Prove that $\sum_{n=-\infty }^{+\infty }\frac{1}{(n+\alpha )^{2}}=\frac{\pi ^{2}}{(\sin\pi \alpha )^{2}}$ with poisson summation formula

Question

I want to show that $\sum_{n=-\infty }^{+\infty }\frac{1}{(n+\alpha )^{2}}=\frac{\pi ^{2}}{(\sin\pi \alpha )^{2}}$, the introduction to the Poisson summation formula is in this link https://en.wikipedia.org/wiki/Poisson_summation_formula, I want to show this by using Poisson summation formula, the hint of this exercise says we can check the function $g$ such that $g=1-|x|$when $|x|\leq 1$ and $g=0$ otherwise, the Fourier transform of g is $F(g)(\xi)=(\frac{\sin(\pi \xi)}{\pi \xi })^{2}$, so how to use g and Poisson summation formula to prove it ? Can anyone help me, thank you in advance

Answer 1

You want to prove that $$\sum_{n=-\infty }^{+\infty }\frac{\sin^2(\pi (n+\alpha))}{\pi^2(n+\alpha )^{2}}=1=\sum_k e^{2i\pi \alpha k}g(k)$$