\chapter[Fourier and Laplace]{Fourier applications and the Laplace transform}
\label{lapl}
\index{transform}
\index{Fourier transform!applications of}
\index{Laplace transform}
\index{Laplace, Pierre-Simon (1749--1827)}
% diagn: this chapter has been pretty heavily revised; it's long, so
% one might not review it line by line, but it should probably be read
% once more sometime, end to end.

The last chapter has unveiled the Fourier transform.  This chapter
pursues some Fourier applications.  The chapter also introduces the
Laplace transform, a Fourier variant used to solve ordinary linear
differential equations constrained by boundary conditions.

%There is a way to Fourier, though. There is a style.

The chapter begins with a review of convolution.

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

\section{Convolution}
\label{lapl:140}
\index{convolution}

\index{convolution!balanced expression of}
\index{convolution!unbalanced expression of}
\index{balanced expression of convolution}
\index{unbalanced expression of convolution}
Section~\ref{fouri:110.40} and its~(\ref{fouri:110:50}) have
mentioned the \emph{convolution} of two functions,
\bq{fouri:convolbal}
  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
and have expressed this convolution also in unbalanced style as
\bq{fouri:convolunbal}
  h(t) \ast f(t) \equiv
    \int_{-\infty}^{\infty} h(t-\tau)
    f(\tau) \,d\tau,
\eq
the two styles---whether~(\ref{fouri:convolbal})
or~(\ref{fouri:convolunbal})---being equivalent. The two styles
express the same operation and yield the same result.

\index{convolution!commutivity of}
\index{convolution!associativity of}
\index{commutivity!of convolution}
\index{associativity!of convolution}
Convolution is commutative and associative
as~(\ref{fouri:110:51}) has observed:
\[
  \begin{split}
    f(t) \ast h(t) &= h(t) \ast f(t); \\
    f(t) \ast [g(t) \ast h(t)] &= [f(t) \ast g(t)] \ast h(t).
  \end{split}
\]
Despite exploring a few properties like these, though, and despite
giving a formal definition, we have developed little insight into the
convolution operation.  What \emph{is} convolution, really?

\subsection{What convolution is}
\label{lapl:140.10}

\index{convolution}
\index{convolution!definition of}
Several answers are possible.  In this book, our answer will chiefly be
this:
\begin{quote}
  {\scshape
    Convolution is the superposition of repetition with various scale
    and delay.
  }
\end{quote}
Well, that's fine.  The answer is dense with multisyllabic words, at
any rate.  What does it mean?

\index{superposition}
\index{Latin}
\index{sum of functions}
\index{function!sum of}
\index{brick}
\index{wall}
To begin, a \emph{superposition} is just a sum, especially a sum of
functions.  The word comes from the Latin\footnote{\cite{Webster1913}}
$\text{\emph{super}} + \text{\emph{{p}{o}{s}{i}{t}{i}{o}}}$, ``above
placement,'' and connotes a laying of one atop another, as bricks in a
wall.  Figure~\ref{fours:000:fig20} of an earlier chapter has plotted a
typical case.

\index{sifting property}
\index{Dirac delta function!sifting property of}
\index{delta function, Dirac!sifting property of}
\index{repetition}
\index{scale}
\index{delay}
Regarding ``repetition with various scale and delay,'' an
example:\footnote{Patterned after~\cite{JamesPeterson-lecture}.}
\[
  \begin{split}
    f(t) &= \left(1\right)\delta\left[t-\left(-\frac{\mbox{0x18}}{5}\right)\right]
    + \left(1\right)\delta\left[t-\left(-\frac 5 2\right)\right]
    + \left(\frac 3 2\right) \delta\left[t-0\rule{0em}{3.1ex}\right]
    \\&\qquad\mbox{} + \left(\frac 2 3\right) \delta\left[t - \frac 1 2\right]
      + \left(-\frac 8 3\right)\delta\left[t-\frac{\mbox{0xB}} 4\right];
    \\
    h(t) &= \Lambda(t).
  \end{split}
\]
Here, the~$1$, $1$, $\frac 3 2$, $\frac 2 3$ and~$-\frac 8 3$ give scale, while
the~$-\frac{\text{0x18}}{5}$, $-\frac 5 2$, $0$, $\frac 1 2$ and~$\frac{\text{0xB}} 4$ give delay.
Convolving according to~(\ref{fouri:convolunbal}) and sifting by~(\ref{integ:670:sift}),
\[
  \begin{split}
    h(t) \ast f(t) &= \left(1\right)\Lambda\left[t-\left(-\frac{\mbox{0x18}}{5}\right)\right]
    + \left(1\right)\Lambda\left[t-\left(-\frac 5 2\right)\right]
    + \left(\frac 3 2\right) \Lambda\left[t-0\rule{0em}{3.1ex}\right]
    \\&\qquad\mbox{} + \left(\frac 2 3\right) \Lambda\left[t - \frac 1 2\right]
      + \left(-\frac 8 3\right)\Lambda\left[t-\frac{\mbox{0xB}} 4\right].
  \end{split}
\]
Figure~\ref{lapl:140:fig10} plots.
\begin{figure}
  \caption{Convolution.}
  \label{lapl:140:fig10}
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      %\localscalebox{1.0}{1.0}
      {
        \small
        \psset{dimen=middle}
        \rput(0,\fyc){%
          \xxaxes{$h(t)$}%
          % the lambda
          {
            \psset{linewidth=2.0pt}
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                 (0.5\tlb,1.4166666666666667\tlb)(1\tlb,0.3333333333333333\tlb)(1.5\tlb,0)%
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               (1.35\tla,0)%
          }%
        }%
      }
    \end{pspicture}
  \ec
\end{figure}%

\index{Dirac delta function}
\index{delta function, Dirac}
But what if $f(t)$ is no mere sum of Dirac deltas\footnote{Mere!
How use accustoms one to the ephemeral!} but a more
general function?  In that case, one can approximate that
\bq{lapl:140:20}
   f(t) = \lim_{\Delta t\ra 0} \Delta t %\, space omitted, doesn't look good here
     \sum_{k=-\infty}^\infty f(k\,\Delta t)
     \delta(t-k\,\Delta t);
\eq
or, if unsure that one can trust~(\ref{lapl:140:20}) as written,
that
\[
   f(t) = \lim_{\Delta t\ra 0}
     \sum_{k=-\infty}^\infty f(k\,\Delta t)
     \Pi\left(\frac{t-k\,\Delta t}{\Delta t}\right),
\]
which, in view of \S~\ref{integ:670}, says the same thing.
Processing~(\ref{lapl:140:20}) by Fig.~\ref{lapl:140:fig10}'s method,
\[
  h(t) \ast f(t) = \lim_{\Delta t\ra 0}
     \Delta t \sum_{k=-\infty}^\infty f(k\,\Delta t)
     h(t-k\,\Delta t).
\]

\subsection{Series coefficients}
\label{lapl:140.20}
\index{Fourier coefficient}
\index{coefficient!Fourier}

\index{time-limited pulse}%
\index{pulse!time-limited}%
Given a time-limited pulse
%\footnote{%
%  This subsection chooses the~$\ge$ sign
%  rather than the~$>$ sign but the choice is arbitrary.  If
%  you have a problem in which the choice matters---such as, for
%  example, a problem in which a Dirac delta discontinuity
%  (\S~\ref{integ:670}) arises just on the edge of the
%  interval~$T_1$---then you will probably need to pose the problem more
%  precisely before attacking it.%
%}
\[
  f(t) = 0\ \mbox{for all $\ds|t-t_o| \ge \frac{T_1}{2}$},
\]
one can write that
\[
    \int_{-\infty}^{\infty} e^{-i\omega t} f(t)\,dt
    =
    \int_{t_o-T_1/2}^{t_o+T_1/2} e^{-i\omega t} f(t)\,dt,
\]
after which comparison of the Fourier transform~(\ref{fouri:byu2})
against the coefficients~(\ref{fours:100:15}) of the Fourier
\emph{series} reveals that
\bq{lapl:140:22}
  a_j = \frac{\Delta\omega}{\sqrt{2\pi}} F(j\,\Delta\omega),
\eq
where $\Delta\omega\, T_1 = 2\pi$ as in~(\ref{fours:080:08}).
If a time-limited pulse repeats at an interval~$T_1$ to form a wave
and the interval is long enough that the wave's pulses do not overlap,
then~(\ref{lapl:140:22}) extracts the coefficients of the waveform's
Fourier series from the transform of the pulse.

