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So, there is the following exercise at the end of a lecture script

By using the function $f(x)=e^{-a|x|}$ ($x \in \mathbb{R}, a>0$) evaluate $$\sum_{n \in \mathbb{N}} \frac{1}{n^2+a^2}$$ and by letting $a \to 0$ find $\sum_{n \in \mathbb{N}} \frac{1}{n^2}$.

Okay, this task wants me to use the Poisson summation formula:

$$\sum_{k \in \mathbb{Z}} \hat{f}(k)=2\pi \sum_{k \in \mathbb{Z}} f(2\pi k)$$

As hinted I computed $\hat{f}(k)=\frac{2a}{a^2+k^2}$. Therefore

$$\sum_{n \in \mathbb{N}}\frac{1}{n^2+a^2}=\frac{1}{2a}\sum_{n \in \mathbb{N}} \hat{f}(n) $$

Further I should decompose the sum since I don't know if Poisson's formula is valid if the sum only runs over the natural numbers. We have $$\sum_{n \in \mathbb{Z}}\frac{1}{n^2+a^2}=2\sum_{n \in \mathbb{N}}\frac{1}{n^2+a^2}+\frac{1}{a^2}$$

All in all I get

$$\begin{align}\sum_{n \in \mathbb{N}}\frac{1}{n^2+a^2}&=-\frac{1}{2a^2}+\frac{1}{2} \sum_{n \in \mathbb{Z}}\frac{1}{n^2+a^2}\\ &=-\frac{1}{2a^2}+\frac{1}{4a} \sum_{n \in \mathbb{Z}} \hat{f}(n)\\ &=-\frac{1}{2a^2}+\frac{\pi}{2a} \sum_{n \in \mathbb{Z}} f(2\pi n) \\ &=-\frac{1}{2a^2}+\frac{\pi}{2a} \sum_{n \in \mathbb{Z}} e^{-2\pi a |n|}\end{align}$$

But I am not sure how to apply the limit $a \to 0$ now. I should get $\frac{\pi^2}{6}$ I guess.

EDIT: As hinted I continue

$$\begin{align}\sum_{n \in \mathbb{N}}\frac{1}{n^2+a^2}&=-\frac{1}{2a^2}+\frac{\pi}{2a} \left(-1+2\sum_{n \in \mathbb{N_0}} e^{-2\pi a n} \right) \\ &=-\frac{1}{2a^2}+\frac{\pi}{2a} \left(-1+ \frac{2}{1-e^{-2\pi a}} \right) \\ &=-\frac{1}{2a^2}+\frac{a\pi}{2a^2} \left(-\frac{1-e^{-2\pi a}}{1-e^{-2\pi a}}+ \frac{2}{1-e^{-2\pi a}} \right) \\ &=-\frac{1}{2a^2}+\frac{a\pi}{2a^2} \frac{1+e^{-2\pi a}}{1-e^{-2\pi a}} \\&=-\frac{1-e^{-2\pi a}}{2a^2 (1-e^{-2\pi a})}+\frac{a\pi}{2a^2} \frac{1+e^{-2\pi a}}{1-e^{-2\pi a}} \\&=\frac{-1+e^{-2\pi a}+a\pi+a\pi e^{-2\pi a}}{2a^2-2a^2e^{-2\pi a}}\end{align}$$

Now, using a few times L'Hospital I indeed got $\frac{\pi^2}{6}$. Surely not the most elegant way. Thanks a lot for all your help and hints.

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    Isn't that last one a geometric series? (and as a side note, you can differentiate both sides with respect to $a$ to evaluate $\sum\frac1{n^4}$ and stuff) – 2017-02-04
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    what a nice solution to the basel problem..wasn't aware of that – 2017-02-04
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    @SimplyBeautifulArt Thank you for your hint about the geometrical series; didn't see that. I could evaluate it this way :) – 2017-02-04
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    @MarvinF Preferably, I would hyperbolic cotangent... it simplifies nicely here with all the fractions. – 2017-02-04

2 Answers 2

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Hint: You should be able to compute $$\sum_{n\in\mathbb Z}e^{-2\pi a|n|}=-1 + 2\sum_{n=0}^{\infty} \left(e^{-2\pi a}\right)^n$$

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    Thanks a lot for your help. With this way I could evaluate the limit, you can have a look at my edit if you want. – 2017-02-04
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The Poisson summation formula leads to: $$ \sum_{n\geq 0}\frac{1}{n^2+a^2}=\frac{1+\pi a \coth(\pi a)}{2a^2}\tag{1}$$ and $\lim_{a\to 0^+}\sum_{n\geq\color{red}{ 1}}\frac{1}{n^2+a^2}$ can be computed through de l'Hospital theorem or just by recalling that in a neighbourhood of the origin $$ z\coth(z)=1 + \frac{z^2}{3}+O\left(z^4\right).\tag{2}$$

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    Oh, I didn't see that. Really elegant (+1), thank you. – 2017-02-04