<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
\parindent=0pt
\begin{document}

{\bf Question}

Evaluate the following integrals using the method of residues.

\begin{itemize}
\item[i)]
$\ds\int_0^{2\pi}\frac{d\theta}{1+a\cos\theta} \hspace{0.3in}
-1&lt;a&lt;1$


\item[ii)]
$\ds\int_0^\infty\frac{\cos bxdx}{(1+x^2)^2} \hspace{0.5in} b&gt;0$.

\end{itemize}

In (ii) describe briefly any convergence properties that you use.


\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[i)]
Let $z=e^{i\theta}, \hspace{0.3in} dz=izd\theta$

$\cos\theta=\frac{1}{2}(z+\frac{1}{z}) \hspace{0.5in} C$ - the
unit circle.

$\ds I=\int_0^2\pi\frac{d\theta}{1+a\cos\theta}=
\int_C\frac{dz}{iz\left(1+\frac{a}{z}\left(z+\frac{1}{z}\right)\right)}$

$\ds\hspace{0.2in}=
\frac{1}{i}\int_C\frac{dz}{z+\frac{a}{2}z^2+\frac{a}{2}}=
\frac{2}{ia}\int_C\frac{dz}{z^2+\frac{2}{a}z+1}$

The denominator has roots
$\ds-\frac{1}{a}\pm\frac{1}{a}\sqrt{1-a^2}$, so the integrand has
a simple pole inside $C$ at $\ds
z=-\frac{1}{a}+\frac{1}{a}\sqrt{1-a^2}$, with residue

$\ds\left.\frac{1}{z-\left(-\frac{1}{a}-\frac{1}{a}\sqrt{1-a^2}\right)}
\right|_{z=-\frac{1}{a}+\frac{1}{a}\sqrt{1-a^2}}=
\frac{1}{\frac{2}{a}\sqrt{1-a^2}}=\frac{a}{2\sqrt{1-a^2}}$

So $\ds I=2\pi
i\frac{2}{ia}\frac{a}{2\sqrt{1-a^2}}=\frac{2\pi}{\sqrt{1-a^2}}$


\item[ii)]
$\ds I=\int_0^\infty\frac{\cos bx}{(1+x^2)^2}dx \hspace{0.5in}
\left|\frac{\cos bx}{(1+x^2)^2}\right|\leq\frac{1}{x^4}$ and
$\ds\int^\infty\frac{1}{x^4}$ converges.

and $\ds2I=\int_{-\infty}^\infty\frac{\cos bx}{(1+x^2)^2}$.

Let $\ds f(z)=\frac{e^{ibz}}{(1+z^2)^2}$, then $f(z)$ has poles of
order 2 at $z=\pm i$.

\newpage

We integrate $f(z)$ round $\Gamma$


DIAGRAM


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

On $C$, $z=Re^{it}$, and $|e^{ibz}|=|e^{-bR\sin t}|\leq1$ and
$\ds|f(z)|\leq\frac{1}{(R^2-1)^2}$

so $\ds\left|\int_C f(z)dz\right|\leq\frac{\pi R}{(R^2-1)^2}\to0$
as $R\to\infty$

So $\ds\int_{\Gamma}f(z)dz=2\pi
i\frac{-i(b+1)e^{-b}}{4}=\frac{\pi(b+1)e^{-b}}{2}$

letting $R\to\infty$ then gives $\ds I=\frac{\pi(b+1)e^{-b}}{4}$

\end{itemize}

\end{document}
</pre></body></html>