# Closed form for $\sum{\frac{\zeta(2n)-1}{(2n)^2}}$

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.

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

Using the infinite product for the sine, for $$0 we have
\begin{align*} \log\frac{x\pi}{\sin x\pi}&=-\sum_{n=1}^\infty\log\left(1-\frac{x^2}{n^2}\right)=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1k\left(\frac{x^2}{n^2}\right)^k \\&=\sum_{k=1}^\infty\frac{x^{2k}}{k}\sum_{n=1}^\infty\frac{1}{n^{2k}}=\sum_{k=1}^\infty\frac{\zeta(2k)}{k}x^{2k}. \end{align*}
Dividing by $$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$$.)