Proving identity regarding cosecant function How to prove following identity: $$\csc^2\pi x=\frac{1}{\pi ^2}\sum_{k=-\infty}^{\infty}\frac{1}{(x-k)^2}$$
The only thing that I can see in this identity is that pole structure matches on both side and nothing else so how should I proceed?
 A: As pointed out by Jack D'Aurizio using the Weierstraß Product of the sine function yields the result by taking the second derivative of the logarithm of the product. So the essential direction is given and I will only fill in the details. First, observe that
\begin{align*}
\frac{\sin(\pi x)}{\pi x}&=\prod_{n\geq 1}\left(1-\frac{x^2}{n^2}\right)\\
\log\left[\frac{\sin(\pi x)}{\pi x}\right]&=\log\left[\prod_{n\geq 1}\left(1-\frac{x^2}{n^2}\right)\right]\\
\frac{\rm d}{{\rm d}x}[\log(\sin(\pi x))-\log(\pi x)]&=\frac{\rm d}{{\rm d}x}\left[\sum_{n\geq1}\log\left(1-\frac{x^2}{n^2}\right)\right]\\
\pi\frac{\cos(\pi x)}{\sin(\pi x)}-\frac1x&=\sum_{n\geq1}\frac{-2x}{n^2-x^2}
\end{align*}

$$\therefore \pi\cot(\pi x)~=~\frac1x+\sum_{n\geq1}\frac{2x}{x^2-n^2}$$

From hereon we can go further to obtain
\begin{align*}
\frac{\rm d}{{\rm d}x}[\pi\cot(\pi x)]=&\frac{\rm d}{{\rm d}x}\left[\frac1x+\sum_{n\geq1}\frac{2x}{x^2-n^2}\right]\\
-\pi^2(1+\cot^2(\pi x))&=-\frac1{x^2}-\sum_{n\geq1}\frac{2(x^2+n^2)}{(x^2-n^2)^2}\\
-\pi^2\csc^2(\pi x)&=-\frac1{x^2}-\sum_{n\geq1}\left[\frac1{(x+n)^2}+\frac1{(x-n)^2}\right]
\end{align*}

$$\therefore~\csc^2(\pi x)~=~\frac1{\pi^2}\sum_{n\in\Bbb Z}\frac1{(x-n)^2}$$

A litte side note concerning the crucial product formula of the sine function. Make sure to check if you can actually interchange the logarithm and the infinite product aswell as if you are allowed to take the termwise derivative of the occuring series. Beside these issues this problem illustrates nicely how to derive new representations from old ones.
