Integral $\int_0^{\infty} \frac{x\cos^2 x}{e^x-1}dx$ I have encountered this integral and I am stuck evaluating it:$I=\int_0^{\infty} \frac{x\cos^2 x}{e^x-1}dx$ 
My try was to expand the numerator into power series, indeed: $$x\cos^2x=\frac{x}{2}(1+\cos(2x)) =\frac{x}{2} +\sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n-1} x^{2n+1}}{(2n)!}$$ And using  $\zeta{(z)} \Gamma{(z)} =\int_0^{\infty} \frac{x^{z-1}}{e^x-1}dx$ gives: $$I=\frac{1}{2}\zeta{(2)}+ \sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n-1}}{(2n)!} \zeta{(2n+2)} \Gamma{(2n+2)}=\frac{1}{2}\zeta{(2)}+ 2 \sum_{n=0}^{\infty} (-1)^n \zeta{(2n+2)}$$ Is there a way to simplify this? Or maybe another approach to this integral? 
Edit: According to the answer in the comment, would this show that $\sum_{n=1}^{\infty} (-1)^{n-1} \zeta{(2n)}=\frac{\pi^2}{6}(2-3\text{ csch}^2(2\pi))+\frac{1}{16} $  ?

Hint. Another way is to use the residue theorem. Since, for a suitable function, we have $$\sum_{n\in\mathbb{Z}}f\left(n\right)=-\sum\left\{ \textrm{Residues of }\pi\cot\left(\pi z\right)f\left(z\right)\textrm{ at }f\left(z\right)\textrm{'s poles}\right\} $$ we get $$\sum_{n\geq1}\frac{1}{2n^{2}}+\sum_{n\geq1}\left(\frac{1}{2\left(4+n^{2}\right)}-\frac{4}{\left(4+n^{2}\right)^{2}}\right)$$ $$=\frac{\pi^{2}}{12}+\frac{1}{16}+\frac{1}{2}\sum_{n\in\mathbb{Z}}\left(\frac{1}{2\left(4+n^{2}\right)}-\frac{4}{\left(4+n^{2}\right)^{2}}\right).$$ Now you have just to check that $f\left(z\right)=\frac{1}{2\left(4+z^{2}\right)}-\frac{4}{\left(4+z^{2}\right)^{2}}$ is an admissible function and $$\mathrm{Res}_{z=2i}\pi\cot\left(\pi z\right)f\left(z\right)=\mathrm{Res}_{z=-2i}\pi\cot\left(\pi z\right)f\left(z\right)=\frac{1}{4}\pi^{2}\mathrm{csch}^{2}\left(2\pi\right).$$
