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\begin{document}

{\bf Question}

\begin{itemize}
\item[a)]
Show that

$$\int_0^{2\pi}\frac{d\theta}{a+b\cos\theta}=\frac{2\pi}{\sqrt{(a^2-b^2)}},$$

where $a>b>0.$

\item[b)]
Evaluate the sum of the series

$$\sum_{n=1}^\infty\frac{1}{n^2-a^2},$$

where $a$ is not an integer.  Give detailed justification of your
method.

\end{itemize}


\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
Let $C$ be the unit circle, $z=e^{i\theta}$,
$d\theta=\frac{dz}{iz}$, $\cos\theta=\frac{1}{2}(z+\frac{1}{z})$.
Thus

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

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 $\ds I=\frac{2}{ib}2\pi i.{\rm
res}(\alpha_1)=\frac{2\pi}{\sqrt{(a^2-b^2)}}$


\item[b)]
The function $\ds f(z)=\frac{\pi\cot\pi z}{z^2-a^2}$ has a simple
pole at $z=n$ with residue $\ds\frac{1}{n^2-a^2}$ for $n\in{\bf
Z}$, and simple poles at $z=\pm a$ with residue
$\ds\frac{\pi\cot\pi a}{2a}$ at each.

Now let $C_N$ be the square with vertices $\pm(N+\frac{1}{2})(1\pm
i) \hspace{0.2in} N\in{\bf N}$

On the upper edge $z=x+(N+\frac{1}{2})i$, and so

$\ds|\cot\pi z|=\left|\frac{\cos\pi z}{\sin\pi z}\right|=
\left|i\frac{e^{i\pi z}+e^{-i\pi z}}{e^{i\pi z}-e^{-i\pi
z}}\right|=\left|\frac{e^{2i\pi z}+1}{e^{2i\pi z}-1}\right|$

$\ds\leq \frac{1+|e^{2\pi iz|}}{1-|e^{2\pi iz|}}=
\frac{1+e^{-2\pi(N+\frac{1}{2})}}{1-e^{-2\pi(N+\frac{1}{2})}}\leq
\frac{2}{1-e^{-\pi}}$ for all $N\in{\bf N}$.

Since $|\cot\pi z|=|\cot\pi(-z)|$ the same bound serves on the
lower side of the square.

On the sides parallel to the imaginary axis

$|\cot\pi z|=|\cot\pi(\pm N+\frac{1}{2}+iy)|$

$=|\cot\pi(\frac{1}{2}+iy)|=|-\tan\pi iy|=|\tanh y|\leq 1$

So $\exists K \,\, \forall N\in{\bf N}, \,\, \forall z\in C_N,
\,\, |\pi\cot\pi z|\leq K$

Now provided $N\geq|a|$, we have

$\ds\int_{C_N}f(z)dz=2\pi i\left\{\sum_{n=-N}^N\frac{1}{n^2-a^2}+
\frac{2\pi\cot\pi a}{2a}\right\}$

Now $\ds\left|\int_{C_N}\frac{\pi\cot\pi z}{z^2-a^2}dz\right|\leq
\frac{K8\left(N+\frac{1}{2}\right)}
{\left(N+\frac{1}{2}\right)^2-|a|^2}\rightarrow 0$ as
$N\to\infty$.

since $|z|\geq N+\frac{1}{2}$ on $C_N$.

Letting $N\to\infty$ therefore gives

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

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

\end{itemize}

\end{document}