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\begin{document}

{\bf Question}

For example 3 in the lectures, show that by collapsing the
integration contour of the Bromwich inversion integral onto the
cut, $f(t)$ can be represented as

$$f(t)=\ds\frac{A}{\pi}\ds\int_{-1}^{1}dy\ds\frac{\cos(ty)}{\sqrt{1-y^2}}.$$


\medskip

{\bf Answer}

We have from lectures:

$$f(t)=\ds\frac{A}{2\pi i}\ds\int
\ds\frac{e^{pt}}{(p^2+1)^{\frac{1}{2}}}dp$$

\hspace{2.35in} \setlength{\unitlength}{.1in}\begin{picture}(0,2)

\put(0,0){\vector(0,1){1}}

\put(0,1){\line(0,1){1}}

\end{picture}

where the contour is given by:

PICTURE \vspace{2in}

Now for $t>0$ we can make a closed contour with a left hand
semicircle, the semicircle contributing 0 in the limit as its
radius $\to \infty$. This closed curve can thus be deformed as
shown.

PICTURE \vspace{2in}

\newpage
(Since PICTURE contributes zero the integral
\setlength{\unitlength}{.1in}\begin{picture}(0,2)

\put(0,0){\vector(0,1){1}}

\put(0,1){\line(0,1){1}}

\end{picture} is unchanged by completing the contour. Hence any
deformation of the closed contour$=\ds\int$).

\hspace{4.72in}
\setlength{\unitlength}{.075in}\begin{picture}(0,2)

\put(0,0){\vector(0,1){1}}

\put(0,1){\line(0,1){1}}

\end{picture}

This leaves four pieces of curve to consider:

\begin{description}
\item[(i)]
Circle of small radius near $p=+i$. Put $p=i+\zeta=i+\epsilon
e^{i\theta}$ to show that as $\epsilon \to 0$ this gives zero
contribution.

\item[(ii)]
Circle near $p=-i$. Do similar analysis ($p=i-\zeta=i-\epsilon
e^{i\theta}$) to show it gives zero contribution.

\item[(iii)]
The line to the right of the cut. As $\epsilon \to 0$ with $p=iy$
we get

$$i\ds\int_{-1}^{+1}\ds\frac{e^{ity}}{\sqrt{1-y^2}} dy$$

Why $iy$? Cut is along imaginary axis. Therefore expect
discontinuity in $\arg(p)$ when $g$ is totally imaginary. So:
$\rightarrow$

\item[(iv)]
On left hand side put $p=-iy$ to get

$$-i\ds\int_{+1}^{-1}\ds\frac{e^{-ity}}{\sqrt{1-y^2}} dy$$

So total is:

\begin{eqnarray*} f(t) & = & \ds\frac{A}{2\pi i}
\left\{i\ds\int_{-1}^{+1}\ds\frac{e^{ity}}{\sqrt{1-y^2}}
dy-i\ds\int_{+1}^{-1}\ds\frac{e^{-ity}}{\sqrt{1-y^2}} dy\right\}\\
& = &
\ds\frac{A}{2\pi}\ds\int_{-1}^{+1}\ds\frac{dy}{\sqrt{1-y^2}}\left[e^{ity}+e^{-ity}\right]\\
& = &
\ds\frac{A}{\pi}\ds\int_{-1}^{+1}\ds\frac{dy}{\sqrt{-y^2}}\cos ty
\end{eqnarray*}
\end{description}

\end{document}