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

{\bf Question}

\begin{itemize}
\item[a)]
The function $Q(z)$ is a rational function such that
$\ds\lim_{z\to\infty}zQ(z)=0$, and the curve $\Gamma$ is the upper
half of the circle $|z|=R$.  Prove that

$$\lim_{R\to\infty}\int_{\Gamma}Q(z)e^{{\rm im}z}dz=0$$

where $m\geq0$.


\item[b)]
Use the result in part (a) and the calculus of residues to show
that

\begin{itemize}
\item[i)]
$\ds\int_0^\infty\frac{dx}{(1+x^2)^2}=\frac{\pi}{4}$

\item[ii)]
$\ds\int_0^\infty\frac{\cos x}{1+x^2}dx=\frac{\pi}{2e}$.

\end{itemize}

\end{itemize}


\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
$Q(z)$ is a rational function and $\ds\lim_{|z|\to\infty}zQ(z)=0$.

Let $\ds Q(z)=\frac{A(z)}{B(z)}$ where $A$ and $B$ are polynomials
of degrees $a$ and $b$.

$\ds zQ(z)=\frac{zA(z)}{B(z)}\sim\frac{{\rm degree\ }a+1}{{\rm
degree\ }b}$

so $a+1<b$ since $zQ(z)\to0$ as $|z|\to\infty$

i.e. $a\leq b-2$, so $\exists K,H$ such that for $|z|>H
\hspace{0.5in} |q(z)|<\frac{B}{|z|^2}$

Now for $z$ on $\Gamma \hspace{0.3in}
|e^{imz}|=e^{-mR\sin\theta}\leq1$ for $\theta\in[0,\pi]$

So for $R>H$

$\ds\left|\int_{\Gamma}Q(Z)e^{imz}dz\right|\leq\frac{K}{R^2}\pi
R\to0$ as $R\to\infty$

\newpage

\item[b)]
\begin{itemize}
\item[i)]
$\ds\int_0^\infty\frac{dx}{(1+x^2)^2}=
\frac{1}{2}\int_{-\infty}^\infty\frac{dx}{(1+x^2)^2}$

We integrate $\ds f(z)\frac{1}{(1+z^2)^2}$ around $C$.

This satisfies the conditions of (a) with $m=0$ and

$\ds q(z)=\frac{1}{(1+z^2)^2}$

$f(z)$ has a pole of order 2 at $z=i$ inside $C$ with residue

$\ds\left.\frac{d}{dz}(z-i)^2f(z)\right|_{z=i}=
\frac{d}{dz}\frac{1}{(z+i)^2}=
\left.\frac{-2}{(z+i)^3}\right|_{z=i}=-\frac{i}{4}$

$\ds\int_C f(z)dz=2\pi i\left(-\frac{i}{4}\right)=\frac{\pi}{2}$

so $\ds\int_0^\infty f(x)dx=\frac{\pi}{4}$


\item[ii)]
We integrate $\ds f(z)=\frac{e^{iz}}{1+z^2}$ around $C$, the
result of (a) applies with $m=1$ and $\ds q(z)=\frac{1}{1+z^2}$

$f(z)$ has a simple pole at $z=i$ inside $\Gamma$ with residue
given by $\ds\lim_{z\to i}\frac{e^iz}{z+i}=\frac{e^{-1}}{2i}$

So $\ds\int_C\frac{e^{iz}}{1+z^2}dz=2\pi
i\frac{e^{-1}}{2i}=\frac{\pi}{e}$

So $\ds\int_{-\infty}^\infty\frac{\cos x}{1+x^2}dx=\frac{\pi}{e}$
and $\ds\int_0^\infty\frac{\cos x}{1+x^2}=\frac{\pi}{2e}$

\end{itemize}

\end{itemize}

\end{document}