies 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 case.
}
\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.}

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
\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} will further improve eqn.~\ref{fouri:110:15},
incidentally.)

\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 dual 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}.%
}%
\[
  \begin{split}
    F(w) &\equiv \mathcal F \{f(u)\}, \\
    \Phi(u) &\equiv \mathcal F \{\phi(w)\}, \\
    \phi(w) &\equiv F(w),
  \end{split}
\]
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.

Let the delay property~(\ref{fouri:110:25}) be styled as
\[
  \gamma(v) \fouripair \Gamma(v),
\]
where
\[
  \begin{split}
    \gamma(v) &\equiv \phi(v-a), \\
    \Gamma(v) &= e^{-iav}\Phi(v), \\
    \gamma(-v) &= \phi(-v-a), \\
    \Gamma(-v) &= e^{iav}\Phi(-v), \\
    w &\equiv v-a, \\
    u &\equiv v.
  \end{split}
\]
[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 $-v \la v$.]
Changing symbols downward a line with respect to each of
(\ref{fouri:110:27f}) and~(\ref{fouri:110:27g}) yields
\[
  G(v) \fouripair g(-v),
\]
where
\[
  \begin{split}
    G(v) &= F(v-a), \\
    g(-v) &= e^{-iav}f(-v), \\
    G(-v) &= F(-v-a), \\
    g(v) &= e^{iav}f(v).
  \end{split}
\]
Since $g(v) \fouripair G(v)$, these mean that
\[
  e^{iav}f(v) \fouripair F(v-a),
\]
which is~(\ref{fouri:110:26}).  Apparently, the frequency-shifting
property is the dual of the delay property.

Besides~(\ref{fouri:110:27f}) and~(\ref{fouri:110:27g}), the reverse
skew is just as possible:
\bq{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
\eq
and
\bq{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
\eq
though the 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~$\chi \mathrm X h H(w_\chi u_\chi)$ 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 dual (or, if you prefer, \emph{metadual}) of a
transform \emph{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}
  \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)
      \ \ \mbox{if $\Im(\alpha)=0$, $\Re(\alpha)\neq 0$} \\
    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)
  \eqb
\end{table}

\index{scaling property of the Fourier transform}
\index{Fourier transform!scaling 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}).
The table's fifth property is proved trivially.

The table's sixth through ninth properties begin from the derivative of
the inverse Fourier transform; that is, of (\ref{fouri:byu})'s second
line.  This derivative is
\[
  \frac{d}{dv}f(v) =
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty}
        e^{iv\theta}
        [i\theta F(\theta)]
        \,d\theta
  = \mathcal{F}^{-1}\{ivF(v)\},
\]
which implies that
\[
  \mathcal{F}\left\{\frac{d}{dv}f(v) \right\} = ivF(v),
\]
the table's sixth property.  The metadual (\S~\ref{fouri:110.27}) of the
sixth property is the table's seventh property, during the
computation of which 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 eighth and ninth properties come by repeated application of the
sixth and seventh.

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

% diagn:  the last time the following subsection was checked, errors
% were found and corrected; therefore, the subsection wants checking
% again sometime
\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}
\index{delta function, Dirac}
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} $h(t)$.  To explore the
mechanical origin and engineering application of the transfer function
would exceed the book's; 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{
    \mathcal{F}\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{
    \mathcal{F}\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{
    \mathcal{F}\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
\[
  \begin{split}
    \gamma(v) &\equiv \int_{-\infty}^{\infty}
    \chi\left(\frac{v}{2}-\psi\right)
    \phi\left(\frac{v}{2}+\psi\right)\,d\psi,\\
    \Gamma(v) &= \left(\sqrt{2\pi}\right) X(v) \Phi(v),\\
    \gamma(-v) &= \int_{-\infty}^{\infty}
    \chi\left(-\frac{v}{2}-\psi\right)
    \phi\left(-\frac{v}{2}+\psi\right)\,d\psi,\\
    \Gamma(-v) &= \left(\sqrt{2\pi}\right) X(-v) \Phi(-v),\\
     w_\chi &\equiv \frac{v}{2} - \psi,\\
     w_\phi &\equiv \frac{v}{2} + \psi,\\
     u &\equiv v,
  \end{split}
\]
or, after changing symbols downward a line 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 in the last paragraph
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{
      \mathcal{F}\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{
      \mathcal{F}\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{
      \mathcal{F}\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
  \[
    \begin{split}
      \gamma(v) &\equiv \int_{-\infty}^{\infty}
      \chi\left(\psi-\frac{v}{2}\right)
      \phi\left(\psi+\frac{v}{2}\right)\,d\psi,\\
      \Gamma(v) &= \left(\sqrt{2\pi}\right) X(-v) \Phi(v),\\
      \gamma(-v) &= \int_{-\infty}^{\infty}
      \chi\left(\psi+\frac{v}{2}\right)
      \phi\left(\psi-\frac{v}{2}\right)\,d\psi,\\
      \Gamma(-v) &= \left(\sqrt{2\pi}\right) X(v) \Phi(-v),\\
       w_\chi &\equiv \psi - \frac{v}{2},\\
       w_\phi &\equiv \psi + \frac{v}{2},\\
       u &\equiv v,
    \end{split}
  \]
  or, after changing symbols downward a line 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
% diagn: new parenthetical note wants review
(The~$v^{*}$ of eqns.~\ref{fouri:110:48} and~\ref{fouri:110:49} seems
to imply that the argument~$v$ might be complex.  Though the 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.)

