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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.

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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.

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  • $\begingroup$ 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. $\endgroup$ – Sam Wong Apr 19 at 9:15
  • $\begingroup$ 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! $\endgroup$ – Sam Wong Apr 19 at 9:55
  • $\begingroup$ 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. $\endgroup$ – Sam Wong Apr 19 at 10:29
  • $\begingroup$ @SamWong Hint: this is not the usual function space we are used to working on. Its domain is the integers, and it is uniformly convergent on that domain with the restriction given above. $\endgroup$ – Ninad Munshi Apr 19 at 19:33

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