\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\pl}{\partial}
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\begin{document}

{\bf Question}

\begin{description}
\item[(a)]
If the Laplace transform of a suitable function $f(t)$ is defined
to be

$$\bar{f}(p)=\ds\int_0^{\infty}dt\ f(t)\exp(-pt),\ Re\ p>0$$

state \emph{carefully} the inverse transform in terms of a contour
integral.

\item[(b)]
\begin{description}
\item[(i)]
Using this integral representation, defining $-\pi<\arg\ p \leq
\pi$, and suitably closing the contour, show that the inverse
Laplace transform of $\bar{f}_1(p)=\log p$ is
$f_1(t)=-\ds\frac{1}{t}$. (You may assume without proof that all
integrals over infinite or vanishing arcs give a zero
contribution.)

\item[(ii)]
Hence show that the inverse Laplace transform of
$\bar{f}_2(p)=\log(p+1)$ is $f_2(t)=-\ds\frac{e^{-t}}{t}$.

\item[(iii)]
Using parts (i) and (ii), show that the inverse Laplace transform
of $\bar{f}_3=\log\left[\ds\frac{(p+1)}{p}\right]$ is
$f_3(t)=\ds\frac{(1-e^{-t})}{t}$.
\end{description}

\item[(c)]
Show that the following equation

$$t\ds\frac{d^2f}{dt^2}+2\ds\frac{df}{dt}+tf=1-2e^{-t}$$

$$f(0)=1,\ f'(0)=-\ds\frac{1}{2}.$$

can be Laplace-transformed to give

$$\ds\frac{d\bar{f}}{dp}=-\ds\frac{1}{p(p+1)}.$$

Consequently, using the results of part b, solve for $f(t)$.
\end{description}

\medskip

\newpage
{\bf Answer}

\begin{description}
\item[(a)]
$\hat{f}(p)=\ds\int_0^{\infty} dt\ f(t)e^{-pt},\ Re(p)>0$

$\Rightarrow f(t)=\ds\frac{1}{2\pi i}\ds\int_C dp\ e^{pt}
\hat{f}(p),\ t>0$

\un{Bromwich Contour}: C:

(p)

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\begin{picture}(2,5)
\put(2,0){\vector(0,1){3}}

\put(2,3){\line(0,1){2}}

\put(2,5){\makebox(0,0)[b]{$c+i\infty$}}

\put(2,0){\makebox(0,0)[t]{$c-i\infty$}}

\end{picture}

All singularities to left of $c$.

\item[(b)]
\begin{description}
\item[(i)]
Want $f_1(t)=\ds\frac{1}{2\pi i}\ds\int_{c-i\infty}^{c+i\infty}dp\
\log p e^{pt}$

$p$ has branch point at $p=0$, define branch on $-\pi<\arg\ p\leq
\pi$.

Complete to \un{left}.

PICTURE \vspace{1in}

Close $p$ to left and form keyhole contour.

Integrals (2),\ (4) and (6) vanish by hint.

\newpage
Therefore

\begin{eqnarray*} (1) & = & \ds\frac{1}{2\pi i}\ds\int_{\infty}^0
dx\ e^{i\pi}\log(xe^{i\pi}e^{-xt}\\ & & (3): p=xe^{i\pi}\\ & &
+\ds\frac{1}{2\pi i}\ds\int_0^{\infty}dx\
e^{i\pi}\log(xe^{-i\pi})e^{-xt}\\ & & (5): p=xe^{-i\pi}\\ & = &
+\ds\frac{1}{2\pi i}\ds\int_0^{\infty}dx (\log x+i\pi)e^{-xt}\\ &
& -\ds\frac{1}{2\pi i}\ds\int_0^{\infty}dx (\log x-i\pi)e^{-xt}\\
& = & \ds\frac{2\pi i}{2\pi i}\ds\int_0^{\infty}dx\ e^{-xt}\\ & =
& -\ds\frac{1}{t} \end{eqnarray*}

Therefore \un{$\hat{f}_1(p)=\log p \Leftrightarrow
f_1(t)=-\ds\frac{1}{t}$}.