\index{convolution!commutivity of}
\index{convolution!associativity of}
\index{commutivity!of convolution}
\index{associativity!of convolution}
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,} 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.

\index{autocorrelation}
Nothing prevents one from correlating a function with
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}
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(t)\right|^{2},\ \ \Im(t) = 0.
\eq
Electrical engineers call the quantity $\left|F(t)\right|^2$
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(t) F(t)
    \\
      h(t) f(t)
    &\fouripair&
      \frac{1}{\sqrt{2\pi}}[H(t) \ast F(t)]
    \\
      R_{fh}(t)
    &\fouripair&
      \left(\sqrt{2\pi}\right) H^{*}(t) F(t)
    \\
      h^{*}(t) f(t)
    &\fouripair&
      \frac{1}{\sqrt{2\pi}}R_{FH}(t)
    \\
      R_{f\!f}(t)
    &\fouripair&
      \left(\sqrt{2\pi}\right) \left|F(t)\right|^{2}
    \\
    f(t) \ast [g(t) \ast h(t)] &=& [f(t) \ast g(t)] \ast h(t)
  \eqb
\end{table}

\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}.

\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 if $f(-v) = -f(v)$ for all~$v$, then $F(-v) = -F(v)$;
  \item 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
\[
  \Lambda(v) \fouripair \frac{\sinarg^2(v/2)}{\sqrt{2\pi}}
\]
is, according to~(\ref{fouri:110:15}),
\[
  \frac{\sinarg^2(v/2)}{\sqrt{2\pi}} \fouripair \Lambda(v).
\]
The scaling property of~(\ref{fouri:110:33}) and
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 next sentence, revised, wants review
The tables lists 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 this
section's subsections to 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}
  \bqb
    1 &\fouripair& \left(\sqrt{2\pi}\right) \delta(v) \\
    u(v) &\fouripair& \frac{1}{\left(\sqrt{2\pi}\right)iv} + \frac{\sqrt{2\pi}}{2} \delta(v) \\
    \delta(v) &\fouripair& \frac{1}{\sqrt{2\pi}} \\
    \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]} \\
    u(v)e^{-av} &\fouripair& \frac{1}{\left(\sqrt{2\pi}\right)(a+iv)},
      \ \ \Re(a) > 0 \\
    u(v)e^{-av}v^n &\fouripair& \frac{n!}{\left(\sqrt{2\pi}\right)(a+iv)^{n+1}}, \\
      &&\ \ \ \ \ \ \Re(a) > 0,\ n \in \mathbb Z, \ n \ge 0 \\
    e^{iav} &\fouripair& \left(\sqrt{2\pi}\right) \delta(v-a) \\
    \sin av &\fouripair& \frac{\sqrt{2\pi}}{j2}\left[ \delta(v-a) - \delta(v+a) \right] \\
    \cos av &\fouripair& \frac{\sqrt{2\pi}}{2}\left[ \delta(v-a) + \delta(v+a) \right] \\
    \sinarg(v) &\fouripair& \frac{\sqrt{2\pi}}{2}\Pi\left(\frac v2\right) \\
    \sinarg^2(v) &\fouripair& \frac{\sqrt{2\pi}}{2}\Lambda\left(\frac v2\right) \\
    \frac{\sinarg v}{\left[1-(v/\pi)^2\right]} &\fouripair& \left(\sqrt{2\pi}\right)\Psi(v) \\
    \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)
  \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}

