alternate proof for $\sum_{n\ge 1}\frac{1}{n^2}=\frac{\pi^2}{6}$

Question

There are many proofs out there for $\sum_{n\in N} n^{-2}=\frac{\pi^2}{6}$. I am looking for a proof/reference of the following way that uses convergence Fourier series. First using the Poisson kernel on $R$, given by $k(x)=\frac{1}{\pi(1+x^2)}$ we get derive an identity $$\sum_{n\in Z}(n^2+t^2)^{-1}=\frac{\pi}{t}\frac{1+e^{-2\pi t}}{1-e^{-2\pi t}}$$ and then taking limits of the above as $t\to 0$ we deduce the result. However I dont understand where this identity comes from and the limit part, what technique is used.

Answer 1

Let $f\in\mathcal{S}(\mathbb{R})$. Then $$\sum_{n\in\mathbb{Z}}f(n) = \sum_{k\in\mathbb{Z}}\hat f(k),$$ where $\hat f$ is the Fourier transformation of $f$.

By Cauchy’s integral formula, $$\mathcal{F}\left\{(n^2+t^2)^{-1}\right\}(\omega)=\int_\mathbb{R}\frac{1}{x^2+t^2}e^{-2\pi i\omega x}\,\mathrm{d}x=\frac{\pi}{t}e^{-2\pi tt\lvert \omega\rvert}$$ By the Poisson summation formula, we obtain $$\sum_{n\in \mathbb{Z}}(n^2+t^2)^{-1}=\sum_{n\in \mathbb{Z}}\frac{\pi}{t}e^{-2\pi tt\lvert \omega\rvert}=\frac{\pi }{t}\frac{1+e^{2\pi t}}{1-e^{2\pi t}}$$

As $t\to0$ note that the sum diverges, since it goes over $\frac{1}{0^2}$.

Reference: https://blog.zilin.one/2014/12/05/poisson-summation-formula-and-basel-problem/

Answer 2

The simplest way to establish the equality:

$$ \sum_{n\in \mathbb Z}\frac1{n^2+t^2}=\frac{\pi\coth\pi t}t\tag1 $$ is to apply Mittag-Leffler's theorem.

From $(1)$ we derive: $$ 2\sum_{n=1}^\infty\frac1{n^2+t^2}=\frac{\pi\coth\pi t}t-\frac1{t^2}\tag2 $$ which in the limit $t\to0$ gives: $$ 2\sum_{n=1}^\infty\frac1{n^2}=\lim_{t\to 0}\left[\frac{\pi\coth\pi t}t-\frac1{t^2}\right]=\frac{\pi^2}3\tag3 $$ where we used the well-known Laurent series: $$ \coth x=\frac1x+\frac x3-\mathcal O(x^3) $$