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For this nice proof I have to use the partial fraction of the meromorphic function $$\frac{\pi^2}{\sin^2{z\pi}}$$ I assume that after the partial fraction is found there is a easy way to show the equation but my main problem is to construct these partial fractions and hope for some general explanatory.

thanks

Chrissoderstirb Huber
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  • Google for the Basel problem – Dr. Sonnhard Graubner Jun 08 '17 at 18:10
  • There are a few answers to the canonical $\pi^2/6$ question on this site which might be relevant: https://math.stackexchange.com/a/324093/1242, https://math.stackexchange.com/a/8773/1242 – Hans Lundmark Jun 08 '17 at 18:43
  • In general we prove $\displaystyle\frac{\pi^2}{\sin^2(\pi z)} = \sum_{n=-\infty}^\infty \frac{1}{(z-n)^2}$ before all, showing $\displaystyle f(z) = \frac{\pi^2}{\sin^2(\pi z)} - \sum_{n=-\infty}^\infty \frac{1}{(z-n)^2}$ is a bounded entire function (thus constant) and vanishing at $i\infty$ – reuns Jun 10 '17 at 08:08

2 Answers2

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HINT:

The infinite product representation of the sine function is given by

$$\sin(\pi z)=\pi z\prod_{n=1}^\infty\left(1-\frac{z^2}{n^2}\right)$$

Note that $\frac{\pi^2}{\sin^2(\pi z)}=-\frac{d^2\,\log(\sin(\pi z))}{dz^2}$?

Now, use $$\frac{\pi^2}{\sin^2(\pi z)}=-\frac{d^2}{dz^2}\log\left(\pi z\prod_{n=1}^\infty\left(1-\frac{z^2}{n^2}\right)\right)$$

Mark Viola
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  • In general we prove $\displaystyle\frac{\pi^2}{\sin^2(\pi z)} = \sum_{n=-\infty}^\infty \frac{1}{(z-n)^2}$ before all, showing $\displaystyle f(z) = \frac{\pi^2}{\sin^2(\pi z)} - \sum_{n=-\infty}^\infty \frac{1}{(z-n)^2}$ is a bounded entire function (thus constant) and vanishing at $i\infty$ – reuns Jun 10 '17 at 08:07
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Restating Mark's approach, if we apply $\frac{d^2}{dz^2}\log(\cdot)$ to both sides of

$$ \frac{\sin(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right) \tag{1}$$ we get: $$ \frac{1}{z^2}-\frac{\pi^2}{\sin^2(\pi z)} =- \sum_{n\geq 1}\left(\frac{1}{(n-z)^2}+\frac{1}{(n+z)^2}\right)\tag{2}$$ and by considering the limit of both sides as $z\to 0$ we get: $$ \zeta(2)=\sum_{n\geq 1}\frac{1}{n^2} = \frac{1}{2}\lim_{z\to 0}\left(\frac{\pi^2}{\sin^2(\pi z)}-\frac{1}{z^2}\right)\stackrel{d.H.}{=}\color{red}{\frac{\pi^2}{6}}.\tag{3} $$

Jack D'Aurizio
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