\subsection{Exponential decay and the Heaviside 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
\[
  \mathcal F \left\{u(v)e^{-av}\right\} =
      \frac{1}{\sqrt{2\pi}}
      \int_{0}^{\infty}
        e^{-(a+iv)\theta}
        \,d\theta
  =
      \frac{1}{\sqrt{2\pi}}
      \left[
        \frac{e^{-(a+iv)\theta}}{-(a+iv)}
      \right]_{\theta=0}^{\infty},
\]
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
$\alpha \la -(a+iv)$,
\[
    u(v)e^{-av}v^n
  \fouripair -\frac{e^{-(a+iv)\theta}}{\sqrt{2\pi}} \sum_{k=0}^n \left.
    \frac{ (n!/k!)\theta^k }{ (a+iv)^{n-k+1} }
  \right|_0^\infty.
\]
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}}{j2}\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}
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}

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.

As we study the Fourier transform, a similar question arises: can any
function be its own \emph{transform?}  Well, 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}
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:fig1} above, respectively in
Figs.~\ref{prob:normdist-fig} and~\ref{fours:095:fig1}.
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}),
\[
  \mathcal F\{ \Omega(v) \}
  = \frac{1}{2\pi} \int_{-\infty}^{\infty}
      \exp\left( -\frac{\theta^2}{2} - iv \theta \right)
    \,d\theta.
\]
Completing the square (\S~\ref{alggeo:240}),%
\footnote{\cite{Davis-Westin-conversation}}
\bqb
  \mathcal F\{ \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 $\xi \la \theta + iv$,
\[
  \mathcal F\{ \Omega(v) \}
  = \frac{\exp\left(-v^2/2\right)}{2\pi}
    \int_{-\infty+iv}^{\infty+iv}
      \exp\left( -\frac{\xi^2}{2} \right)
    \,d\xi.
\]
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}
  \mathcal F\{ \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}
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_{-2\pi/2}^{2\pi/2} \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
\[
  \mathcal F\{ \Omega(v) \}
  = \frac{\exp\left(-v^2/2\right)}{\sqrt{2\pi}}.
\]
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
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 $\Pi(t)$, $\Lambda(t)$ 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.

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
one 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) \mathcal F\{u(v)\} F(v) =
        \left[\frac{1}{iv} + \pi \delta(v)\right] F(v)
\eqb
in which $\mathcal F\{u(v)\}$ 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.

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.

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

\section{The Laplace transform}
\label{fouri:200}
\index{Laplace transform}
\index{transform!Laplace}
\index{Laplace, Pierre-Simon (1749--1827)}

Fourier straightforwardly transforms pulses like those of
Figs.~\ref{fouri:100:fig} (page
% bad break
\pageref{fouri:100:fig})
and~\ref{fours:095:fig1} (page~\pageref{fours:095:fig1}) but stumbles
on time-unlimited functions like $f(t) =
\cos \omega_o t$ or even the ridiculously simple $f(t)=1$.  Only by the
clever, indirect techniques of \S\S~\ref{fouri:110} and~\ref{fouri:120}
has Fourier been able to transform such functions at all.  Fourier's
clever, indirect techniques are valid and even interesting but can still
prompt a mathematician to wonder whether a simpler alternative to Fourier
did not exist.

At the sometimes acceptable cost of omitting one of the Fourier
integral's two tails,%
\footnote{
  There has been invented a version of the Laplace transform which omits
  no tail~\cite[chapter~3]{Hsu:sig}.  This book does not treat it.
}
the \emph{Laplace transform}
\bq{fouri:laplace}
  F(s) = \mathcal L\{ f(t) \} \equiv
    \int_{0^{-}}^\infty e^{-st} f(t) \,dt
\eq
offers such an alternative.  Here, $s=j\omega$ is the transform
variable and, when~$s$ is purely imaginary, the Laplace transform is
very like the Fourier; but Laplace's advantage lies in that it
encourages the use of a complex~$s$, usually with a negative real part,
which in~(\ref{fouri:laplace})'s integrand tends to suppress even the
tail not omitted, thus effectively converting a time-unlimited function
to an integrable pulse---and Laplace does all this without resort to
indirect techniques.%
\footnote{\cite[chapter~3]{Hsu:sig}\cite[chapter~7]{Phillips/Parr}\cite[chapter~19]{JJH}}

