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

{\bf Question}

Use the calculus of residues to show that

\begin{itemize}
\item[a)]
$\ds\int_0^{2\pi}\frac{d\theta}{2-\sin\theta}=\frac{2\pi}{\sqrt3}$

\item[b)]
$\ds\sum_{r=-\infty}^\infty\frac{1}{\left(r+\frac{1}{2}\right)^2+1}=
\pi\tanh\pi$.

\end{itemize}



\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
Let $\ds z=e^{i\theta} \hspace{0.3in} d\theta=\frac{dz}{iz}
\hspace{0.3in} \sin\theta=\frac{1}{2i}\left(z-\frac{1}{z}\right)
\hspace{0.3in} C$ - the unit circle

$\ds I=\int_0^{2\pi}\frac{d\theta}{2-\sin\theta}=
\int_C\frac{dz}{iz\left(2-\frac{1}{2i}\left(z-\frac{1}{z}\right)\right)}=
\int_C\frac{-2dz}{z^2-4iz-1}$

The denominator has roots at $z=i(2\pm\sqrt3)$, so the integrand
has a simple pole at $z=i(2-\sqrt3)$ inside $C$ with residue

$\ds\frac{1}{i(2-\sqrt3)-i(2+\sqrt3)}=\frac{1}{-2i\sqrt3}$

So $\ds I=-2.2\pi i\frac{-1}{2i\sqrt3}=\frac{2\pi}{\sqrt3}$


\item[b)]
Let $\ds f(z)= \frac{1}{\left(z+\frac{1}{2}\right)^2+1}=
\frac{1}{\left(z+\frac{1}{2}+i\right)\left(z+\frac{1}{2}-i\right)}$

This has simple poles at $\ds-\frac{1}{2}-i$ and
$\ds-\frac{1}{2}+i$.

Res$\ds\left(-\frac{1}{2}-i\right)=
\frac{1}{-\frac{1}{2}-i+\frac{1}{2}-i}=-\frac{1}{2i}$

Res$\ds\left(-\frac{1}{2}+i\right)=
\frac{1}{-\frac{1}{2}+i+\frac{1}{2}+i}=\frac{1}{2i}$

Consider $\pi\cot\pi zf(z)=g(z)$

Res$\ds\left(g,-\frac{1}{2}-i\right)=
-\frac{\pi\cot\pi\left(-\frac{1}{2}-i\right)}{2i}=
-\frac{\pi\tanh\pi}{2}$

Res$\ds\left(g,-\frac{1}{2}+i\right)=
\frac{\pi\cot\pi\left(-\frac{1}{2}+i\right)}{2i}=
-\frac{\pi\tanh\pi}{2}$

Let $\Gamma_N$ be the square with vertices
$\ds\pm\left(N+\frac{1}{2}\right)\left(1\pm i\right)$ on $\Gamma_N
\,\,\, \pi\cot\pi z$ is bounded uniformly and

$\ds\int_{\Gamma_N}g(z)dz\to0$ as $N\to\infty$

$\ds\int_{\Gamma_N}=2\pi
i\left(\sum_{r=-N}^Nf(r)-\pi\tanh\pi\right)$

Letting $N\to\infty$ gives

$\ds\sum_{r=-\infty}^\infty\frac{1}{\left(r+\frac{1}{2}\right)^2+1}=
\pi\tanh\pi$

\end{itemize}

\end{document}