Find the sum of $\sum \frac{1}{k^2 - a^2}$ when $0So I have been trying for a few days to figure out the sum of 
$$ S = \sum_{k=1}^\infty \frac{1}{k^2 - a^2} $$ where $a \in (0,1)$. So far from my nummerical
analysis and CAS that this sum equals
$$ S = \frac{1}{2a} \left[ \frac{1}{a} \, - \, \pi \cot(a\pi) \right] $$
But I have not been able to prove this yet. Anyone know how? My guess is that the 
sum of this series is related to fourier-series but nothing particalr comes to mind. 
For the easy values, I have been able to use telescopic series, and a bit of algebraic magic, but for the general case I am stumpled. Anyone have any ideas or hints? Cheers.
 A: You may prove this by expanding $\cos(zx)$ in Fourier series as shown here.
This paper could help too as well as articles in SE dealing with evaluation of $\zeta(n)$ with $n$ even.
A: This question was settled in the Mathematics chatroom, but I'll put up the solution here for reference.
Starting with the infinite product
$$\frac{\sin\,\pi x}{\pi x}=\prod_{k=1}^\infty \left(1-\frac{x^2}{k^2}\right)$$
taking the logarithm of both sides gives
$$\log\left(\frac{\sin\,\pi x}{\pi x}\right)=\log\left(\prod_{k=1}^\infty \left(1-\frac{x^2}{k^2}\right)\right)=\sum_{k=1}^\infty \log\left(1-\frac{x^2}{k^2}\right)$$
Differentiation gives
$$\frac{\pi x}{\sin\,\pi x}\left(\frac{\cos\,\pi x}{x}-\frac{\sin\,\pi x}{\pi x^2}\right)=\sum_{k=1}^\infty \frac{-2x}{k^2\left(1-\frac{x^2}{k^2}\right)}$$
which simplifies to
$$\pi\cot\,\pi x-\frac1{x}=-2x\sum_{k=1}^\infty \frac1{k^2-x^2}$$
or
$$\sum_{k=1}^\infty \frac1{k^2-x^2}=\frac1{2x^2}-\frac{\pi\cot\,\pi x}{2x}$$
A: Let us consider the principal value of the conditionally convergent infinite harmonic series
$$
\begin{align}
f(z)
&=\sum_{k=-\infty}^\infty\frac{1}{z+k}\\
&=\lim_{n\to\infty}\sum_{k=-n}^n\frac{1}{z+k}\tag{1a}\\
&=\lim_{n\to\infty}\frac1z+\sum_{k=1}^n\frac{1}{z-k}+\frac{1}{z+k}\tag{1b}\\
&=\frac1z+\sum_{k=1}^\infty\frac{2z}{z^2-k^2}\tag{1c}
\end{align}
$$
The series in $(1c)$ converges absolutely for all non-integer $z$.
Each of the terms in $(1c)$ is odd, so $f(-z)=-f(z)$.
The series in $(1a)$ shows that $f$ has a simple pole with residue $1$ at each integer.
$f$ has period $1$:
$$
\begin{align}
f(z+1)-f(z)
&=\lim_{n\to\infty}\sum_{k=-n}^n\frac{1}{z+k+1}-\frac{1}{z+k}\\
&=\lim_{n\to\infty}\frac{1}{z+n+1}-\frac{1}{z-n}\\
&=0\tag{2}
\end{align}
$$
$f(1/2)=0$:
$$
\begin{align}
f(1/2)
&=\lim_{n\to\infty}\sum_{k=-n}^n\frac{1}{k+1/2}\\
&=\lim_{n\to\infty}\frac{1}{n+1/2}\\
&=0\tag{3}
\end{align}
$$
Take the derivative of $f$:
$$
f'(z)=\lim_{n\to\infty}\sum_{k=-n}^n\frac{-1}{(z+k)^2}\tag{4}
$$
This series converges absolutely. and the terms monotonically go to $0$ as $|\Im(z)|\to\infty$.
Let's consider $if(iy)$ as $y\to\infty$. Using $(1c)$, we get
$$
\begin{align}
if(iy)
&=\frac1y+\sum_{k=1}^\infty\frac{2y}{y^2+k^2}\\
&=\frac1y+\sum_{k=1}^\infty\frac{2/y}{1+(k/y)^2}\tag{5}
\end{align}
$$
As $y\to\infty$, the summation in $(5)$ is a Riemann sum for the integral
$$
\int_0^\infty\frac{2\mathrm{d}x}{1+x^2}=\pi\tag{6}
$$
Thus, $if(iy)\to\pi$ as $y\to\infty$ and $if(iy)\to-\pi$ as $y\to-\infty$.
Since $f$ has period $1$ and $f'(z)\to0$ as $|\Im(z)|\to\infty$, it is evident that $f(z)\to-i\pi$ as $\Im(z)\to\infty$ and $f(z)\to i\pi$ as $\Im(z)\to-\infty$. This means that $f$ is bounded when away from the real axis.
The functions $f$ and $\pi\cot(\pi z)$ have the same poles, with identical
residues, and both are bounded when away from the real axis.  Thus,
their difference is bounded for all $z$.  Since their difference is
analytic and bounded, it must be constant.  This difference is $0$ at
$1/2$, so it must be $0$ everywhere.  Therefore, the principal value of
$$
\sum_{k=-\infty}^\infty\frac{1}{z+k}=\pi\cot(\pi z)\tag{7}
$$
for all $z$.

Combining $(1c)$ and $(7)$ yields
$$
\frac1z+\sum_{k=1}^\infty\frac{2z}{z^2-k^2}=\pi\cot(\pi z)\tag{8}
$$
Therefore,
$$
\begin{align}
\sum_{k=1}^\infty\frac{1}{k^2-z^2}
&=\frac{1}{2z^2}-\frac{\pi\cot(\pi z)}{2z}\\
&=\frac{1}{2z}\left[\frac1z-\pi\cot(\pi z)\right]\tag{9}
\end{align}
$$
A: @Kobe found here by means of residue calculus
$$\sum_{k = 1}^\infty \frac{1}{k^2 + a^2} = \frac{\pi\coth(\pi a)}{2a} - \frac{1}{2a^2}$$
To get your result, replace $a$ by $ia$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
S & \equiv \bbox[5px,#ffd]{\sum_{k = 1}^{\infty}
{1 \over k^{2} - a^{2}}} =
\sum_{k = 0}^{\infty}
{1 \over \pars{k + 1 - a}\pars{k + 1 + a}}
\\[5mm] & =
{\Psi\pars{1 - a} - \Psi\pars{1 + a} \over -2a} =
{\Psi\pars{1 - a} - \bracks{\Psi\pars{a} + 1/a} \over -2a} 
\\[5mm] & =
{1 \over -2a}\bracks{\pi\cot\pars{\pi a} - {1 \over a}} =
\bbx{{1 \over 2a}\bracks{{1 \over a} - \pi\cot\pars{\pi a}}}
\\ &
\end{align}