\item[(ii)]
Use Laplace's transform property that

$$L\left[e^{at}f(t)\right]=\hat{f}(p)$$

$\Rightarrow L\left[e^{-t}f(t)\right]=\hat{f}(p+1)$

\begin{eqnarray*} \rm{Therefore\ if}\ \hat{f}(p+1) = \hat{f}_2(p) & = &
\log(p+1)\\ \Rightarrow \hat{f}_2(p) & = & \hat{f}_1(p+1)\\
\Rightarrow f_2(t) & = & e^{-t}f_1(t)\\ \Rightarrow
\un{f_2(t)=-\ds\frac{e^{-t}}{t}} \end{eqnarray*}

\newpage
\item[(iii)]
Laplace transforms are linear, so

$$L[f_1(t)+f_2(t)]=L[f_2(t)]$$

\begin{eqnarray*} \Leftrightarrow L\left[\ds\frac{1-e^{-t}}{t}\right] & =
&
L\left[\ds\frac{1}{t}\right]+L\left[-\ds\frac{e^{-t}}{t}\right]\\
& = &
-L\left[\ds\frac{1}{t}\right]+L\left[-\ds\frac{e^{-t}}{t}\right]\\
& = & \log(p+1)-\log\ p\\ & = & \log\left(\ds\frac{p+1}{p}\right)
\end{eqnarray*}

$\Rightarrow
\un{L^{-1}\left[\log\left(\ds\frac{p+1}{p}\right)\right]=\ds\frac{1-e^{-t}}{t}}$
\end{description}

\item[(c)]
\begin{eqnarray*} L[tf''] & = & -\pl pL[f'']\\ & = &
-\ds\frac{pl}{\pl
p}[p^2\bar{f}(p)-p\undb{f(0_+)}+\undb{f'(0_+)}]\\ & & \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ =+1\ \ \ \ \ \ =-\ds\frac{1}{2}\\ & =
& -\ds\frac{\pl}{\pl p}\left[p^2\bar{f}-p-\ds\frac{1}{2}\right]\\
& = & -2p\bar{f}-p^2\bar{f}'+1\\ L[2f'] & = &
2[p\bar{f}-\undb{f(0_+)}]\\ & & \ \ \ \ \ \ \ \ \ \ \ =+1\\ & = &
2p\bar{f}-2\\ L[tf] & = & -\pl_p \bar{f}(p)=-\bar{f}'(p)\\ L[1] &
= & \ds\int_0^{\infty} dt\ e^{-pt}=\ds\frac{1}{p}\\ L[-2e^{-t}] &
= & -2\ds\int_0^{\infty} dt\ e^{-(p+1)t} = \ds\frac{-2}{(p+1)}
\end{eqnarray*}


Therefore $L$[equation] is

$-2p\bar{f}-p^2\bar{f}'+1+2p\bar{f}-2-f'=\ds\frac{1}{p}-\ds\frac{2}{p+1}$

Therefore

\begin{eqnarray*} -(p^2+1)\bar{f}' & = & 1+\ds\frac{1}{p}-\ds\frac{2}{p+1}\\
& = & \ds\frac{p(p+1)+(p+1)-2p}{p(p+1)}\\ & = &
\ds\frac{p^2+p+p+1-2p}{p(p+1)}\\ \rm{Therefore}\ -(p^2+1)\bar{f}'
& = & \ds\frac{(p^2+1)}{p(p+1)}\\ \rm{Therefore}\ \bar{f}' & = &
\un{-\ds\frac{1}{p(p+1)}}\ \rm{as\ required} \end{eqnarray*}

\begin{eqnarray*} \bar{f}' & = &
-\ds\frac{1}{p}+\ds\frac{1}{p+1}\\ \Rightarrow \bar{f} & = &
-\log\ p+\log(p+1)+\log\ c\ \ c=const\\ \bar{f} & = &
\log\left\{\left[\ds\frac{(p+1)}{p}\right]c\right\}\\ \Rightarrow
f(t) & = & \un{D\ds\frac{(1-e^{-t})}{t}} \end{eqnarray*}

By previous bits of question and standard results. $D=const$

But if $f(0)=1 \Rightarrow D=1$ since

$$f(t)=\left(\ds\frac{1-e^{-t}}{t}\right)$$

and $\lim_{t \to 0}f(t)=\lim_{t \to 0}+\ds\frac{e^{-t}}{1}=1$ by
L'Hopital

Hence

$$\un{f(t)=\ds\frac{1-e^{-t}}{t}}$$
\end{description}
\end{document}