\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}

{\bf Question}

Solve the following integral equations for $\phi(x)$ by Laplace
transformations.

\begin{description}
\item[(i)]
$f(x)=\ds\int_0^x du\ (x-u)^{-\frac{1}{2}}\phi(u)$

\item[(ii)]
$\phi(t)=\alpha t+\ds\int_0^1 du\ \sin(t-u)\phi(u)$

\end{description}

\medskip

{\bf Answer}

Solve integral equations by Laplace transforming:

\begin{description}
\item[(i)]
\begin{eqnarray*} L[f] & = & L\left[\undb{\ds\int_0^x du\
(u-x)^{-\frac{1}{2}}\phi(u)} \right]\\ & & \hspace{.65in}
convolution\\ & & \hspace{.95in} \Downarrow\\ L[f(x)] & = &
L[x^{-\frac{1}{2}} \times L[\phi(x)]\\ \bar{f}(p) & = &
\ds\frac{(-\frac{1}{2})!}{p^{\frac{1}{2}}} \times \bar\phi(p)\
\rm{from\ standard\ result}\end{eqnarray*}

$\ds\frac{(-\frac{1}{2})!}{p^{\frac{1}{2}}}=\ds\int_0^{\infty}dx\
e^{-px}e^{-\frac{1}{2}}$

Thus
$\bar\phi(p)=\ds\frac{\sqrt{p}\bar{f}(p)}{(-\frac{1}{2})!}=\sqrt{\ds\frac{p}{\pi}}\bar{f}(p)$

so \begin{eqnarray*} \phi(x) & = &
L^{-1}\left[\ds\frac{\sqrt{p}\bar{f}(p)}{\sqrt{\pi}}\right]\\ & =
& L^{-1}[\bar{a}\bar{b}]\\ & = & a \ast b\
\un{\rm{convolution}}\end{eqnarray*}

where $\bar{a}=\ds\frac{\sqrt{p}}{\sqrt{\pi}},\
\bar{b}=\bar{f}(p)$

Require $L^{-1}\left[\ds\frac{\sqrt{p}}{\sqrt{\pi}}\right]$ and
$L^{-1}[\bar{f}(p)] \stackrel{\rm{easy}}{\longrightarrow}
\un{f(x)}$

\newpage
By standard result or by contour integral

$L^{-1}\left[\ds\frac{p^{\frac{1}{2}}}{\sqrt{\pi}}\right]
=\ds\frac{t^{-\frac{3}{2}}}{\Gamma(-\frac{1}{2})\sqrt{\pi}} =
\ds\frac{-t^{-\frac{3}{2}}}{2\sqrt{\pi}\sqrt{\pi}}
=\ds\frac{-t^{-\frac{3}{2}}}{2\pi}$

$\Rightarrow \phi(x)=-\ds\frac{1}{2\pi}\ds\int_0^x d\xi\
f(x-\xi)\xi^{-\frac{3}{2}}$ is a solution of the integral
equation.

\item[(ii)]
Again Laplace transform:

\begin{eqnarray*} L[\phi(t)] & = & L[\alpha t]=L[\sin t \ast
\phi(t)]\\ \bar\phi(p) & = & \ds\frac{\alpha}{p^2}+L[\sin t]
\times L[\phi(t)]\\ & & \rm{where}\ \ds\int_0^{\infty}
\ds\frac{dt\ e^{(p+i)t}}{2i}-\ds\int_0^{\infty} \ds\frac{dt\
e^{(p-i)t}}{2i}\\ \bar\phi(p) & = &
\ds\frac{\alpha}{p^2}+\ds\frac{\bar\phi(p)}{(1+p^2)}\\ \Rightarrow
\bar\phi(p) & = & \ds\frac{\alpha(1+p^2)}{p^4}\\ & = &
\ds\frac{\alpha}{p^4}+\ds\frac{\alpha}{p^2} \end{eqnarray*}

so by contour integral inversion with 2 integrals which contain a
fourth order pole at $p=0$ and a second order pole at $p=0$
respectively, or by standard methods,

$\phi(t)=\alpha\left[\ds\frac{t^3}{3!}+\ds\frac{t}{1!}\right]
\Rightarrow \phi(t)=\alpha t\left(1+\ds\frac{t^2}{6}\right)$
\end{description}

\end{document}