\index{collated pulse}%
\index{pulse!collated}%
\index{overlap}%
\index{pulse!overlapping}%
If pulses overlap then one must do extra work to reach the same
result, representing the repeating waveform
via~(\ref{fouri:convolunbal}) and~(\ref{integ:670:sift}) as
\bq{lapl:140:23}
  \hat{f}(t) \equiv
  \sum_{k=-\infty}^\infty f(t-kT_1) =
  \left[ \sum_{k=-\infty}^\infty \delta(t-kT_1) \right]
  \ast f(t).
\eq
The summation in~(\ref{lapl:140:23})
might diverge, in which case the repeating waveform would not exist,
but if the summation does not diverge then~(\ref{lapl:140:22}) still
obtains despite the overlap.  To prove it, beginning
from~(\ref{fours:100:15}) by successive steps,\footnote{G.~Andrew Walls
has kindly provided the technique at the author's request.
Walls' original is~\cite{GAWalls}.}
\[
  \begin{split}
    a_j &=
      \frac{1}{T_1} \int_{t_o-T_1/2}^{t_o+T_1/2}
      e^{-ij\,\Delta\omega\,\tau} \hat f(\tau) \,d\tau
    \\&=
      \frac{1}{T_1} \int_{-\infty}^{\infty}
      \Pi\left(\frac{\tau-t_o}{T_1}\right)
      e^{-ij\,\Delta\omega\,\tau} \hat f(\tau) \,d\tau
    \\&=
      \frac{\sqrt{2\pi}}{T_1} \left[ \fourf \bigg\{
      \Pi\left(\frac{t-t_o}{T_1}\right) \hat f(t)
      \bigg\}\right]_{\omega=j\,\Delta\omega}
    \\&=
      \frac{1}{T_1} \left[ \fourf \bigg\{
      \Pi\left(\frac{t-t_o}{T_1}\right)
      \bigg\} \ast \fourf \bigg\{ \hat f(t)
      \bigg\} \right]_{\omega=j\,\Delta\omega}
    \\&=
      \frac{1}{T_1}
      \left[ \left( e^{-i\omega t_o}
        \fourf \bigg\{ \Pi\left(\frac{t}{T_1}\right) \bigg\} \right)
      \ast \fourf \bigg\{ \hat f(t) \bigg\}
      \right]_{\omega=j\,\Delta\omega},
  \end{split}
\]
in which we have used the square pulse of \S~\ref{fours:095}
and some properties of Tables~\ref{fouri:110:tbl20}
and~\ref{fouri:110:tbl41}.
Finding in Table~\ref{fouri:120:tbl20} the transform pair $\Pi(v)
\fouripair \sinarg(v/2)/\sqrt{2\pi}$, applying~(\ref{fouri:110:33})
to this pair with $\alpha = 1/T_1$, observing that
$\Delta\omega\,T_1=2\pi$, and changing $\fourf_{vv} \la
\fourf_{\omega t}$ yields the modified pair
\bq{lapl:140:24}
  \Pi\left(\frac{t}{T_1}\right)
  \fouripair
  \frac{\left(\!\sqrt{2\pi}\right)\sinarg(2\pi\omega /2\,\Delta\omega)}{\Delta\omega},
\eq
which, when used against the last expression for~$a_j$, gives that
\[
  \begin{split}
    a_j &=
      \frac{1}{T_1}
      \left[
        \frac{e^{-i\omega t_o}
        \left(\!\sqrt{2\pi}\right)\sinarg(2\pi\omega/2\,\Delta\omega)}{\Delta\omega}
      \ast \hat F(\omega)
      \right]_{\omega=j\,\Delta\omega}
    \\&=
      \frac{1}{\sqrt{2\pi}}
      \left\{
        \left[ e^{-i\omega t_o}
        \sinarg\left(\frac{2\pi\omega}{2\,\Delta\omega}\right) \right]
        \ast \hat F(\omega)
      \right\}_{\omega=j\,\Delta\omega}
    \\&=
      \frac{1}{\sqrt{2\pi}}
      \left\{\int_{-\infty}^{\infty}
        e^{-i(\omega-\eta)t_o}
        \sinarg\left[\frac{2\pi(\omega-\eta)}{2\,\Delta\omega}\right] \hat F(\eta)
      \,d\eta\right\}_{\omega=j\,\Delta\omega}
    \\&=
      \frac{e^{-ij\,\Delta\omega\,t_o}}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty}
        e^{i\eta t_o}
        \sinarg\left[\frac{2\pi(j\,\Delta\omega-\eta)}{2\,\Delta\omega}\right]
        \hat F(\eta)
      \,d\eta.
  \end{split}
\]
%% superdiagn: there should not be any break here, neither a paragraph
%% break nor a page break; once prose earlier in the chapter has
%% changed, altering the layout's positioning, these breaks should
%% probably be commented out (and perhaps deleted).
%\pagebreak
%
%\noindent
Transforming~(\ref{lapl:140:23}),
\bq{lapl:140:23t}
  \hat{F}(\omega) =
  \left[ \sum_{k=-\infty}^\infty \delta(\omega-k\,\Delta\omega) \right]
  F(\omega)\,\Delta\omega.
\eq
Using~(\ref{lapl:140:23t}) against the last expression for~$a_j$,
\[
  \begin{split}
    a_j &=
      \frac{e^{-ij\,\Delta\omega\,t_o}\,\Delta\omega}{\sqrt{2\pi}}
    \\&\qquad\mbox{}\times
      \int_{-\infty}^{\infty}
        e^{i\eta t_o}
        \sinarg\left[\frac{2\pi(j\,\Delta\omega-\eta)}{2\,\Delta\omega}\right]
        \left[ \sum_{k=-\infty}^\infty \delta(\eta-k\,\Delta\omega) \right]
        F(\eta)
      \,d\eta
    \\&=
      \frac{e^{-ij\,\Delta\omega\,t_o}\,\Delta\omega}{\sqrt{2\pi}}
    \\&\qquad\mbox{}\times
      \sum_{k=-\infty}^\infty \int_{-\infty}^{\infty}
        e^{i\eta t_o}
        \sinarg\left[\frac{2\pi(j\,\Delta\omega-\eta)}{2\,\Delta\omega}\right]
        \left[ \delta(\eta-k\,\Delta\omega) \right]
        F(\eta)
      \,d\eta.
  \end{split}
\]
%Here we will briefly pause to consider the last integral's convergence
%(though convergence is, as \S\S~\ref{taylor:314.25}
%and~\ref{conclu:convergence} mention, admittedly a point at which the
%book seldom otherwise tarries).
%Insofar as Dirac's delta~$\delta(\cdot)$ is understood to be a pulse of
%arbitrarily large but finite amplitude~$1/\ep$ and arbitrarily small but finite
%width\footnote{One could quibble over the width if the delta were
%held to be Gaussian in shape as in chapter~\ref{integ}'s
%footnote~\ref{integ:670:fn}.  However, a Gaussian pulse diminishes so
%aggressively (\S~\ref{fouri:130}) away from its central region that, to
%the applicationist, the question is more a formal concern than a
%plausible threat.  This book will pursue the question no
%further.}~$\ep$, the
%product of the complex-exponential and sine-argument
%functions within the integrand does not exceed~$1$ in magnitude.  Therefore,
%the last integral converges if $\int_{-\infty}^\infty|F(\eta)|\,d\eta$ does
%(see also \S~\ref{fouri:100.40} and its eqn.~\ref{fouri:100:46}).
The integral might not converge but if it does, sifting,
\[
  \begin{split}
    a_j &=
      \frac{e^{-ij\,\Delta\omega\,t_o}\,\Delta\omega}{\sqrt{2\pi}}
        \sum_{k=-\infty}^\infty
        e^{ik\,\Delta\omega\,t_o}
        \sinarg\left[\frac{2\pi(j\,\Delta\omega-k\,\Delta\omega)}{2\,\Delta\omega}\right]
        F(k\,\Delta\omega)
    \\&=
      \frac{\Delta\omega}{\sqrt{2\pi}}
        \sum_{k=-\infty}^\infty
        e^{i(k-j)\,\Delta\omega\,t_o}
        \sinarg[\pi(j-k)]
        F(k\,\Delta\omega).
  \end{split}
\]
The indices~$j$ and~$k$ being integers,
\[
  e^{i(k-j)\,\Delta\omega\,t_o} \sinarg[\pi(j-k)]
  = \begin{cases}
    1 & \mbox{if $j=k$,} \\
    0 & \mbox{otherwise,}
  \end{cases}
\]
by which the last expression for~$a_j$ implies~(\ref{lapl:140:22}).

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

\section{Constructive functions}
\label{lapl:123}
\index{constructive function}
\index{function!constructive}

Section~\ref{fouri:100.90} has introduced the approximation and
construction, and subsequent transformation, of an arbitrary pulse.
That subsection wants evenly spaced samples, though.  What if the
samples are unevenly spaced?

This section develops constructive functions to represent among others
samples that are unevenly spaced.

\subsection{The irregular triangular pulse}
\label{lapl:123.20}
\index{Fourier transform!of an irregular triangular pulse}
\index{irregular triangular pulse!Fourier transform of}
\index{triangular pulse!irregular, Fourier transform of}
\index{pulse!irregular triangular}

