I was looking at this proof And I thought there should be a relation between $\int_{0}^{\infty}\frac{x^2}{\sinh^2(x)}dx$ and ${\pi}^2$. After dividing by ${\pi}^2$ I noticed that it is the sum of the square of natural numbers. Looks like there is a relation between $\int_{0}^{\infty}\frac{x^k}{\sinh^2(x)}dx$ and $\sum\limits_{n=1}^{\infty}\frac{1}{n^k}$ and I have no clue how to prove it.
I appreciate any help in this regard.
As mentioned by Metamorphy You can use $1/\sinh^2 x=4e^{-2x}/(1-e^{-2x})^2 $ and $ (1-z)^{-2}=\sum_{n=0}^\infty(n+1)z^n $ to write the integral as
$$\int_{0}^{\infty}\frac{x^k}{\sinh^2(x)}dx=\int_{0}^{\infty}x^k\frac{4e^{-2x}}{(1-e^{-2x})^2}dx=\int_{0}^{\infty}4e^{-2x}x^k\frac{1}{(1-e^{-2x})^2}dx$$ $$=\int_{0}^{\infty}4e^{-2x}x^k\sum_{n=0}^{\infty}(n+1)e^{-2nx}dx$$ Which is:
$$=\int_{0}^{\infty}\sum_{n=0}^{\infty}4e^{-2x}x^k(n+1)e^{-2nx}dx =\int_{0}^{\infty}\sum_{n=0}^{\infty}4x^k(n+1)e^{-(2n+2)x}dx$$ $$=\sum_{n=0}^{\infty}4(n+1)\int_{0}^{\infty}x^ke^{-(2n+2)x}dx$$ Now by using properties of Gamma function mentioned above $$\int_{0}^{\infty}x^{s-1}e^{-ax}\,\mathrm{d}x=\Gamma(s)a^{-s} $$ You can write: $$=\sum_{n=0}^{\infty}4(n+1)\Gamma(k+1)(2n+2)^{-(k+1)} =\sum_{n=0}^{\infty}k!\frac{4(n+1)}{(n+1)^{k+1}2^{k+1}} =\frac{4\cdot k!}{2^{k+1}}\sum_{n=0}^{\infty}\frac{1}{(n+1)^{k}}\\\\ =\frac{ k!}{2^{k-1}}\sum_{n=1}^{\infty}\frac{1}{n^{k}}$$