\index{limit of integration}
\index{$0^{-}$ and~$0^{+}$ as limits of integration}
The lower limit~$0^{-}$ of $\int_{0^{-}}^\infty$ integration means that
the integration includes $t=0$.  In other words,
$\int_{0^{-}}^\infty \delta(t)dt = 1$ whereas
$\int_{0^{+}}^\infty \delta(t)dt = 0$.

\index{pattern!filling of}
\index{filling of a pattern}
\index{Laplace transform!of an integral}
\index{integral!Laplace transform of}
\index{Laplace transform!of a derivative}
\index{derivative!Laplace transform of}
Time-unlimited functions Laplace can straightforwardly transform
by
% bad break
(\ref{fouri:laplace})---without resort to clever, indirect
techniques---include
\[
  1 \laplair \frac{1}{s}.
\]
Others are listed in Table~\ref{fouri:laplace-properties}.
Laplace \emph{properties}, some of which are derived in the same easy
way, are listed in Table~\ref{fouri:laplace-pairs}\@.  Further
Laplace properties, also listed in the table, want some technique to
derive, for instance the differentiation property, which comes by
\[
  \begin{split}
    \mathcal L\left\{ \frac{d}{dt} f(t) \right\}
      &= \int_{0^{-}}^\infty e^{-st} \frac{d}{dt} f(t) \,dt
      = \int_{t=0^{-}}^\infty e^{-st} \,d\left[f(t)\right]
      \\&=
        \left. e^{-st} f(t) \right|_{0^{-}}^\infty
        + s \int_{0^{-}}^\infty e^{-st} f(t)\,dt
      \\&= -f(0^{-}) + sF(s)
  \end{split}
\]
via the by-parts method of \S~\ref{inttx:230}.
The integration property merely reverses the differentiation property on the
function $g(t) \equiv \int_{0^{-}}^t f(\tau) \,d\tau,$
for which $g(0^{-}) = 0$, filling the~``?'' with $F(s)/s$ in this pattern:%
{
  \settowidth\tla{$\ds\int_{0^{-}}^t f(\tau) \,d\tau$}
  \settowidth\tlb{$\ds 0 + sG(s)$}
  \bqb
    g(t) = \int_{0^{-}}^t f(\tau) \,d\tau &\fouripair& \makebox[\tlb][l]{$\ds G(s)$} = \ ?\ ; \\
    \frac{dg}{dt} = \makebox[\tla][r]{$\ds f(t)$} &\fouripair& 0 + sG(s) = F(s).
  \eqb
}%
The ramping property comes by
differentiating and negating~(\ref{fouri:laplace}) as
\[
  -\frac{d}{ds} F(s)
    = -\frac{d}{ds} \int_{0^{-}}^\infty e^{-st} f(t) \,dt
    = \int_{0^{-}}^\infty e^{-st} [tf(t)] \,dt
    = \mathcal L \{tf(t)\}.
\]
Higher-order properties come by repeated
application.  The convolution property comes as it did in
\S~\ref{fouri:110.40} except that here we take advantage of the
presence of Heaviside's unit step $u(\cdot)$ to manipulate the limits
of integration, beginning
\bqb
  \lefteqn{
    \mathcal{L}\left\{
      \int_{-\infty}^{\infty}
        u\left(\frac{t}{2}-\psi\right)
        h\left(\frac{t}{2}-\psi\right)
        u\left(\frac{t}{2}+\psi\right)
        f\left(\frac{t}{2}+\psi\right)
      \,d\psi
    \right\}
  }
  &&\\&=&
    \int_{-\infty}^{\infty}
      e^{-st}
    \int_{-\infty}^{\infty}
        u\left(\frac{t}{2}-\psi\right) h\left(\frac{t}{2}-\psi\right)
        u\left(\frac{t}{2}+\psi\right) f\left(\frac{t}{2}+\psi\right)
      \,d\psi
      \,dt,
\eqb
wherein evidently $u(t/2-\psi)u(t/2+\psi)=0$ for all $t < 0$, regardless
of the value of~$\psi$.  As in \S~\ref{fouri:110.40}, here also we
change $\phi \la t/2 + \psi$ and $\mu \la \phi - 2\psi$, eventually
reaching the form
\bqb
  \lefteqn{
    \mathcal{L}\left\{
      \int_{-\infty}^{\infty}
        u\left(\frac{t}{2}-\psi\right)
        h\left(\frac{t}{2}-\psi\right)
        u\left(\frac{t}{2}+\psi\right)
        f\left(\frac{t}{2}+\psi\right)
      \,d\psi
    \right\}
  }
  &&\\&=&
    \left[
      \int_{-\infty}^{\infty} e^{-s\mu} u(\mu)h(\mu) \,d\mu
    \right]
    \left[
      \int_{-\infty}^{\infty} e^{-s\phi} u(\phi)f(\phi) \,d\phi
    \right],
\eqb
after which once more we take advantage of Heaviside, this time to
curtail each integration to begin at~$0^{-}$ rather than at~$-\infty$,
thus completing the convolution property's proof.
\begin{table}
  \caption{Properties of the Laplace transform.}
  \label{fouri:laplace-properties}
  \bqb
    u(t-t_o)f(t-t_o) &\laplair& e^{-st_o}F(s) \\
    e^{-at}f(t) &\laplair& F(s+a) \\
    Af(\alpha t) &\laplair& \frac{A}{\alpha}F\left(\frac{s}{\alpha}\right),
      \ \Im(\alpha)=0,\ \Re(\alpha) > 0 \\
    A_1f_1(t) + A_2f_2(t) &\laplair& A_1F_1(t) + A_2F_2(t) \\
    \frac{d}{dt}f(t) &\laplair& sF(s) - f(0^{-}) \\
    \frac{d^2}{dt^2}f(t) &\laplair& s^2F(s) - sf(0^{-}) -\left[\frac{df}{dt}\right]_{t=0^{-}} \\
    \frac{d^n}{dt^n}f(t) &\laplair& s^nF(s)
      - \sum_{k=0}^{n-1}\left\{s^k\left[\frac{d^{n-1-k}}{dt^{n-1-k}}f(t)\right]_{t=0^{-}}\right\} \\
    \int_{0^{-}}^t f(\tau) \,d\tau &\laplair& \frac{F(s)}{s} \\
    tf(t) &\laplair& -\frac{d}{ds}F(s) \\
    t^nf(t) &\laplair& (-)^n\frac{d^n}{ds^n}F(s) \\
    \left[u(t)h(t)\right] \ast \left[u(t)f(t)\right] &\laplair& H(s) F(s)
  \eqb
\end{table}