The Fourier transform of the irregular triangular pulse of
Fig.~\ref{lapl:123:fig20}, in which $B > 0$ and $C > 0$,
\begin{figure}
  \caption{An irregular triangular pulse.}
  \label{lapl:123:fig20}
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\end{figure}
is by
\[
  \begin{split}
    \frac{\left(\!\sqrt{2\pi}\right)F_{\text{irr triangular}}(v)}{A}
    &=
      \int_{-B}^0 e^{-iv\theta} \left( 1 + \frac{\theta}{B} \right) d\theta
      + \int_0^C e^{-iv\theta} \left( 1 - \frac{\theta}{C} \right) d\theta
    \\&=
      \left[
        e^{-iv\theta} \left( -\frac{1}{iv} - \frac{\theta}{iBv} + \frac{1}{Bv^2} \right)
      \right]_{\theta=-B}^0
    \\&\qquad\mbox{} +
      \left[
        e^{-iv\theta} \left( -\frac{1}{iv} + \frac{\theta}{iCv} - \frac{1}{Cv^2} \right)
      \right]_{\theta=0}^C
    \\&=
      \left[
        \frac{1-e^{iBv}}{Bv^2} - \frac{1}{iv}
      \right]
      +
      \left[
        \frac{1-e^{-iCv}}{Cv^2} + \frac{1}{iv}
      \right]
    \\&=
        \frac{1-e^{iBv}}{Bv^2} + \frac{1-e^{-iCv}}{Cv^2}.
  \end{split}
\]
Thus,
\bq{lapl:123:25}
  f_{\text{irr triangular}}(v) \fouripair \frac{A}{\sqrt{2\pi}} \left(
    \frac{1-e^{iBv}}{Bv^2} + \frac{1-e^{-iCv}}{Cv^2}
  \right),\ \ \mbox{$B > 0,\ C > 0$}.
\eq
As~$v$ vanishes, $1-e^{\alpha v} \ra -\alpha v - \alpha^2v^2/2 - \cdots$, so
\[
  \begin{split}
    \lim_{v\ra 0}F_{\text{irr triangular}}(v)
    &= \lim_{v\ra 0}\frac{A}{\sqrt{2\pi}} \bigg(
      \frac{-iBv+B^2v^2/2+\cdots}{Bv^2}
    \\&\qquad\qquad\qquad\quad\mbox{} + \frac{iCv+C^2v^2/2+\cdots}{Cv^2}
    \bigg),
  \end{split}
\]
which implies that
\bq{lapl:123:26}
  F_{\text{irr triangular}}(0) = \frac{(A)(B+C)}{2\sqrt{2\pi}},
\eq
a result that matches the result direct application
of~(\ref{fouri:byu}) at $v=0$ yields.

\subsection{The ramp and level (failing attempt)}
\label{lapl:123.30}
\index{Fourier transform!of a ramp and level}
\index{ramp and level!Fourier transform of}
\index{level, ramp and}
\index{failed attempt}
\index{failing attempt}
\index{attempt!failing}

This book like other mathematical books seldom explores failing
attempts, successful attempts wanting pages enough.  Nevertheless, the
applied mathematician in practice, scratching paper with his pencil,
spends so much time in failing attempts that the occasional review of a
failing attempt can be instructive, as follows.

If $C \ra \infty$ in Fig.~\ref{lapl:123:fig20} then the function of
Fig.~\ref{lapl:123:fig30} emerges, a function we shall call the
\emph{ramp and level.}
\begin{figure}
  \caption{A ramp and level.}
  \label{lapl:123:fig30}
  \bc
    \nc\xxxab{4.3}
    \nc\xxyab{1.8}
    \nc\xxyac{0.4}
    \setlength\tla{3.0cm}
    \setlength\tlb{1.2cm}
    \nc\xxl{0.15}
    \nc\xxa{1.0}\nc\xxb{1.2}\nc\xxc{1.8}
    \nc\fxa{-5.0} \nc\fxb{5.0}
    \nc\fya{-0.8} \nc\fyb{2.6}
    \begin{pspicture}(\fxa,\fya)(\fxb,\fyb)
      %\psframe[linewidth=0.5pt,dimen=outer](\fxa,\fya)(\fxb,\fyb)
      \localscalebox{1.0}{1.0}
      {
        \small
        \psset{dimen=middle}
        \rput(0,0){%
          {
            \psset{linewidth=0.5pt}
            \settowidth\tlc{$B$}
            \psline(-\xxxab,0)(\xxxab,0)
            \psline(0,-\xxyac)(0,\xxyab)
            \uput[r](\xxxab,0){$v$}
            \uput[u](0,\xxyab){$f_{\text{ramp and level}}(v)$}
            \psline(-\xxl,\xxa\tlb)( \xxl,\xxa\tlb)
            \uput[ul](0,\xxa\tlb){$A$}
            \psline(-\xxb\tlb, \xxl)(-\xxb\tlb,-\xxl)
              \uput[d](-\xxb\tlb,-\xxl){\makebox[\tlc][r]{$-B$}}
          }%
          {
            \psset{linewidth=2.0pt}
            \psline%
               (-1.35\tla,0)%
               (-\xxb\tlb,0)(0,\xxa\tlb)%
               ( 1.35\tla,\xxa\tlb)
          }%
        }%
      }
    \end{pspicture}
  \ec
\end{figure}

The function's Fourier transform would be hard to compute directly but,
in the limit, one can carefully extract the transform
from~(\ref{lapl:123:25}) by letting $C = 1/\ep$---the $\ep\ra 0^{+}$
being a positive infinitesimal---after which~(\ref{lapl:123:25}) has
that
\[
    f_{\text{ramp and level}}(v) \fouripair \lim_{\ep\ra 0^{+}}
      \frac{A}{\sqrt{2\pi}} \left[
      \frac{1-e^{iBv}}{Bv^2}
      + \frac{\big(\ep\big)\big(1-e^{-iv/\ep}\big)}{v^2}
    \right].
\]
% diagn: this sentence looks good but its prose wants one more quick
% review.
That is,
\bq{lapl:123:33}
    f_{\text{ramp and level}}(v) \fouripair
      \frac{\big(A\big)\big(1-e^{iBv}\big)}{\left(\!\sqrt{2\pi}\right)Bv^2},
    \ v\neq 0,
\eq
in which the condition on the right that $v\neq 0$ does not here forbid
the~$v$ on the \emph{left} from vanishing.

So far, so good.  However, the reasoning \S~\ref{lapl:123.20}
has used to reach~(\ref{lapl:123:26}) for vanishing~$v$ is
insufficiently precise for the present problem.  The trouble is that, in
\S~\ref{lapl:123.20}, only~$v$ vanished, whereas now $1/C$ vanishes
also.  One must ask, is the product~$Cv$ to vanish?  And if it is not,
then is the inverse product $1/Cv$ alternately to vanish?  We must
decide.

\index{infinitude}
\index{finitude}
\index{infinitesimality}
\index{failure}
Noticing the exponential factor~$e^{-iCv}$ in~(\ref{lapl:123:25}), we
might try letting the product~$Cv$ vanish, defining the finite
$\rho \equiv v/\ep^2$---where by \emph{finite} we here mean that~$\rho$ is
neither infinite nor infinitesimal (nor zero), and therefore that~$v$
is infinitesimal on the order of~$\ep^2$.  Having thus decided and
thus defined, we make the unnumbered equation
preceding~(\ref{lapl:123:33}) to be \bqb
  \lefteqn{\lim_{\ep\ra 0^{+}} F_{\text{ramp and level}}(\ep^2\rho)}
  \\&=& \lim_{\ep\ra 0^{+}}
    \frac{A}{\sqrt{2\pi}} \left(
      \frac{1-e^{iB\ep^2\rho}}{B\ep^4\rho^2}
      + \frac{1-e^{-i\ep\rho}}{\ep^3\rho^2}
    \right)
  \\&=& \lim_{\ep\ra 0^{+}}
    \frac{A}{\sqrt{2\pi}} \bigg(
      \frac{-iB\ep^2\rho+B^2\ep^4\rho^2/2+\cdots}{B\ep^4\rho^2}
  \\&&\qquad\qquad\qquad\mbox{}
      + \frac{i\ep\rho+\ep^2\rho^2/2-i\ep^3\rho^3/6+\cdots}{\ep^3\rho^2}
    \bigg)
  \\&=& \lim_{\ep\ra 0^{+}}
    \frac{A}{\sqrt{2\pi}} \left(
      \frac{1}{2\ep} + \frac{3B-i\rho}{6}
    \right),
\eqb
a result which might be interesting but which is not obviously helpful.

The technique used---having already prospered in \S\S~\ref{cexp:230},
\ref{inttx:260} and \ref{inttx:460} among others---remains reasonable,
so we shall retain the technique in our toolbox as it were for future use.
Notwithstanding, even a reasonable technique can fail to solve some
particular problem.  Whether \emph{this} technique could with extra effort
eventually solve the present problem is not a question we will further
pursue (for example, one could alternately try letting $1/Cv$ vanish
instead of~$Cv$).  We will merely rather observe that, in the present
instance, the technique does not seem to have prospered yet.

\index{shelf}
Let us shelve the problem for the moment.  We will retrieve it from
the shelf, working it via a different technique, in
\S~\ref{lapl:123.60}.

