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{\bf Question}

Use the Bromwich inversion integral to find the following:

\begin{description}
\item[(i)]
$L^{-1}\left[b\exp\ds\frac{(-ap)}{(p^2+b^2)}\right],\ a&gt;0,\ t \ne
a$.

What value do you get when $t=a$?

\item[(ii)]
$L^{-1}[p^n],\ n=$positive integer.

\item[(iii)]
$L^{-1}[p^{\frac{7}{2}}]$

\item[(iv)]
$L^{-1}[p^{-\alpha}],\ a&gt;0.$

NB Consider al types of $\alpha$.

\item[(v)]
$L^{-1}[\exp(-p)\cosh(p)]$

\item[(vi)]
$L^{-1}\left[\ds\frac{\sqrt{p}}{(1-p^{\frac{3}{2}})}\right]$

(The following results should be useful

Gamma function:
$L[t^z]=\ds\frac{z!}{p^{(z+1)}}=\ds\frac{\Gamma(z+1)}{p^{(z+1)}},\
\Gamma(-n) \to \infty,\ n=0$ or positive integer

Delta function: $L[\delta(t-a)]=\exp(-pa),\ L[\delta(t)]=1$)
\end{description}


\medskip

{\bf Answer}

\begin{description}
\item[(i)]
$f(t)=\ds\frac{1}{2\pi i} \ds\int dp\
\ds\frac{de^{-ap}e^{pt}}{p^2+b^2}$

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\vspace{2in}

Simple poles at $\pm bi$.

First take $t&gt;a$. Then you have $e^{p(t-a)}$ as a factor and you
need to use a left hand semicircle (where $e^{p(t-a)}$ is exp.
small) to complete the contour.

This now includes 2 poles with residues:

$$\left\{\begin{array} {rrcl} \rm{at}\ p=ib &amp;
\ds\frac{be^{ib(t-a)}}{2ib} &amp; = &amp; \ds\frac{e^{ib(t-a)}}{2i}\\ &amp; &amp;
\\ \rm{at}\ p=-ib &amp; \ds\frac{be^{-ib(t-a)}}{-2ib} &amp; = &amp;
\ds\frac{-e^{-ib(t-a)}}{2i} \end{array} \right.$$

Hence $f(t)=\sin b(t-a)$ by residue calculus.

Now take $t&gt;a$. Must complete with right hand semicircle. No poles
included now, to $f(t)=0$.

If $t=a$ need to evaluate $J=\ds\int\ds\frac{dp}{p^2+b^2}=f(a)$

Write it as:

\begin{eqnarray*} J &amp; = &amp; \ds\frac{1}{2ib}
\ds\int\left\{\ds\frac{1}{p-ib}-\ds\frac{1}{p+ib}\right\}dp\\ &amp; =
&amp;
\ds\frac{1}{2ib}[\log(p-ib)-\log(p+ib)]_{c-i\infty}^{c+i\infty}\\
&amp; = &amp; \ds\frac{1}{2ib}\left[\log\left|\ds\frac{p-ib}{p+ib}\right|
+i\arg(p-ib)-i\arg(p+ib)\right]_{c-i\infty}^{c+i\infty}\\ &amp; = &amp; 0
\end{eqnarray*}

(since $\arg(p-ib)$ with $p=c+i\infty$ is $\ds\frac{\pi}{2}$ etc
$\cdots$)

so finally:
$L^{-1}\left[\ds\frac{be^{-ap}}{(p^2+b^2)}\right]=u(t-a)\sin
b(t-a)$

\item[(ii)]
Two methods:
\begin{description}
\item[(a)]
Use Gamma result that $L[t^z]=\ds\frac{\Gamma(z+1)}{p^{(z+1)}}$

Then $\ds\frac{t^z}{\Gamma(z+1)}=L^{-1}\left[p^{-(z+1)}\right]$ so
putting $z+1=-n$ we have

$L^{-1}[p^n]=\ds\frac{t^{-n-1}}{\Gamma(-n)}=\ds\frac{t^{-n-1}}{\infty}=0$

\item[(b)]
Use contour:

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

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\vspace{2in}

But no poles \un{anywhere} so if you complete to the \un{left} the
closed contour encloses no poles. The integral therefore $=0$.
However we must check the semicircle contribution vanishes:

$$0=\ds\int+\ds\int \Rightarrow \ds\int=-\ds\int$$

\hspace{1in} PICTURE\ \ \ \
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\end{picture}\ \ \ PICTURE

This does in the left hand part of the plane, since $p^n e^{pt}
\rightarrow 0$ as $p=Re^{i\theta}\ \ R \to \infty\ \ \theta \in
\left(\frac{\pi}{2},\ \frac{3\pi}{2}\right)$ i.e., $Re(pt)&lt;0$ so
$f(t)=0$

This is a \lq\lq non-sensical" result, since obviously $L[0]=0$
and not $p^n$!!!Rather it's a $2 \to 1$ mapping. But if we \un{do}
see $p^n$ where $n$ is a positive integer in a Laplace transform,
we know it vanishes in the inverse transform.

\newpage
\item[(iii)]
$f(t)=\ds\frac{1}{2\pi i}\ds\int dp\ p^{+\frac{7}{2}} e^{pt}$

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Here we can use the Gamma function result:

$L[t^z]=\ds\frac{\Gamma(z+1)}{p^{(z+1)}}$ with $z+1
=-\ds\frac{7}{2} \Rightarrow z=-\ds\frac{9}{2}$

Thus $L^{-1}[p^{\frac{7}{2}}]=
\ds\frac{t^{-\frac{9}{2}}}{\Gamma(-\frac{9}{2}+1)}
=\ds\frac{t^{-\frac{9}{2}}}{\Gamma(-\frac{7}{2})}$

What's $\Gamma(-\frac{7}{2})$? I's \un{not} zero ($\Gamma$ is only
zero when $\Gamma(-n)\ n$ integer $&gt;0$)

Well $\Gamma(z+1)=z\Gamma(z)$ (know this!)