\begin{table}
  \caption{Laplace transform pairs.}
  \label{fouri:laplace-pairs}
  \[
    \renewcommand\arraystretch{2.0}
    \br{rclcrcl}
      &&&&
      \ds e^{-at}t\sin\omega_o t &\laplair& \ds \frac{2(s+a)\omega_o}{[(s+a)^2+\omega_o^2]^2} \\
      \delta(t) &\laplair& 1 &&
      % diagn: the pair on the next line has been corrected; check it
      \ds e^{-at}t\cos\omega_o t &\laplair& \ds \frac{(s+a)^2-\omega_o^2}{[(s+a)^2+\omega_o^2]^2} \\
      \ds 1 &\laplair& \ds \frac{1}{s} &&
      \ds e^{-at} &\laplair& \ds \frac{1}{s+a} \\
      \ds t &\laplair& \ds \frac{1}{s^2} &&
      \ds e^{-at}t &\laplair& \ds \frac{1}{(s+a)^2} \\
      \ds t^n &\laplair& \ds \frac{n!}{s^{n+1}} &&
      \ds e^{-at}t^n &\laplair& \ds \frac{n!}{(s+a)^{n+1}} \\
      \ds \sin\omega_o t &\laplair& \ds \frac{\omega_o}{s^2+\omega_o^2} &&
      \ds e^{-at}\sin\omega_o t &\laplair& \ds \frac{\omega_o}{(s+a)^2+\omega_o^2} \\
      \ds \cos\omega_o t &\laplair& \ds \frac{s}{s^2+\omega_o^2} &&
      \ds e^{-at}\cos\omega_o t &\laplair& \ds \frac{s+a}{(s+a)^2+\omega_o^2}
    \er
  \]
\end{table}
Splitting the sine and cosine functions into their complex exponential
components according to Table~\ref{cexp:tbl-prop}, application of
Laplace's definition
\linebreak % bad break
(\ref{fouri:laplace})
to each component yields
Table~\ref{fouri:laplace-pairs}'s sine and cosine pairs.
The pair $t \laplair 1/s^2$ of the latter table
comes by application of the property that
$tf(t) \laplair -(d/ds)F(s)$ to the pair $1 \laplair 1/s$, and the pair
$t^n \laplair n!/s^{n+1}$ comes by repeated application of the same
property.  The pairs transforming $e^{-at}t\sin\omega_o t$,
$e^{-at}t\cos\omega_o t$, $e^{-at}t$ and $e^{-at}t^n$ come similarly.

