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I am interested in finding a closed form that involve some known constants to express following infinite sum:

$$\sum_{n=1}^{\infty}{\frac{\zeta(2n)-1}{(2n)^2}}$$

you can find out some similar results when denominator exponent is 0 or 1 in wikipedia zeta function entry, but I was not able to find anything related to $n^2$ denominator.

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  • $\begingroup$ This smells of PolyLog, which is not a nice thing to handle. $\endgroup$ Commented Feb 17, 2021 at 17:00
  • $\begingroup$ $\displaystyle\sum_{n=1}^\infty\frac{\zeta(2n)}{2n^2}=\int_0^\pi\log\frac{x}{\sin x}\frac{dx}{x}$, doesn't look familiar... $\endgroup$
    – metamorphy
    Commented Feb 18, 2021 at 2:39
  • $\begingroup$ @metamorphy, where can I find a proof of that? $\endgroup$
    – Dabed
    Commented Mar 4, 2021 at 16:24
  • $\begingroup$ @DanielD., posted it as an "answer". $\endgroup$
    – metamorphy
    Commented Mar 4, 2021 at 17:27

1 Answer 1

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This not-an-answer explains the integral representation given in my comment.

Using the infinite product for the sine, for $0<x<1$ we have

\begin{align*} \log\frac{x\pi}{\sin x\pi}&=-\sum_{n=1}^\infty\log\left(1-\frac{x^2}{n^2}\right)=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1k\left(\frac{x^2}{n^2}\right)^k \\&=\sum_{k=1}^\infty\frac{x^{2k}}{k}\sum_{n=1}^\infty\frac{1}{n^{2k}}=\sum_{k=1}^\infty\frac{\zeta(2k)}{k}x^{2k}. \end{align*}

Dividing by $x$ and integrating, we get the desired: $$\sum_{k=1}^\infty\frac{\zeta(2k)}{2k^2}=\int_0^1\log\frac{x\pi}{\sin x\pi}\frac{dx}{x}\underset{\color{gray}{x\pi=t}}{\phantom{\big[}=\phantom{\big]}}\int_0^\pi\log\frac{t}{\sin t}\frac{dt}{t}.$$

(If we denote this by $S$, the sum in question is $S/2-\pi^2/24$.)

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  • $\begingroup$ thank you very much, pretty nice! $\endgroup$
    – Dabed
    Commented Mar 4, 2021 at 17:49

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