\subsection{The right-triangular pulse}
\label{lapl:123.40}
\index{Fourier transform!of an right-triangular pulse}
\index{right-triangular pulse!Fourier transform of}
\index{triangular pulse!right-, Fourier transform of}
\index{pulse!right-triangular}

\index{integral!absolute}
\index{absolute integral}
If $B=0$ or $C=0$ (but not both) in the irregular triangular pulse of
\S~\ref{lapl:123.20} and Fig.~\ref{lapl:123:fig20}, then one of the
\emph{right-triangular pulses}
%\footnote{The writer is unaware of a conventional name for this pulse.
%The name \emph{right-triangular} seems as serviceable as any.}
of Fig.~\ref{lapl:123:fig40} results.
\begin{figure}
  \caption{Right-triangular pulses.}
  \label{lapl:123:fig40}
  \bc
    \nc\leftname{\ensuremath f_{\text{leftward}}(v)}
    \nc\rightname{\ensuremath f_{\text{rightward}}(v)}
    \settowidth\tld{$\leftname$}
    \nc\xxxab{4.3}
    \nc\xxyab{1.8}
    \nc\xxyac{0.4}
    \setlength\tla{3.0cm}
    \setlength\tlb{1.2cm}
    \nc\xxl{0.15}
    \nc\xxa{1.0}\nc\xxb{1.2}\nc\xxc{1.8}
    \nc\fyc{3.6}
    \nc\fxa{-5.0} \nc\fxb{5.0}
    \nc\fya{-0.8} \nc\fyb{6.2}
    \begin{pspicture}(\fxa,\fya)(\fxb,\fyb)
      %\psframe[linewidth=0.5pt,dimen=outer](\fxa,\fya)(\fxb,\fyb)
      \localscalebox{1.0}{1.0}
      {
        \small
        \psset{dimen=middle}
        \rput(0,\fyc){%
          {
            \psset{linewidth=0.5pt}
            \settowidth\tlc{$B$}
            \psline(-\xxxab,0)(\xxxab,0)
            \psline(0,-\xxyac)(0,\xxyab)
            \uput[r](\xxxab,0){$v$}
            \uput[u](0,\xxyab){\makebox[\tld][r]{$\leftname$}}
            \psline(0,\xxa\tlb)( \xxl,\xxa\tlb)
            \uput[r](\xxl,\xxa\tlb){$A$}
            \psline(-\xxb\tlb, \xxl)(-\xxb\tlb,-\xxl)
              \uput[d](-\xxb\tlb,-\xxl){\makebox[\tlc][r]{$-B$}}
          }%
          {
            \psset{linewidth=2.0pt}
            \psline%
               (-1.35\tla,0)%
               (-\xxb\tlb,0)(0,\xxa\tlb)(0,0)%
               ( 1.35\tla,0)
          }%
        }%
        \rput(0,0){%
          {
            \psset{linewidth=0.5pt}
            \settowidth\tlc{$B$}
            \psline(-\xxxab,0)(\xxxab,0)
            \psline(0,-\xxyac)(0,\xxyab)
            \uput[r](\xxxab,0){$v$}
            \uput[u](0,\xxyab){\makebox[\tld][r]{$\rightname$}}
            \psline(-\xxl,\xxa\tlb)(0,\xxa\tlb)
            \uput[l](-\xxl,\xxa\tlb){$A$}
            \psline(\xxc\tlb, \xxl)(\xxc\tlb,-\xxl)
              \uput[d](\xxc\tlb,-\xxl){$C$}
          }%
          {
            \psset{linewidth=2.0pt}
            \psline%
               (-1.35\tla,0)%
               ( 0,0)(0,\xxa\tlb)(\xxc\tlb,0)%
               ( 1.35\tla,0)
          }%
        }%
      }
    \end{pspicture}
  \ec
\end{figure}
Depending on whether the pulse consists of the leftward or the
rightward triangle, the pulse's Fourier transform is computed as in
the earlier subsections to be
\bq{lapl:123:40}
  \begin{split}
    f_{\text{leftward}}(v) &\fouripair
      \frac{A}{\sqrt{2\pi}}\left(
        \frac{1-e^{iBv}}{Bv^2} - \frac{1}{iv}
      \right),
    \\
    f_{\text{rightward}}(v) &\fouripair
      \frac{A}{\sqrt{2\pi}}\left(
        \frac{1-e^{-iCv}}{Cv^2} + \frac{1}{iv}
      \right),
  \end{split}
\eq
with
\bq{lapl:123:41}
  \begin{split}
    F_{\text{leftward}}(0) &= \frac{AB}{2\sqrt{2\pi}}, \\
    F_{\text{rightward}}(0) &= \frac{AC}{2\sqrt{2\pi}}.
  \end{split}
\eq
%diagn: review the rest of this recently abbreviated paragraph.
The transform's slow decay, going only on the order of $1/v$, may
be worth noticing, incidentally---the slow decay presumably
owing\footnote{This book will not attempt a general proof of the
point.} to the pulse's discontinuity.
%
% Version 1:
%
%Whether leftward or rightward, this transform carries a potential
%disadvantage, though.  The magnitude of its Fourier transform decays
%slowly for large~$v$, going only on the order of $1/v$.  This is not
%necessarily a problem in all cases but can complicate cases in
%which the transform is to be integrated or otherwise summed, for
%$\int_{(\cdot)}^\infty |dv/v|$ diverges.  Comparing the transform of the
%present subsection's right-triangular pulse against the transform of
%the full irregular triangular pulse of \S~\ref{lapl:123.20} and
%Fig.~\ref{lapl:123:fig20} suggests that the right-triangular's slow
%decay (to the extent to which it matters) were an
%artifact\footnote{This book will not attempt a general proof of the
%point.} of its discontinuity, a discontinuity the full irregular
%triangular pulse avoids.

\subsection{The irregular step}
\label{lapl:123.50}
\index{Fourier transform!of an irregular step}
\index{irregular step!Fourier transform of}
\index{step, irregular}

Section~\ref{fouri:120.30} has taken the Fourier
transform~(\ref{fouri:120:22}) of Heaviside's unit step.  Scaling
Heaviside vertically by a factor of
$A_{\text{rightward}}-A_{\text{leftward}}$ and then,
using~(\ref{fouri:110:22}), shifting the scaled Heaviside vertically by
an offset of $A_{\text{leftward}}$ yields the transform pair
\bq{fouri:120:50}
  f_{\text{irr step}}(v) \fouripair
  \frac{A_{\text{rightward}}-A_{\text{leftward}}}{\left(\!\sqrt{2\pi}\right)iv} +
  \frac{(A_{\text{rightward}}+A_{\text{leftward}})\sqrt{2\pi}}{2} \delta(v)
\eq
of the irregular step of Fig.~\ref{lapl:123:fig50}.
\begin{figure}
  \caption{An irregular step.}
  \label{lapl:123:fig50}
  \bc
    \nc\xxxab{4.3}
    \nc\xxyab{1.8}
    \nc\xxyac{0.4}
    \setlength\tla{3.0cm}
    \setlength\tlb{1.2cm}
    \nc\xxl{0.15}
    \nc\xxa{0.3}\nc\xxb{1.0}
    \nc\fxa{-5.0} \nc\fxb{5.0}
    \nc\fya{-0.8} \nc\fyb{2.6}
    \begin{pspicture}(\fxa,\fya)(\fxb,\fyb)
      %\psframe[linewidth=0.5pt,dimen=outer](\fxa,\fya)(\fxb,\fyb)
      \localscalebox{1.0}{1.0}
      {
        \small
        \psset{dimen=middle}
        \rput(0,0){%
          {
            \psset{linewidth=0.5pt}
            \psline(-\xxxab,0)(\xxxab,0)
            \psline(0,-\xxyac)(0,\xxyab)
            \uput[r](\xxxab,0){$v$}
            \uput[u](0,\xxyab){$f_{\text{irr step}}(v)$}
            \psline(-\xxl,\xxa\tlb)( \xxl,\xxa\tlb)
              \uput[r](0,\xxa\tlb){$A_{\text{leftward}}$}
            \psline(-\xxl,\xxb\tlb)( \xxl,\xxb\tlb)
              \uput[l](0,\xxb\tlb){$A_{\text{rightward}}$}
          }%
          {
            \psset{linewidth=2.0pt}
            \psline%
               (-1.35\tla,\xxa\tlb)%
               ( 0,\xxa\tlb)(0,\xxb\tlb)%
               ( 1.35\tla,\xxb\tlb)
          }%
        }%
      }
    \end{pspicture}
  \ec
\end{figure}

\subsection{The ramp and level}
\label{lapl:123.60}
\index{Fourier transform!of a ramp and level}
\index{ramp and level!Fourier transform of}
\index{level, ramp and}
\index{attempt!successful}

% diagn: For some indefinable reason, the writer remains vaguely uneasy
% with the first sentence of this paragraph.  Still, maybe it's good.
% Reread the paragraph sometime, and then maybe reconsider.
Sections~\ref{lapl:123.40} and~\ref{lapl:123.50} have drawn their
results from a lengthy sequence of application-level logic, running
through chapters~\ref{fouri} and~\ref{lapl},
of abstruse Fourier mathematics.  Preceding
\S\S~\ref{lapl:123.40} and~\ref{lapl:123.50} however was
\S~\ref{lapl:123.30} which, as you will recall, has left one unsolved
problem on the shelf.  The unsolved problem is the ramp and level,
Fig.~\ref{lapl:123:fig30}.

\index{superposition}
Superimposing the leftward right-triangular pulse of \S~\ref{lapl:123.40}
upon the irregular step of \S~\ref{lapl:123.50}, the latter with
$A_{\text{leftward}}=0$ and $A_{\text{rightward}}=A$, now solves the
shelved problem immediately:
\bq{lapl:123:50}
  f_{\text{ramp and level}} \fouripair
  \frac{A}{\sqrt{2\pi}}\left[
      \frac{1-e^{iBv}}{Bv^2} + \pi\delta(v)
  \right].
\eq
%(Note incidentally that this transform decays as
%$1/v^2$, a property the irregular triangular pulse of
%Fig.~\ref{lapl:123:fig20} has but the
%right-triangular pulse and the irregular step separately lack.
%Note also that it is the transforms of the discontinuous functions that
%go slowly as $1/v$.)

