<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}

Use the calculus of residues to show that

\begin{itemize}
\item[a)]
$\ds\int_0^{2\pi}\frac{d\theta}{a+b\cos\theta}=
\frac{2\pi}{\sqrt{a^2-b^2}}$ when $a&gt;b&gt;0$, and


\item[ii)]
$\ds\sum_{n=1}^\infty\frac{1}{n^2-a^2}=\frac{1}{2a^2}-
\frac{\pi}{2a}\cot\pi a$ when $a$ is not an integer.

\end{itemize}


\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
Let $\ds z=e^{i\theta} \hspace{0.3in} d\theta=\frac{dz}{iz}
\hspace{0.3in}$ $\cos\theta=\frac{1}{2}(z+\frac{1}{z})$

$C$ is the unit circle.

$\ds I=\int_0^2\pi\frac{d\theta}{a+b\cos\theta} \hspace{0.2in}
a&gt;b&gt;0$

$\ds\hspace{0.2in}=\int_C\frac{dz}
{iz\left(a+\frac{b}{2}\left(z+\frac{1}{z}\right)\right)}=
\frac{2}{ib}\int_C\frac{dz}{z^2+\frac{2a}{b}z+1}$

The integrand has simple poles at

$\ds z=-\frac{a}{b}+\sqrt{\frac{a^2}{b^2}-1}=\alpha_1$ - inside
$C$

$\ds z=-\frac{a}{b}-\sqrt{\frac{a^2}{b^2}-1}=\alpha_2$ - outside
$C$

The residue at $z=\alpha_1$ is
$\ds\frac{1}{\alpha_1-\alpha_2}=\frac{b}{2\sqrt{a^2-b^2}}$

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

\newpage

\item[b)]
Let $\ds f(z)=\frac{1}{z^2-a^2} \hspace{0.3in} a\not\in Z$

Apply the usual method with $\pi\cot\pi zf(z)$, which has simple
poles at $z=\pm a$.

res$\ds(a)=\frac{\pi\cot\pi a}{2a}={\rm res}(-a)$

So $\ds\sum_{n=-\infty}^\infty\frac{1}{n^2-a^2}=-\frac{\pi\cot\pi
a}{a}$

So $\ds2\sum_{n=1}^\infty\frac{1}{n^2-a^2}+\frac{1}{-a^2}=
-\frac{\pi\cot\pi a}{a}$

Hence $\ds \sum_{n=1}^\infty\frac{1}{n^2-a^2}=
\frac{1}{2a^2}-\frac{\pi\cot\pi a}{2a}$

\end{itemize}

\end{document}
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