This post (Closed form for $\sum_{n=-\infty}^{\infty}\frac{1}{(n-a)^2+b^2}$.) gives a closed form for $\sum_{n=-\infty}^{\infty}\frac{1}{(n-a)^2+b^2}$ with $b\gt0.$
And the result is $$\sum_{n\in\mathbb{Z}}\frac{1}{(n-a)^2+b^2}=\frac{\pi}{b}\sum_{k\in\mathbb{Z}}e^{-2\pi i k a-2\pi |k| b}=\frac{\pi\sinh2\pi b}{b\left(\cosh2\pi b-\cos2\pi a\right)}.$$
By inspecting the proof, the crucial part is the calculation of the Fourier transform of $f$, where $f(x)=\frac{1}{(x-a)^2+b^2}.$
In our case, in order to use the Poisson summation formula, we need to let $f(x)=\frac{1}{(x-a)^2}.$
My first question is, can we still use the residue theorem to calculate the Fourier transform of $f$ now? (I am not so familiar with complex analysis..)
My second question is, does the sum of Fourier series, $\sum_{k\in\mathbb{Z}}\hat{f}(k)$, still converge?
I think the $\sum_{k\in\mathbb{Z}}\hat{f}(k)$ should be like $\sum_{k\in\mathbb{Z}}e^{-2\pi i k a}$, which is not convergent...
Oh, I do know the closed form is $$\sum_{n=-\infty}^{\infty}\frac{1}{(n-a)^2}=\frac{\pi^2}{(\sin\pi\alpha)^2}$$
What I am doing is to prove this closed form with Poisson summation formula.
Thanks for help.
Assuming $a\notin \Bbb{Z}$ we can take the limit as $b\to 0^+$:
$$\lim_{b\to0^+} \frac{\pi \sinh 2\pi b}{b(\cosh 2\pi b - \cos 2\pi \alpha)} = \lim_{b\to0^+} \frac{2\pi^2 \sinh 2\pi b}{2\pi b(\cosh 2\pi b - \cos 2\pi \alpha)}$$
$$ = \frac{2\pi^2}{1-\cos2\pi\alpha} = \frac{\pi^2}{\sin^2(\pi\alpha)}$$
since Fourier transforms have a continuity condition for absolutely summable functions.