\index{Fourier transform!of a level and ramp}
\index{level and ramp!Fourier transform of}
\index{ramp, level and}
As well as the ramp and level, one could likewise define a \emph{level
and ramp,} the level and ramp mirroring the ramp and
level, its level extending not to the right but to the left as in
Fig.~\ref{lapl:123:fig55}.
\begin{figure}
  \caption{A level and ramp.}
  \label{lapl:123:fig55}
  \bc
    \nc\xxxab{4.3}
    \nc\xxyab{1.8}
    \nc\xxyac{0.4}
    \setlength\tla{3.0cm}
    \setlength\tlb{1.2cm}
    \nc\xxl{0.15}
    \nc\xxa{1.0}\nc\xxb{1.2}\nc\xxc{1.8}
    \nc\fxa{-5.0} \nc\fxb{5.0}
    \nc\fya{-0.8} \nc\fyb{2.6}
    \begin{pspicture}(\fxa,\fya)(\fxb,\fyb)
      %\psframe[linewidth=0.5pt,dimen=outer](\fxa,\fya)(\fxb,\fyb)
      \localscalebox{1.0}{1.0}
      {
        \small
        \psset{dimen=middle}
        \rput(0,0){%
          {
            \psset{linewidth=0.5pt}
            \settowidth\tlc{$C$}
            \psline(-\xxxab,0)(\xxxab,0)
            \psline(0,-\xxyac)(0,\xxyab)
            \uput[r](\xxxab,0){$v$}
            \uput[u](0,\xxyab){$f_{\text{level and ramp}}(v)$}
            \psline(-\xxl,\xxa\tlb)( \xxl,\xxa\tlb)
            \uput[ur](0,\xxa\tlb){$A$}
            \psline(\xxc\tlb, \xxl)(\xxc\tlb,-\xxl)
              \uput[d](\xxc\tlb,-\xxl){\makebox[\tlc][r]{$C$}}
          }%
          {
            \psset{linewidth=2.0pt}
            \psline%
               (-1.35\tla,\xxa\tlb)%
               (0,\xxa\tlb)(\xxc\tlb,0)%
               ( 1.35\tla,0)
          }%
        }%
      }
    \end{pspicture}
  \ec
\end{figure}
Its transform is evidently
\bq{lapl:123:55}
  f_{\text{level and ramp}} \fouripair
  \frac{A}{\sqrt{2\pi}}\left[
      \frac{1-e^{-iCv}}{Cv^2} + \pi\delta(v)
  \right].
\eq

\subsection{Construction}
\label{lapl:123.90}
\index{function!arbitrary}
\index{arbitrary function}
\index{approximation to an arbitrary function}
\index{construction of an arbitrary function}

\index{sample!nonuniform}
\index{nonuniform sample}
\index{tail!of a function}
\index{function!tail of}
Section~\ref{fouri:100.90} has employed regular nonanalytic pulses of
\S~\ref{fours:095} to approximate a more or less arbitrary function,
but the technique of \S~\ref{fouri:100.90} remains
limited in that it
\begin{itemize}
  \item cannot track a discontinuity or other infinite slope,
  \item requires uniform sampling, and
  % diagn: the next item has been revised and wants review.
  \item lacks a practical way to model a function's leftward or
    rightward tail when, as in Figs.~\ref{lapl:123:fig30},
    \ref{lapl:123:fig50} and~\ref{lapl:123:fig55}, such a tail
    exists, $\lim_{v\ra\pm\infty}f(v)$ being nonzero.
\end{itemize}
This section's constructive functions are less limited.  An example to
illustrate their use follows.

\index{observation}
\index{measurement!experimental}
\index{experimental measurement}
\index{function!experimentally measurable}
\index{function!sample of}
\index{sample}
% diagn: review the next sentence and, indeed, all the paragraph's
% prose.
Consider an imprecisely known function $f(t)$ that,
though imperfectly, were nevertheless experimentally measurable.
Suppose that the decimal samples of Table~\ref{lapl:123:tbl90} were
observed.
\begin{table}
  \caption{A sampled function (example).}
  \label{lapl:123:tbl90}
  \begin{center}
    \begin{tabular}{rrr}
      $t$ & $f(t)$ \\
      $ -50.0$ & $0.00$ \\
      $ -10.0$ & $0.15$ \\
      $  -2.0$ & $0.92$ \\
      $   0.0$ & $1.10$ \\
      $   0.5$ & $1.08$ \\
      $   4.0$ & $0.60$ \\
      $  50.0$ & $0.06$
    \end{tabular}
  \end{center}
\end{table}
One might then approximate the function as%
{%
  \nc\xxmp[1]{\ensuremath{#1}}
  \settowidth\tla{\scriptsize$\xxmp{   1.00}$}
  \settowidth\tlb{\scriptsize$\xxmp{ -50.0 }$}
  \settowidth\tlc{\scriptsize$\xxmp{ -10.0 }$}
  \settowidth\tld{\scriptsize$\xxmp{ -10.0 }$}
  \settowidth\tle{$       -10.0  $}
  \settowidth\tlf{$\mbox{}+\mbox{}$}
  \nc\xxa[1]{\ensuremath A= \makebox[\tla][r]{\scriptsize${#1}$},}
  \nc\xxb[2]{\ensuremath B=(\makebox[\tlc][r]{\scriptsize${#2}$})-(\makebox[\tlb][r]{\scriptsize${#1}$}),}
  \nc\xxc[2]{\ensuremath C=(\makebox[\tld][r]{\scriptsize${#2}$})-(\makebox[\tlc][r]{\scriptsize${#1}$})}
  \nc\xxbc[3]{\ensuremath\xxb{#1}{#2}\xxc{#2}{#3}}
  \nc\xxe[1]{\ensuremath [t-(\makebox[\tle][r]{${#1}$})]}
  \nc\xxfit[1]{\ensuremath f_{\text{irr triangular},{#1}}}
  \nc\xxfarg[4]{\ensuremath \xxfit{\xxa{#1}\xxbc{#2}{#3}{#4}}\xxe{#3}}
  \bqb
    \lefteqn{f(t) \approx} &&
      \\&\!\!\makebox[\tlf][l]{}&\!\!\!\!\!
        \xxfarg{0.15}{-50.0}{-10.0}{ -2.0}
      \\&\!\!+&\!\!\!\!\!
        \xxfarg{0.92}{-10.0}{ -2.0}{  0.0}
      \\&\!\!+&\!\!\!\!\!
        \xxfarg{1.10}{ -2.0}{  0.0}{  0.5}
      \\&\!\!+&\!\!\!\!\!
        \xxfarg{1.08}{  0.0}{  0.5}{  4.0}
      \\&\!\!+&\!\!\!\!\!
        \xxfarg{0.60}{  0.5}{  4.0}{ 50.0}
      \\&\!\!+&\!\!\!\!\!
        f_{\text{ramp and level},A=0.06,B=(50.0)-(4.0)}
        [t-(50.0)],
  \eqb%
}%
constructing the approximation from this section's various constructive
functions, each of which is Fourier transformable as the section has shown.

\index{interpolation}
\index{irregular triangular pulse!overlapping}
\index{triangular pulse!overlapping}
\index{overlap}
% diagn: review from here to the end of the subsection.
Note the manner in which the example's several triangular pulses
overlap.  Observe that the example's approximation exactly fits its
experimental samples, interpolating between along straight lines.
Such techniques are generally useful.

\index{pulse!raised-cosine, irregular}
\index{raised-cosine pulse!irregular}
\index{irregular raised cosine}
\index{irregular raised cosine!comparison against the irregular triangular pulse}
\index{irregular triangular pulse!comparison against the irregular\\raised cosine} % bad break
Variations and refinements
%\footnote{%
%  One comparatively obvious prospective refinement might be to use an
%  irregular raised cosine (see \S~\ref{fours:095} for the
%  \emph{regular} raised cosine) in place of the irregular triangle.
%}
are possible, like using an irregular raised cosine (extending
\S~\ref{fours:095}) in place of the irregular triangle, but this book
will pursue the matter no further.