During the application of either
Table~\ref{fouri:laplace-properties} or
Table~\ref{fouri:laplace-pairs},~$a$ may be, and~$s$ usually is,
complex; but~$\alpha$, $\omega_o$ and~$t$ are normally real.

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

\section{Solving differential equations by Laplace}
\label{fouri:250}
\index{differential equation!solution of by the Laplace transform}
\index{Laplace transform!solving a differential equation by}

\index{time domain}
\index{frequency domain}
\index{domain!time and frequency}
\index{transform domain}
\index{domain!transform}
\index{Laplace transform!comparison of against the Fourier transform}
\index{Fourier transform!comparison of against the Laplace transform}
The Laplace transform is curious, but Fourier is admittedly
more straightforward even if it is harder to
analyze.  Though exceptions exist, the Fourier transform is probably
the better general tool for frequency-domain analysis, among other
reasons because Fourier brings an inverse transformation
formula~(\ref{fouri:byu2}) whereas Laplace does
not.\footnote{%
  Actually, formally, Laplace does support an
  inverse transformation formula, $u(t)f(t) = (1/i2\pi)
  \int_{-i\infty}^{i\infty} e^{st} F(s)\,ds$, but to apply this
  inverse requires contour integration~\cite[eqn.~7.2]{Phillips/Parr}\@.
  The writer has no experience with it.  We'll not use it.  It
  comes of changing $s \la i\omega$ in~(\ref{fouri:eqn}).%
}

\index{initial condition}
Laplace excels Fourier in its property of
Table~\ref{fouri:laplace-properties} that $(d/dt)f(t) \laplair sF(s) -
f(0^{-})$.  Fourier's corresponding property of
Table~\ref{fouri:110:tbl20} lacks the $f(0^{-})$, an initial condition.

