Can you help me to evaluate $\int_{-\infty}^\infty\frac{x^2}{-1+\cosh (2x)}dx$ as $\pi^2/6$? And do you find a similar integral for $\zeta(4)$?

Question

I was inspired in the shape of the integrals for $\zeta(2)$ in A. Córdoba, Encounters at the interface between Number Theory and Harmonic Analysis, Proceedings of the Segundas Jornadas de Teoría de Números, page 102 (2007) Biblioteca de la Revista Matemática Iberoamericana, to ask to Wolfram Alpha online calculator about integrals of a different kind, to obtain integrals that get zeta values.

Example 1. After a lot of trials with Wolfram Alpha I've found (here was fixed a typo, see the comments from the users, thanks them) $$\int_{-\infty}^\infty\frac{x^2}{-1+\cosh (2x)}dx=\frac{\pi^2}{6}.$$

When I did the change of variables $u=e^{2x}$ I can show (using the change of variables and the evaluation from Wolfram Alpha) that previous integral is evaluated as $$\int_0^\infty\frac{\log^2 u}{-8u+4u^2+4}du=\frac{\pi^2}{6},$$ see it:

integrate (log x)^2/(-8x+4x^2+4) dx from x=0 to x=infinite

Example 2. After a lot experiments, I would like to get $\frac{\pi^4}{90}$ as an integral of previous kind, I say a double integral, and not a multiple of an integral of this kind, some of my attempts were, for example (in comments you can see an ample variety) this $$\int_{-\infty}^\infty\int_{-\infty}^\infty\frac{x^2y^2}{(-1+\cosh (2x))(1+\cosh(y)))}dxdy=\frac{\pi^4}{9},$$ or this $$\int_{-\infty}^\infty\int_{-\infty}^\infty\frac{x^2y^2}{(1+\cosh (2x))(1+\cosh( \sqrt{5}y)))}dxdy=\frac{\pi^4}{90\sqrt{5}}.$$ In comments are more of my attempts to get a similar integral for $\zeta(4)=\frac{\pi^4}{90}$.

These are the codes of previous Example 2:

integrate x^2y^2/((-1+cosh(2x))(1+cosh(y)))dx dy from x=-infty to x=infty, from y=-infty to y=infty

integrate x^2y^2/((1+cosh(2x))(1+cosh(sqrt(5)y)))dx dy from x=-infty to x=infty, from y=-infty to y=infty

I would like to know

Question. Can you provide to me hints to finish the calculation $\int_0^\infty\frac{\log u}{-8u+4u^2+4}du=\frac{\pi^2}{6}$? Can you obtain an integral, not a multiple of an integral of previous kind (see also the comments) to get $\zeta(4)$? Many thanks.

Answer 1

Can you help me to evaluate $\displaystyle \int_{-\infty}^\infty\frac{x^2}{-1+\cosh (2x)}dx$ as $\pi^2/6$?

And can you find a similar integral for $\zeta(4)$?

Hint. One may observe that, for $p\ge1$, one has $$ \begin{align} \int_{-\infty}^\infty\frac{x^{2p}}{-1+\cosh(2x)}dx&=2\int_0^\infty\frac{x^{2p}}{-1+\cosh(2x)}dx \\\\&=2\int_0^\infty\frac{x^{2p}}{-1+\dfrac{e^{2x}+e^{-2x}}2}dx \\\\&=4\int_0^\infty\frac{x^{2p}\:e^{-2x}}{(1-e^{-2x})^2}dx \\\\&=4\sum_{n=1}^\infty n\int_0^\infty x^{2p}e^{-2nx}dx \\\\&=4\:\sum_{n=1}^\infty \frac{\Gamma(2p+1)}{2^{2p+1}n^{2p}} \\\\&= \frac1{2^{2p-1}}\:\Gamma(2p+1)\zeta(2p). \end{align} $$ Then, taking for example $p=1$ and $p=2$, one gets

$$ \begin{align} \int_{-\infty}^\infty\frac{x^2}{-1+\cosh (2x)}dx&=\zeta(2)=\frac{\pi^2 }6 \\\\ \int_{-\infty}^\infty\frac{x^4}{-1+\cosh(2x)}dx&=3\zeta(4)=\frac{\pi^4 }{30}. \end{align} $$