% There is little wrong with the following section as far as the writer knows
% but it just seems like a bit much.  The book does not need it.  Keep
% it commented, though: don't delete.
%\subsection{The irregular raised cosine}
%\label{lapl:123.95}
%\index{Fourier transform!of an irregular raised cosine}
%\index{irregular raised cosine!Fourier transform of}
%\index{raised-cosine pulse!irregular}
%\index{raised-cosine pulse!irregular, Fourier transform of}
%\index{pulse!irregular raised-cosine}
%
%\index{area!under a pulse}
%\index{pulse!area under}
%One possible refinement to the irregular triangular pulse of
%Fig.~\ref{lapl:123:fig20} is to use instead the irregular raised cosine
%\settowidth\tla{$-B$}
%\settowidth\tlb{$\mbox{}=\mbox{}$}
%\bq{lapl:123:95}
%  \begin{split}
%    f_{\text{irr raised cosine}}(v) &=
%    \begin{cases}
%      [A]\left[1 + \cos (\pi v/B) \right]/2 &\mbox{if $-B \le t \le 0$,} \\
%      [A]\left[1 + \cos (\pi v/C) \right]/2 &\mbox{if $\makebox[\tla][r]{$0$} \le t \le C$,} \\
%      0 &\mbox{otherwise;}
%    \end{cases}
%    \\& \makebox[\tlb][l]{} B > 0,\ C > 0.
%  \end{split}
%\eq
%Its transform is by
%\bqb
%    \lefteqn{\frac{\left(\!\sqrt{2\pi}\right)F_{\text{irr raised cosine}}(v)}{A}} &&
%    \\&=&
%      \frac 1 2 \int_{-B}^0 e^{-iv\theta} \left[1+\cos\left(\frac{\pi\theta}{B}\right) \right] d\theta
%    \\&&\qquad\mbox{} +
%      \frac 1 2 \int_0^C e^{-iv\theta} \left[1+\cos\left(\frac{\pi\theta}{C}\right) \right] d\theta
%    \\&=&
%      -\left[\frac {e^{-iv\theta}}{i4v} \left(2
%      +\frac{ve^{i\pi\theta/B}}{v-\pi/B}
%      +\frac{ve^{-i\pi\theta/B}}{v+\pi/B} \right)\right]_{\theta=-B}^0
%    \\&&\qquad\mbox{} -
%      \left[\frac {e^{-iv\theta}}{i4v} \left(2
%      +\frac{ve^{i\pi\theta/C}}{v-\pi/C}
%      +\frac{ve^{-i\pi\theta/C}}{v+\pi/C} \right)\right]_{\theta=0}^C,
%\eqb
%implying that
%\bq{lapl:123:96}
%  \begin{split}
%  f_{\text{irr raised cosine}}(v)
%    &\fouripair
%      \frac {A}{i2\left(\!\sqrt{2\pi}\right)v} \bigg\{
%        \frac{1}{1-(\pi/Cv)^2}
%        -\frac{1}{1-(\pi/Bv)^2}
%    \\&\qquad\mbox{}
%      +e^{-iCv} \left[
%        \frac{1}{(Cv/\pi)^2-1}
%      \right]
%      -e^{iBv} \left[
%        \frac{1}{(Bv/\pi)^2-1}
%      \right]
%    \bigg\},
%    \\B&>0,\ C>0.
%  \end{split}
%\eq
%One could compute $F_{\text{irr raised cosine}}(0)$ by the method used
%to arrive at~(\ref{lapl:123:26}) but it is easier just to
%apply~(\ref{fouri:byu}).  Either way,
%\bq{lapl:123:97}
%  F_{\text{irr raised cosine}}(0) = \frac{(A)(B+C)}{2\sqrt{2\pi}},
%\eq
%same as~(\ref{lapl:123:26}) because, as~(\ref{fouri:byu}) at $v=0$
%observes, the areas under the pulses are the same.
%At $v=\pm \pi/B$ or $v=\pm \pi/C$, denominators of~(\ref{lapl:123:96})
%vanish, but in the limit,
%\[
%  \begin{split}
%    w^2 &\equiv \left(\frac{Cv}{\pi}\right)^2-1
%    \\
%    \left(\frac{\pi}{C}\right)^2\left(1+w^2\right) &= v^2
%    \\
%    \frac{1}{1-(Cv/\pi)^2}
%    &=
%    \frac{1}{1-(Cv/\pi)^2}
%    +
%    \frac{(Cv/\pi)^2-(Cv/\pi)^2}{1-(Cv/\pi)^2}
%    \\&=
%    \frac{1-(Cv/\pi)^2}{1-(Cv/\pi)^2}
%    +
%    \frac{(Cv/\pi)^2}{1-(Cv/\pi)^2}
%    \\&=
%    1+
%    \frac{1}{(\pi/Cv)^2-1}
%    \\
%    -\frac{1}{w^2}
%    &=
%    1+
%    \frac{1}{(\pi/Cv)^2-1}
%    \\
%    \frac{1}{1-(\pi/Cv)^2}
%    &=
%    1+
%    \frac{1}{w^2}
%    =
%    \frac{1+w^2}{w^2}
%    \\
%    e^{-iCv}
%    &=
%    e^{-i\pi\sqrt{1+w^2}}
%    = e^{-i\pi(1+\sqrt{1+w^2}-1)}
%    = -e^{-i\pi(\!\sqrt{1+w^2}-1)}
%    \\
%    G &\equiv
%      \frac{1}{1-(\pi/Cv)^2}
%      +e^{-iCv} \left[
%        \frac{1}{(Cv/\pi)^2-1}
%      \right]
%    \\&=
%      1+\frac 1{w^2}
%      +e^{-iCv} \left[
%        \frac{1}{w^2}
%      \right]
%    \\&=
%      1 + \frac{1+e^{-i\pi(\!\sqrt{1+w^2}-1)}}{w^2}
%    \\&=
%      1 + \frac{2-i\pi(\!\sqrt{1+w^2}-1)-\pi^2(\!\sqrt{1+w^2}-1)^2/2+\cdots}{w^2}
%    \\
%    \sqrt{1+w^2}-1 &=
%      \frac{w^2}{2} + \cdots
%    \\
%    G &= 1 +\frac{2}{w^2} - \frac{i\pi}{2}
%  \end{split}
%\]
%
%To analyze leftward and rightward raised cosines analogous to the
%leftward and rightward right triangles of Fig.~\ref{lapl:123:fig40} is
%left as an exercise to the interested reader.
%
%\index{irregular raised cosine!comparison against the irregular triangular pulse}
%\index{irregular triangular pulse!comparison against the irregular\\raised cosine} % bad break
%Just because the raised cosine is subtler, or at any rate fancier, than
%the triangle does not necessarily make it \emph{better.}
%A series of triangular pulses can exactly track a slanted straight
%line, for example, whereas a series of raised cosines cannot.  On the
%other hand, a raised cosine's transform decays as $1/v^3$ for
%large~$v$, which some applications might prefer to the triangle's
%slower $1/v^2$.  If you wish to know, when in doubt, this writer
%probably tries the triangle first, but which is finally preferred would
%depend on the application.

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

\section{the Laplace transform}
\label{lapl: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.

\index{\lapll}
\index{tail!omission of}
\index{tail!suppression of}
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) = \lapll\{ f(t) \} \equiv
    \int_{0^{-}}^\infty e^{-st} f(t) \,dt
\eq
offers such an alternative.  Here, $s=i\omega$ 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}
\index{Fourier transform!properties of}
\index{Fourier transform pair}
\index{transform pair}
\index{Laplace transform!of a convolution}
\index{Laplace transform!higher-order properties of}
\index{Laplace transform!ramping property of}
\index{ramping property}
Without resort to clever, indirect techniques,
Laplace's~(\ref{fouri:laplace}) can transform
\[
  1 \laplair \frac{1}{s},
\]
listed among other pairs by Table~\ref{fouri:laplace-pairs}.  Laplace
transform \emph{properties,} some of which are derived in the same easy
way, are listed in Table~\ref{fouri:laplace-properties}\@.  Further
Laplace properties, also listed in the table, want some technique to
derive, for instance the differentiation property, which comes by
\[
  \begin{split}
    \lapll\left\{ \frac{d}{dt} f(t) \right\}
      &= \int_{0^{-}}^\infty e^{-st} \frac{d}{dt} f(t) \,dt
      = \int_{t=0^{-}}^\infty e^{-st} \,d\left[f(t)\right]
      \\&=
        \left. e^{-st} f(t) \right|_{0^{-}}^\infty
        + s \int_{0^{-}}^\infty e^{-st} f(t)\,dt
      \\&= -f(0^{-}) + sF(s)
  \end{split}
\]
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
\[
  -\frac{d}{ds} F(s)
    = -\frac{d}{ds} \int_{0^{-}}^\infty e^{-st} f(t) \,dt
    = \int_{0^{-}}^\infty e^{-st} [tf(t)] \,dt
    = \lapll \{tf(t)\}.
\]
Higher-order properties come by repeated application.
The convolution property comes as it did in
\S~\ref{fouri:110.40}, beginning
\bqb
  \lefteqn{\lapll\left\{\left[u_1(t)h(t)\right] \ast \left[u_1(t)f(t)\right]\right\}}
    &&\\&=&
    \int_{-\infty}^{\infty}
      e^{-st}
    \int_{-\infty}^{\infty}
        u_1\left(\frac{t}{2}-\psi\right) h\left(\frac{t}{2}-\psi\right)
        u_1\left(\frac{t}{2}+\psi\right) f\left(\frac{t}{2}+\psi\right)
      d\psi
      \,dt,
\eqb
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{\lapll\left\{\left[u_1(t)h(t)\right] \ast \left[u_1(t)f(t)\right]\right\}}
  &&\\&=&
    \left[
      \int_{-\infty}^{\infty} e^{-s\mu} u_1(\mu)h(\mu) \,d\mu
    \right]
    \left[
      \int_{-\infty}^{\infty} e^{-s\phi} u_1(\phi)f(\phi) \,d\phi
    \right],
\eqb
after which we take advantage of the
presence of Heaviside's unit step $u_1(\cdot)$ of~(\ref{integ:670:11})
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}
  \index{Fourier transform!properties of}
  \bqb
    %\int_{t_o}^\infty e^{-st} f(t-t_o) \,dt &=& e^{-st_o} \int_{0}^\infty e^{-s\tau}f(\tau) \,d\tau \\
    u_1(t-t_o)f(t-t_o) &\laplair& e^{-st_o}F(s),\ t_o \ge 0 \\
    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_1(t)h(t)\right] \ast \left[u_1(t)f(t)\right] &\laplair& H(s) F(s)
  \eqb
\end{table}