To see why this matters, consider for example the linear differential
equation%
\footnote{\cite[Example~19.31]{JJH}}$\mbox{}^{,}$%
\footnote{\index{state space}%
  It is enlightening to study the same differential equation
  in \emph{state-space} style \cite[chapter~8]{Phillips/Parr},
  \[
    \frac{d}{dt}\ve f(t) = \mf{rr}{ 0 & 1 \\ -3 & -4 } \ve f(t)
      + \mf{c}{0 \\ e^{-2t}},\ \ %
    \ve f(0) = \mf{c}{1 \\ 2},
  \]
  where
  \[
    \ve f(t) \equiv \mf{c}{1\\d/dt}f(t).
  \]
  The effort required to
  assimilate the notation rewards the student with significant insight
  into the manner in which initial conditions---here symbolized $\ve
  f(0)$---determine a system's subsequent evolution.
}
\bqb
  \frac{d^2}{dt^2}f(t) + 4\frac{d}{dt}f(t) + 3f(t) &=& e^{-2t}, \\
  \left. f(t) \right|_{t=0^{-}} &=& 1, \\
  \left. \frac{d}{dt}f(t) \right|_{t=0^{-}} &=& 2.
\eqb
Applying the properties of Table~\ref{fouri:laplace-properties} and
transforms of Table~\ref{fouri:laplace-pairs}, term by term, yields the
transformed equation
\bqb
  \lefteqn{
    \left\{
      s^2F(s) - s\bigg[ f(t) \bigg]_{t=0^{-}} - \bigg[
      \frac{d}{dt}f(t) \bigg]_{t=0^{-}}
    \right\}
  }
  &&\\&&\ \ \ \ \mbox{}
  + 4 \left\{
    sF(s) - \bigg[ f(t) \bigg]_{t=0^{-}}
  \right\}
  + 3F(s)
  = \frac{1}{s+2}.
\eqb
That is,
\[
  (s^2+4s+3)F(s) - (s+4)\bigg[ f(t) \bigg]_{t=0^{-}}
  - \bigg[\frac{d}{dt}f(t) \bigg]_{t=0^{-}} = \frac{1}{s+2}.
\]
Applying the known initial conditions,
\[
  (s^2+4s+3)F(s) - (s+4)[1] - [2]
  = \frac{1}{s+2}.
\]
Combining like terms,
\[
  (s^2+4s+3)F(s) - (s+6) = \frac{1}{s+2}.
\]
Multiplying by $s+2$ and rearranging,
\[
  (s+2)(s^2+4s+3)F(s) = s^2+8s+\mbox{0xD}.
\]
Isolating the heretofore unknown frequency-domain function $F(s)$,
\[
  F(s) = \frac{s^2+8s+\mbox{0xD}}{(s+2)(s^2+4s+3)}.
\]
Factoring the denominator,
\[
  F(s) = \frac{s^2+8s+\mbox{0xD}}{(s+1)(s+2)(s+3)}.
\]
Expanding in partial fractions (this step being the key to the whole
technique: see \S~\ref{inttx:260}),
\[
  F(s) = \frac{3}{s+1} - \frac{1}{s+2} - \frac{1}{s+3}.
\]
Though we lack an inverse transformation formula, it seems that we do
not need one because---having split the frequency-domain equation into
such simple terms---we can just look up the inverse transformation in
Table~\ref{fouri:laplace-pairs}, term by term.  The time-domain solution
\[
  f(t) = 3e^{-t} - e^{-2t} - e^{-3t}
\]
results.
One can verify the solution by substituting it back into
differential equation.
%Laplace has fathomed the differential equation.

Laplace can solve many linear differential equations in this way.

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

\section{Initial and final values by Laplace}
\label{fouri:260}
\index{initial-value theorem}
\index{final-value theorem}
\index{Laplace transform!initial and final values by}

The method of \S~\ref{fouri:250} though effective is sometimes too much
work, when all one wants to know are the initial and/or final values of
a function $f(t)$, when one is not interested in the details between.
The Laplace transform's \emph{initial-} and \emph{final-value theorems,}
\bq{fouri:260:10}
  \begin{split}
    f(0^{+}) &= \lim_{s \ra \infty} sF(s), \\
    \lim_{t\ra\infty} f(t) &= \lim_{s \ra 0} sF(s),
  \end{split}
\eq
meet this want.

One derives the initial-value theorem by the successive steps
\bqb
  \lim_{s \ra \infty} \mathcal L \left\{\frac{d}{dt}f(t)\right\}
    &=& \lim_{s \ra \infty} sF(s) - f(0^{-}), \\
  \lim_{s \ra \infty} \int_{0^{-}}^\infty e^{-st} \frac{d}{dt}f(t) \,dt
    &=& \lim_{s \ra \infty} sF(s) - f(0^{-}), \\
  \int_{0^{-}}^{0^{+}} \frac{d}{dt}f(t) \,dt
    &=& \lim_{s \ra \infty} sF(s) - f(0^{-}), \\
  f(0^{+}) - f(0^{-})
    &=& \lim_{s \ra \infty} sF(s) - f(0^{-}),
\eqb
which invoke the time-differentiation property of
Table~\ref{fouri:laplace-properties} and the last of which implies
(\ref{fouri:260:10})'s first line.  For the final value, one begins
\bqb
  \lim_{s \ra 0} \mathcal L \left\{\frac{d}{dt}f(t)\right\}
    &=& \lim_{s \ra 0} sF(s) - f(0^{-}), \\
  \lim_{s \ra 0} \int_{0^{-}}^\infty e^{-st} \frac{d}{dt}f(t) \,dt
    &=& \lim_{s \ra 0} sF(s) - f(0^{-}), \\
  \int_{0^{-}}^\infty \frac{d}{dt}f(t) \,dt
    &=& \lim_{s \ra 0} sF(s) - f(0^{-}), \\
  \lim_{t\ra\infty} f(t) - f(0^{-})
    &=& \lim_{s \ra 0} sF(s) - f(0^{-}),
\eqb
and (\ref{fouri:260:10})'s second line follows immediately.%
\footnote{\cite[\S~7.5]{Phillips/Parr}}

