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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!

Mittens
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isty43
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    it seems something related to the derivative of the gamma function or so – Masacroso Apr 07 '21 at 13:00
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    Maybe the answer posted to https://math.stackexchange.com/questions/1290096/factorization-of-the-sine will help. Also, https://math.stackexchange.com/questions/286474/evaluating-an-integral-without-using-the-digamma-function-or-complex-analysis and https://math.stackexchange.com/questions/1348737/can-anyone-prove-the-identity-sum-m-infty-infty-z-pi-m-2-sin-z – Gerry Myerson Apr 07 '21 at 13:03
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    And https://math.stackexchange.com/questions/3951093/proving-frac1-ez-1-frac1-z-1-2-2z-sum-n-1-infty-frac and https://math.stackexchange.com/questions/335789/eulers-infinite-product-for-the-sine-function-and-differential-equation-relatio – Gerry Myerson Apr 07 '21 at 13:07
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    https://en.m.wikipedia.org/wiki/Mittag-Leffler%27s_theorem. You will find the series among examples. – user Apr 07 '21 at 13:09
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    @user perhaps you should expand that into an answer! – YiFan Tey Apr 08 '21 at 00:30

1 Answers1

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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). $

NoName
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