TITLE: How to prove this identity $\pi=\sum\limits_{k=-\infty}^{\infty}\left(\frac{\sin(k)}{k}\right)^{2}\;$?
QUESTION [76 upvotes]: How to prove this identity? $$\pi=\sum_{k=-\infty}^{\infty}\left(\dfrac{\sin(k)}{k}\right)^{2}\;$$
I found the above interesting identity in the book $\bf \pi$ Unleashed. 
Does anyone knows how to prove it?
Thanks.

REPLY [6 votes]: This is a great place to apply Abel-Plana's formula
$$\sum_{n\geqslant 0} f(n)=\int_0^\infty f(x)\, dx+\frac{f(0)}{2}+i\int_0^\infty\frac{f(ix)-f(-ix)}{e^{2\pi x}-1}\, dx$$

Allow $f(x)=\left(\frac{\sin x}{x}\right)^2$

We note that, with the fact that $f$ is even $$\sum_{n\geqslant 0} \left(\frac{\sin n}{n}\right)^2=\frac{1}{2}\left(\sum_{n\in\Bbb{Z}}\left(\frac{\sin n}{n}\right)^2-1\right)$$
We already know that $$\int_0^\infty \left(\frac{\sin x}{x}\right)^2\, dx=\frac{\pi}{2};$$ and $$\frac{\sin 0}{0}:=1$$
Also, we easily determine that:$$\left(\frac{\sin ix}{ix}\right)^2=\left(\frac{\sin (-ix)}{-ix}\right)^2=\frac{\sinh^2 x}{x^2}$$
With these, the result is obvious.
I'm sorry, I know this is a bit overpowering of a formula to use on this question, but I thought it was worth noting.