\begin{table}
  \caption{Laplace transform pairs.}
  \label{fouri:laplace-pairs}
  \index{Fourier transform pair}
  \index{transform pair}
  \[
    \renewcommand\arraystretch{2.0}
    \br{rclcrcl}
      &&&&
      \ds e^{-at}t\sin\omega_o t &\laplair& \ds \frac{2(s+a)\omega_o}{[(s+a)^2+\omega_o^2]^2} \\
      \delta(t) &\laplair& 1 &&
      \ds e^{-at}t\cos\omega_o t &\laplair& \ds \frac{(s+a)^2-\omega_o^2}{[(s+a)^2+\omega_o^2]^2} \\
      \ds 1 &\laplair& \ds \frac{1}{s} &&
      \ds e^{-at} &\laplair& \ds \frac{1}{s+a} \\
      \ds t &\laplair& \ds \frac{1}{s^2} &&
      \ds e^{-at}t &\laplair& \ds \frac{1}{(s+a)^2} \\
      \ds t^n &\laplair& \ds \frac{n!}{s^{n+1}} &&
      \ds e^{-at}t^n &\laplair& \ds \frac{n!}{(s+a)^{n+1}} \\
      \ds \sin\omega_o t &\laplair& \ds \frac{\omega_o}{s^2+\omega_o^2} &&
      \ds e^{-at}\sin\omega_o t &\laplair& \ds \frac{\omega_o}{(s+a)^2+\omega_o^2} \\
      \ds \cos\omega_o t &\laplair& \ds \frac{s}{s^2+\omega_o^2} &&
      \ds e^{-at}\cos\omega_o t &\laplair& \ds \frac{s+a}{(s+a)^2+\omega_o^2}
    \er
  \]
\end{table}
\index{sine!Laplace transform of}
\index{cosine!Laplace transform of}
\index{sinusoid!Laplace transform of}
\index{exponential!Laplace transform of}
\index{complex exponential!Laplace transform of}
\index{Laplace transform of a sinusoid}
\index{Laplace transform of an exponential}
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 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{lapl:250}
\index{differential equation!solution of by the Laplace transform}
\index{Laplace transform!solving a differential equation by}

% diagn: this revised paragraph (but not its footnote) wants review.
\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.  One should probably regard the Fourier transform to be
the principal, more 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, $u_1(t)f(t) = (1/i2\pi)
  \int_{-i\infty}^{i\infty} e^{st} F(s)\,ds$, 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 however in its property of
Table~\ref{fouri:laplace-properties} that
\linebreak % bad break
$(d/dt)f(t) \laplair sF(s) - f(0^{-})$.  Fourier's corresponding
property of Table~\ref{fouri:110:tbl20} lacks the $f(0^{-})$, an
initial condition.

\index{verification!of a solution}
\index{solution!verification of}
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},\ \ t\ge 0;\\
  \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{lapl:260}
\index{initial-value theorem}
\index{final-value theorem}
\index{Laplace transform!initial and final values by}

The method of \S~\ref{lapl: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 uninterested in the details between.
The Laplace transform's \emph{initial-} and \emph{final-value theorems,}
\bq{lapl: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.

% diagn: this revised paragraph wants review.
One derives the initial-value theorem via the steps
\bqb
  \lefteqn{\lim_{s \ra \infty} sF(s) - f(0^{-})}
  &&\\&=&
    \lim_{s \ra \infty} \lapll \left\{\frac{d}{dt}f(t)\right\}
  \\&=&
    \lim_{s \ra \infty} \int_{0^{-}}^\infty e^{-st} \frac{d}{dt}f(t) \,dt
  \\&=&
    \lim_{\ep\ra 0^{+}} \left[\int_{0^{-}}^\infty e^{-st} \frac{d}{dt}f(t) \,dt\right]_{s=1/\ep^2}
  \\&=&
  \lim_{\ep\ra 0^{+}} \left[
    \int_{0^{-}}^\ep
    e^{-st} \frac{d}{dt}f(t) \,dt
    + \int_{\ep}^\infty
    e^{-st} \frac{d}{dt}f(t) \,dt
  \right]_{s=1/\ep^2}
  \\&=&
    \lim_{\ep\ra 0^{+}} \int_{0^{-}}^\ep \frac{d}{dt}f(t) \,dt
  =
    \int_{0^{-}}^{0^{+}} \frac{d}{dt}f(t) \,dt
  =
  f(0^{+}) - f(0^{-}),
\eqb
which invoke the time-differentiation property of
Table~\ref{fouri:laplace-properties} and the last of which implies
(\ref{lapl:260:10})'s first line.  For the final value, one begins
\bqb
  \lim_{s \ra 0} sF(s) - f(0^{-})
  &=&
  \lim_{s \ra 0} \lapll \left\{\frac{d}{dt}f(t)\right\}
  \\&=&
  \lim_{s \ra 0} \int_{0^{-}}^\infty e^{-st} \frac{d}{dt}f(t) \,dt
  \\&=&
  \int_{0^{-}}^\infty \frac{d}{dt}f(t) \,dt
  =
  \lim_{t\ra\infty} f(t) - f(0^{-}),
\eqb
and (\ref{lapl:260:10})'s second line follows immediately.%
\footnote{\cite[\S~7.5]{Phillips/Parr}}

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

\section{The spatial Fourier transform}
\label{lapl: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}
\index{physicist}
\index{engineer}
\index{frequency!spatial}
\index{spatial frequency}
\index{propagation vector}
\index{vector, propagation}
\index{triple integral}
\index{dimension}
In the study of wave mechanics,
% diagn: add an appropriate in-book reference?
physicists and engineers sometimes elaborate the Fourier transform'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 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.

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

% diagn: review this revised section and its tables.
\section{The Fourier transform in cyclic frequencies}
\label{lapl:360}
\index{cyclic frequency}
\index{cyclic frequency!Fourier transform in}
\index{frequency!cyclic}
\index{Fourier transform!in cyclic frequencies}

Many applications prefer\footnote{\cite{Couch}} a clever alternate
definition of the Fourier transform,\footnote{The~$\Phi$ of this
section is not the~$\Phi$ of \S~\ref{fouri:110}.  The two sections use
the Greek letter~$\Phi$ for different purposes.}
\bqa
    \Phi(\nu) &\equiv& \int_{-\infty}^{\infty}
        e^{-i2\pi\nu t}
            f(t)
        \,dt,
    \label{lapl:360.10}
    \\
    \nu &\equiv& \frac{\omega}{2\pi},
    \label{lapl:360.20}
\eqa
instead of~(\ref{fouri:byu2}).  In~(\ref{lapl:360.10})
and~(\ref{lapl:360.20}), cyclic frequency~$\nu$ appears rather
than angular frequency~$\omega$.
Comparing against~(\ref{fouri:byu2}),
\bq{lapl:360.30}
  \Phi(\nu) = \left(\!\sqrt{2\pi}\right)F(\omega),
\eq
such that, for example, referring to Table~\ref{fouri:120:tbl20}, if
$f(t) = \Lambda(t)$, then $\Phi(t) = \sinarg^2(\pi\nu)$---or, if you
prefer, letting this section use the symbol~$\rightarrow$ to announce the cyclic
transform~(\ref{lapl:360.10}), $f(t) \rightarrow \sinarg^2(\pi\nu)$.

\index{cyclic frequency!inverse Fourier transform in}
\index{cyclic frequency!convolution in}
\index{convolution!in cyclic frequencies}
Substituting~(\ref{lapl:360.20}) and~(\ref{lapl:360.30})
into~(\ref{fouri:byu2}), the inverse cyclic transform is
%\[
%  f(t) = \frac{1}{\sqrt{2\pi}}
%      \int_{-\infty}^{\infty}
%        e^{i\omega t}
%            \frac{\Phi(\nu)}{\sqrt{2\pi}}
%        \,d(2\pi\nu). \\
%\]
%That is,
\bq{lapl:360.40}
  f(t) =
      \int_{-\infty}^{\infty}
        e^{i2\pi\nu t}
            \Phi(\nu)
        \,d\nu,
\eq
an elegant result.  Even more elegant are the alternate forms of the
convolution properties of Table~\ref{fouri:110:tbl41}.
Defining $s(f) \equiv h(t) \ast f(t)$ and letting $\Sigma(\nu)$,
$\mathrm X(\nu)$ and $\Phi(\nu)$ respectively be the cyclic alternatives to
$S(\omega)$, $H(\omega)$ and $F(\omega)$, we have that
$\Sigma(\nu) = (\!\sqrt{2\pi})S(\omega)
= 2\pi H(\omega)F(\omega) = \mathrm X(\nu)\Phi(\nu)$.
Calculating in such a manner,
\bq{lapl:360.50}
  \begin{split}
      h(t) \ast f(t) &\rightarrow \mathrm X(\nu) \Phi(\nu), \\
      h(t) f(t) &\rightarrow \mathrm X(\nu) \ast \Phi(\nu),
  \end{split}
\eq
the latter because
\[
  \begin{split}
    h(t) f(t)
    &\fouripair \frac{1}{\sqrt{2\pi}}
      [H(\omega) \ast F(\omega)]
    \\&\fouripair \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty} H\left(\frac{\omega}{2}-\eta\right)
      F\left(\frac{\omega}{2}+\eta\right) d\eta
    \\&\fouripair \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty}
      \bigg[\frac{1}{\sqrt{2\pi}} \mathrm X\left(\frac{\nu}{2}-\nu'\right)\bigg]
      \bigg[\frac{1}{\sqrt{2\pi}} \Phi\left(\frac{\nu}{2}+\nu'\right)\bigg]
      2\pi\,d\nu'
    \\&\fouripair \frac{1}{\sqrt{2\pi}}
      \int_{-\infty}^{\infty}
      \mathrm X\left(\frac{\nu}{2}-\nu'\right)
      \Phi\left(\frac{\nu}{2}+\nu'\right)
      d\nu'
    \\&\fouripair \frac{1}{\sqrt{2\pi}}
      \mathrm [X(\nu) \ast \Phi(\nu)].
  \end{split}
\]
Thus we see that, so long as one does not mind forgoing the use of
angular frequencies, the cyclic alternative brings some practical
advantages.%

