\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand\ds{\displaystyle}
\begin{document}
\parindent=0pt

QUESTION

 Use the standard semicircular contour to evaluate
\begin{description}

\item[(a)]
$$\int_0^\infty\frac{dx}{2+x^2}$$

\item[(b)]
$$\int_0^\infty\frac{x^2dx}{\left(x^2+a^2\right)^2}$$

\end{description}

\bigskip

ANSWER
\begin{description}

\item[(a)]
$$ \ $$ $\begin{array}{l} \ds I=\frac{1}{2}\int_{-\infty}^\infty
\frac{d}{2+x^2}
\end{array}
\ \ \
\begin{array}{c}
\epsfig{file=362-2-1.eps, width=50mm}
\end{array}$


This has two simple poles, 1 inside the contour.\\ $\ds
I=\frac{1}{2}2\pi i\frac{1}{2 \sqrt{2}i}=\frac{\pi}{2\sqrt{2}}$


\item[(b)]
$$ \ $$ $\begin{array}{l} \ds I=\frac{1}{2}\int_{-\infty}^\infty
\frac{x^2 dx}{\left(x^2+a^2\right)^2}
\end{array}
\ \ \
\begin{array}{c}
\epsfig{file=362-2-2.eps, width=50mm}
\end{array}$

This has two double poles, 1 inside the contour.
\begin{eqnarray*}
\textrm{Res}\left(\frac{z^2}{\left(z^2+a^2\right)^2},a_i\right)&=&\lim_{z
\to ai} \frac{d}{dz}\frac{z^2}{\left(z+ai\right)^2}\\
&=&\lim_{z\to
a_i}\left(\frac{2z}{\left(z+ai\right)^2}-\frac{2z^2}{\left(z+ai\right)^3}\right)\\
&=&\frac{2ai}{-4a^2}-\frac{-2a^2}{-8a^3i}=-\frac{i}{4a}\\
\textrm{So } I &=& \frac{1}{2}2\pi
i\left(-\frac{1}{4a}\right)=\frac{\pi}{4a}
\end{eqnarray*}


\end{description}

\end{document}