# Evaluate $\sum_{n=-\infty}^{\infty}\frac{1}{(n-a)^2}$ with Poisson summation formula

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 $$b\to 0^+$$:
$$\lim_{b\to0^+} \frac{\pi \sinh 2\pi b}{b(\cosh 2\pi b - \cos 2\pi \alpha)} = \lim_{b\to0^+} \frac{2\pi^2 \sinh 2\pi b}{2\pi b(\cosh 2\pi b - \cos 2\pi \alpha)}$$
$$= \frac{2\pi^2}{1-\cos2\pi\alpha} = \frac{\pi^2}{\sin^2(\pi\alpha)}$$
• Hi Munshi, thanks for your post. I feel like what you are using here is that, $\mathcal{F}(f_n)\to \mathcal{F}{f}$ pointwisely as $f_n\to f$ pointwisely, where $\mathcal{F}(f)$ means the Fourier transform of $f$. It seems not true to me. Can you make the argument of this part more precise? Thank you so much. – Sam Wong Apr 19 at 9:15
• Oh Munshi, I realized the argument is true if both $f_n$ and $f$ are absolutely summable functions. But here, our limit function $f(x)=\frac{1}{(x-a)^2}$ is not absolutely summable. What's worse, the convergence of $f_b(x):=\frac{1}{(x-a)^2+b^2}$ to $f(x)=\frac{1}{(x-a)^2}$ is not uniform.(If the convergence is uniform then we can pass the limit with a simple trick.) Could we manage a way to overcome this? Thanks! – Sam Wong Apr 19 at 9:55
• I still can't work out the problem I mentioned above. But I have figured out an alternative way. Actually we can use the Dominated Convergence Theorem to interchange the limit in $b$ and the infinite sum in $n$, and then the rest follows from your calculation. – Sam Wong Apr 19 at 10:29