Came across this in an old textbook and I'm struggling to simplify this in any way.
I tried to integrate it and write it as a product: $\int y\:=\:\frac{1}{2x}\left(\log\prod \:_{n=1}^{\infty }\left(x^2-n^2\pi ^2\right)\right)$
but that doesn't seem to help and neither did my attempt at partial fraction decomposition.
Any help would be appreciated!
Best Answer
It follows from the Euler/Weierstrass factorisation for sine:
$\displaystyle \sin z =z\prod_{n \ge 1}\left ( 1-\frac{z^2}{n^2 \pi^2}\right ) \implies \log \sin z = \log z +\sum_{n \ge 1} \log\left(1-\frac{z^2}{n^2 \pi^2}\right) \\ \displaystyle \implies \log(\sin z)-\log(z) = \sum_{n \ge 1} \log\left(1-\frac{z^2}{n^2 \pi^2}\right). $
Differentiating both sides:
$\displaystyle \cot z - \frac{1}{z} = \sum_{n \ge 1}\frac{2 z}{z^2-n^2 \pi^2 } \implies \sum_{n \ge 1}\frac{1}{z^2-n^2 \pi^2 } = \frac{1}{2z}\left(\cot z -\frac{1}{z}\right). $