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

\section{The spatial Fourier transform}
\label{fouri:350}
\index{spatial Fourier transform}
\index{Fourier transform!spatial}

\index{time}
\index{space}
\index{wave mechanics}
\index{kernel}
\index{spatiotemporal phase factor}
\index{phase factor!spatiotemporal}
\index{convention}
\index{integration!volume}
\index{integration!triple}
In the study of wave mechanics,
% diagn: add an appropriate in-book reference?
physicists and engineers sometimes elaborate this chapter's
kernel~$e^{iv\theta}$ or~$e^{i\omega t}$, or by whichever pair of
letters is let to represent the complementary variables of
transformation, into the more general, spatiotemporal phase factor%
\footnote{
  The choice of sign here is a matter of convention, which differs by
  discipline.  This book tends to reflect its author's preference for
  $f(\ve r,t) \sim \int e^{i(+\omega t - \ve
  k\cdot\ve r)}F(\ve k, \omega)d\omega\,d\ve k$, convenient in electrical
  modeling but slightly less convenient in quantum-mechanical work.
}
$e^{i(\pm\omega t \mp\ve k\cdot\ve r)}$; where~$\ve k$ and~$\ve r$ are
three-dimensional geometrical vectors and~$\ve r$ in particular
represents a position in space.  To review the general interpretation
and use of such a factor lies beyond the
% diagn: on the next line, chapter's or book's?
book's
scope but the factor's very form,
\[
  e^{i(\pm\omega t \mp\ve k\cdot\ve r)}
  = e^{i(\pm\omega t \mp k_xx \mp k_yy \mp k_zz)},
\]
suggests Fourier transformation with respect not merely to time but also
to space.  There results the \emph{spatial Fourier transform}
\bq{fouri:spatial}
  \begin{split}
    F(\ve k) &=
      \frac{1}{(2\pi)^{3/2}}
      \int_V
        e^{+i\ve k\cdot\ve r}
          f(\ve r)
        \,d\ve r,
    \\
    f(\ve r) &=
      \frac{1}{(2\pi)^{3/2}}
      \int_V
        e^{-i\ve k\cdot\ve r}
          F(\ve k)
        \,d\ve k,
  \end{split}
\eq
analogous to~(\ref{fouri:byu2}) but cubing the $1/\sqrt{2\pi}$ scale
factor for the triple integration and reversing the sign of the kernel's
exponent.  The transform variable~$\ve k$, analogous to~$\omega$, is a
\emph{spatial frequency,} also for other reasons called a
\emph{propagation vector.}

\index{integration!surface}
\index{integration!double}
\index{integration!fourfold}
\index{integration!sixfold}
Nothing prevents one from extending~(\ref{fouri:spatial}) to four
dimensions, including a fourth integration to convert time~$t$ to
temporal frequency~$\omega$ while also converting position~$\ve r$ to
spatial frequency~$\ve k$.  On the other hand, one can restrict it to
two dimensions or even one.  Thus, various plausibly useful Fourier
transforms include
\[
  \begin{split}
    F(\omega) &=
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty} e^{-i\omega t} f(t) \,dt, \\
    F(k_z) &=
      \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty} e^{+ik_zz} f(z) \,dz, \\
    F(\ve k_\rho) &=
      \frac{1}{2\pi}
      \int_S e^{+i\ve k_\rho\cdot\we\rho} f(\we\rho) \,d\we\rho, \\
    F(\ve k) &=
      \frac{1}{(2\pi)^{3/2}}
      \int_V e^{+i\ve k\cdot\ve r} f(\ve r) \,d\ve r, \\
    F(\ve k, \ve \omega) &=
      \frac{1}{(2\pi)^{2}} \int_V \int_{-\infty}^{\infty}
      e^{i(-\omega t+\ve k\cdot\ve r)} f(\ve r,t) \,dt\,d\ve r,
  \end{split}
\]
among others.

