sequences and series - Find the sum of $sum 1/(k^2 - a^2)$ when $0

So 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.

Answer

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

$$\sum_{k=1}^\infty \frac1{k^2-x^2}=\frac1{2x^2}-\frac{\pi\cot\,\pi x}{2x}$$