% diagn: check this paragraph and its tables. They are new.
Table~\ref{lapl:360:tbl10} lists some notable Fourier properties in the cyclic
style.
\begin{table}
  \caption{Properties of the cyclic Fourier transform.}
  \label{lapl:360:tbl10}
  \settowidth\tla{$\ds\int_{-\infty}^{\infty} H^{*}(v) F(v) \,dv$}
  \bqb
    \Phi(\nu) &=& \left(\!\sqrt{2\pi}\right)F(\omega) \\
    f(v-a) &\ra& e^{-ia2\pi v}\Phi(v) \\
    e^{iav}f(v) &\ra& \Phi\left(v-\frac{a}{2\pi}\right) \\
    \left(\!\sqrt{|\alpha|}\right)Af(\alpha v) &\ra&
      \frac{A}{\sqrt{\left|\alpha\right|}}\Phi\left(\frac{v}{\alpha}\right) \\
    Af[(\alpha)(v-a)] &\ra&
      \frac{Ae^{-ia2\pi v}}{\left|\alpha\right|}\Phi\left(\frac{v}{\alpha}\right) \\
    &&\quad \Im(\alpha)=0,\ \Re(\alpha)\neq 0\\
    \frac{d^n}{dv^n}f(v) &\ra& (i2\pi v)^n\Phi(v) \\
    (-iv)^nf(v) &\ra& \frac{d^n}{(2\pi)^n\,dv^n}\Phi(v) \\
    &&\quad n\in\mathbb Z,\ n\ge 0\\
    \int_{-\infty}^v f(\tau) \,d\tau &\ra&
      \frac{\Phi(v)}{i2\pi v} + \frac{\Phi(0)\delta(v)}{2} \\
    \int_{-\infty}^{\infty} h^{*}(v) f(v) \,dv &=&
      \int_{-\infty}^{\infty} \mathrm{X}^{*}(v) \Phi(v) \,dv
      % The next qualification should be kept, but probably commented out.
      %\ \ \mbox{if $\Im(v)=0$}
      \\
    \int_{-\infty}^{\infty} \left|f(v)\right|^2 \,dv &=&
      \makebox[\tla][l]{$\ds\int_{-\infty}^{\infty} \left|\Phi(v)\right|^2 \,dv$}
      %\ \ \mbox{if $\Im(v)=0$}
      \\
    a_j &=& \Delta\nu\, \Phi(j\,\Delta\nu) \\
    %
    %
    %
      h(t) \ast f(t)
    &\ra&
      \mathrm{X}(\nu) \Phi(\nu)
    \\
      h(t) f(t)
    &\ra&
      \mathrm{X}(\nu) \ast \Phi(\nu)
    \\
      R_{fh}(t)
    &\ra&
      {\mathrm{X}}^{*}(\nu) \Phi(\nu)
    \\
      h^{*}(t) f(t)
    &\ra&
      R_{\Phi\mathrm{X}}(\nu)
    \\
      R_{f\!f}(t)
    &\ra&
      \left|\Phi(\nu)\right|^{2}
  \eqb
\end{table}
Table~\ref{lapl:360:tbl20} lists cyclic Fourier transform pairs.
\begin{table}
  \caption{Cyclic Fourier transform pairs.}
  \label{lapl:360:tbl20}
  \index{Fourier transform pair}
  \index{transform pair}
  \setlength\tabcolsep{0.3333333333333333\arraycolsep}
  \renewcommand\arraystretch{2.00}
  \nc\psirv{\ensuremath{\ds%
    \frac{\cos(r\pi v)\sinarg(\pi v)}{\left[1-(2rv)^2\right]}}}
  \nc\fourpsirv{\ensuremath{\ds \fourf\{\Psi_r(v)\}_{rv=\pm\pi}}}
  \nc\expra{\ensuremath{\ds \frac{n!}{(a+i2\pi v)^{n+1}},}}
  \settowidth\tla{$\ds\psirv$}
  \settowidth\tlb{$\ds\fourpsirv$}
  \settowidth\tlc{$\Psi_r$}
  \settowidth\tld{$\ds\fouripair$}
  \settowidth\tle{$\ds\expra$}
  \settowidth\tlg{$\ds\frac{1}{i2}$}
  \nc\wideequal{\ensuremath{\ds\makebox[\tld][c]{$\ds =$}}}
  \begin{center}
    \begin{tabular}{rclcrcl}
      $1$ &$\ra$& $\ds \delta(v)$
      &\mbox{\ \ }&
      $\delta(v)$ &$\ra$& $1$ \\
      $\Lambda(v)$ &$\ra$& $\ds\sinarg^2(\pi v)$ &&
      $u(v)$ &$\ra$& $\ds\frac{1}{i2\pi v}
        + \frac{\delta(v)}{2}$ \\
      $\Lambda_r(v)$ &$\ra$& $\ds\sinarg(\pi rv)\sinarg(\pi v)$ &&
      $\Pi(v)$ &$\ra$& $\ds\sinarg \pi v$ \\
      $\Psi(v)$ &$\ra$& $\ds\frac{\sinarg 2\pi v}{\left[1-(2v)^2\right]}$ &&
      $\ds\fourf\{\Psi(v)\}_{v=\pm 1/2}$ &$\wideequal$& $\ds\frac{1}{2}$ \\
      $\Psi_r(v)$ &$\ra$& $\ds\psirv$ &&
      $\ds\fourf\{\Psi_r(v)\}_{rv=\pm 1/2}$ &$\wideequal$&
        $\ds\frac{2\pi\sinarg(\pi v)}{8}$ \\
      &&&&
      &$\wideequal$& $\ds\frac{2\pi\sinarg(2\pi/4r)}{8}$ \\
      $\sinarg^2(v)$ &$\ra$& $\ds\pi\Lambda(\pi v)$ &&
      $\sinarg(v)$ &$\ra$& $\ds\pi\Pi(\pi v)$
    \end{tabular}
  \end{center}
  \bqb
    u(v)e^{-av} &\ra& \makebox[\tle][l]{$\ds\frac{1}{a+i2\pi v},$}
      \ \ \Re(a) > 0 \\
    u(v)e^{-av}v^n &\ra& \makebox[\tle][l]{$\expra$}\ \
      \Re(a) > 0,\ n \in \mathbb Z, \ n \ge 0 \\
    e^{iav} &\ra& \makebox[\tle][l]{$\ds \delta\left(v-\frac{a}{2\pi}\right),$}
      \ \ \Im(a) = 0\\
    \sin av &\ra& \frac{1}{i2}\left[ \delta\left(v-\frac{a}{2\pi}\right) - \delta\left(v+\frac{a}{2\pi}\right) \right],
      \ \,\Im(a) = 0\\
    \cos av &\ra& \makebox[\tlg][c]{$\ds\frac{1}{2}$}\left[ \delta\left(v-\frac{a}{2\pi}\right) + \delta\left(v+\frac{a}{2\pi}\right) \right],
      \ \,\Im(a) = 0\\
    \makebox[0em][r]{$\sqrt{T_1}$}\sum_{j=-\infty}^{\infty} \delta(v-jT_1) &\ra&
        \sqrt{\Delta\nu} \sum_{j=-\infty}^{\infty} \delta(v-j\,\Delta\nu), \ \ \Delta\nu\,T_1 = 1 \\
     \Omega\left[\left(\!\sqrt{2\pi}\right) v\right] &\ra& \Omega\left[\left(\!\sqrt{2\pi}\right) v\right]
  \eqb
\end{table}
Among the properties of Table~\ref{lapl:360:tbl10}, the integration property is
derived as in \S~\ref{fouri:125} using the cyclic transform of $u(v)$, listed
in Table~\ref{lapl:360:tbl20}; and the properties involving $\delta(v)$ appeal
to~(\ref{integ:670:23}).

The book you are reading lacks a strong preference between the angular
transform of the rest of the chapter and this section's cyclic
alternative.  The book has nonetheless chosen to standardize on the
angular transform because
angular frequencies better comport with expressions like $\sin \omega
t$ that appear in chapters such as~\ref{cexp} and~\ref{vcalc},
yet the way the cyclic alternative dispenses with so many stray factors
of $\sqrt{2\pi}$ is indeed attractive.
% diagn: review the next sentence, which is new; Thad does not think it too
% colloquial, but still it bears a further reading for review again
On the other hand, $\Omega(t) \ra (\!\sqrt{2\pi})\Omega(\nu)$, whereas
$\Omega(t) \fouripair \Omega(\omega)$, so the angular transform wins
\emph{that} contest.  Applied mathematicians can put both options to good use.%
\footnote{%
  \index{engineer}%
  \index{symmetry!of the Fourier transform}%
  \index{Fourier transform!symmetry of}%
  In some countries including the writer's, engineers have tended to
  use a third definition of the transform,
  $F_{\text{engineer}}(\omega) = (\!\sqrt{2\pi})F(\omega)$.
  Working with engineers (the writer is one), the writer has had
  occasion to employ the third definition many times. Engineers are
  fine people but the writer cannot especially recommend their third
  definition, which obscures the natural, mutual
  symmetry of the transform and its inverse.%
}