(cf. factorial: $n!=n(n-1)!$)

Thus $\left.\begin{array}{rcl}
\left(-\ds\frac{7}{2}\right)\Gamma\left(-\ds\frac{7}{2}\right) &amp; =
&amp; \Gamma\left(-\ds\frac{5}{2}\right)\\ \\
\left(-\ds\frac{5}{2}\right)\Gamma\left(-\ds\frac{5}{2}\right) &amp; =
&amp;  \Gamma\left(-\ds\frac{3}{2}\right)\\ \\
\left(-\ds\frac{3}{2}\right)\Gamma\left(-\ds\frac{3}{2}\right)
 &amp; = &amp; \Gamma\left(-\ds\frac{1}{2}\right)\\ \\
\left(-\ds\frac{1}{2}\right)\Gamma\left(-\ds\frac{1}{2}\right) &amp; =
&amp; \Gamma\left(\ \ \ds\frac{1}{2}\right) \end{array} \right\}$


$\Rightarrow \Gamma\left(-\ds\frac{7}{2}\right) =
\left(-\ds\frac{2}{7}\right)\left(-\ds\frac{2}{5}\right)
\left(-\ds\frac{2}{3}\right)\left(-\ds\frac{2}{1}\right)
\Gamma\left(\ds\frac{1}{2}\right) = \ds\frac{16}{105} \times
\sqrt{\pi}$

\un{Spin off result}:

Contour integral method shows that by completing to the right,

$$\ds\frac{t^-\frac{9}{2}}{\Gamma\left(-\ds\frac{7}{2}\right)}=\ds\frac{1}{2\pi
i}\ds\int_{PICTURE} dp\ p^{\frac{7}{2}}e^{pt}$$

PICTURE \vspace{2in}

Hence a contour integral representation for
$\ds\frac{1}{\Gamma(-z)}$ is

$$\ds\frac{1}{\Gamma(-z)}=\ds\frac{1}{2\pi i}\ds\int_{PICTURE} dp\
p^z e^p$$

\item[(iv)]

$L^{-1}[p^{-\alpha}]\ \alpha&gt;0$

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

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If $\alpha$ were an integer $\Rightarrow$ pole of order $n$ at
$p=0$, complete by a left semicircle where $Re(pt)&lt;0$. The residue
gives $f(t)=\ds\frac{t^{n-1}}{(n-1)!}$ (from standard residue of
higher order pole formula)

If $0&lt;\alpha&lt;1$ move the integral to the contour

PICTURE \vspace{2in}

The small circle around zero gives zero as it shrinks leaving

\begin{eqnarray*} f(t) &amp; = &amp; \frac{1}{2\pi i}\left[\int_{\infty
e^{-i\pi}}^0+\ds\int_0^{\infty
e^{+i\pi}}\right]\frac{e^{pt}}{p^{\alpha}}dp\\ &amp; = &amp; \frac{1}{2\pi
i}\int_0^{\infty}\ds\frac{e^{-xt}}{x^{\alpha}}dx
\left[e^{+i\alpha\pi}-e^{-i\alpha\pi}\right]\\ &amp; = &amp;
\ds\frac{1}{2\pi} \alpha! 2i\sin \alpha \pi\\ &amp; = &amp;
\ds\frac{\alpha! \sin\alpha \pi}{\pi}\\ &amp; = &amp;
\ds\frac{\Gamma(\alpha+1)\sin \alpha \pi}{\pi} \end{eqnarray*}
Result.

\un{Spin off}: Thus we deduce from (iii) that the identity

$\ds\frac{1}{\Gamma(-z)}=\Gamma(z+1)\ds\frac{\sin z\pi}{\pi}
\Rightarrow \un{\Gamma(z)\Gamma(1-z)=\ds\frac{\pi}{\sin \pi z}}$

The \lq\lq reflection formula" for the $\Gamma$-reduction.

\item[(v)]
$L^{-1}\left[e^{-p}\cosh p\right]$

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

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$\ \ =\ds\frac{1}{2\pi i}\ds\int
e^{p(t-1)}\left(\ds\frac{e^p+e^{-p}}{2}\right)=\ds\frac{1}{4\pi
i}\ds\int e^{pt} dp+\ds\frac{1}{4\pi i}\ds\int e^{p(t-2)}$

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Now use $\delta$-function result:

$$L[\delta(t-a)]=e^{-pa}$$

$\Rightarrow$

$L^{-1}\left[e^{-p}\cosh
p\right]=\ds\frac{1}{2}[\delta(t=0)]+\ds\frac{1}{2}[\delta(t-2)]
=\ds\frac{1}{2}\delta(t)+\ds\frac{1}{2}\delta(t-2)$

This result follows from the contour integral method by deforming
contours to complete where $Re(pt)$ and $Re(p(t-2))&lt;0$
respectively. The integrals enclose \un{no singularities} for $t
\ne 0,\ t \ne 2$ and the semicircle contributions vanish.

Thus the $\ds\int e^{pt} dp$ and $\ds\int e^{p(t-2)} dp$ both $=0$

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unless $t=0$ or $t=2$.

When $t=0$ or $t=2$ they \un{don't} converge [$\int dp$ is
infinite]

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so we have $\delta$-function spikes there.
\end{description}
\end{description}
\end{document}
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