Q:

Computing the integral $x^2\textrm{sech}^2(x)$

I'm trying to compute the integral $$\int_{0}^{\infty}dx \, x^{2}\operatorname{sech}^{2}(x)=\frac{\pi^{2}}{12}.$$ Manually, one obtains, quite naively, $$\int dx \, x^{2}\operatorname{sech}^{2}(x)=\operatorname{Li}_{2}(-e^{-2x})-x^{2}-2x\log(e^{-2x}+1)+x^{2}\tanh(x),$$ which of course is not possible to evaluate at infinity in this form. 
I was trying to turn this thing into a series, but I'm having trouble. 

A:

Since $\operatorname{sech}x=\frac{2e^{-x}}{1+e^{-2x}}$, your integral is $$\int_0^\infty\frac{4x^2e^{-2x}dx}{(1+e^{-2x})^2}=4\sum_{n\ge 1}(-1)^{n-1}n\int_0^\infty x^2e^{-2nx}dx=\sum_{n\ge 1}\frac{(-1)^{n-1}}{n^2}=\eta(2)=\frac{\pi^2}{12}.$$

A:

Sort of hopping off of @J.G.'s shoulders, we may evaluate 
$$f(s)=\int_0^\infty x^s\text{sech}^2(x) dx$$
using the same trick, namely
$$f(s)=4\int_0^\infty \frac{x^s e^{-2x}dx}{(1+e^{-2x})^2}=4\sum_{n\geq1}(-1)^{n-1}n\int_0^\infty x^s e^{-2nx}dx$$
This integral is rather easy:
$$\int_0^\infty x^se^{-2nx}dx=\frac{\Gamma(s+1)}{(2n)^{s+1}}$$
So 
$$f(s)=\frac{2^2}{2^{s+1}}\Gamma(s+1)\sum_{n\geq1}\frac{(-1)^{n-1}}{n^s}\\
f(s)= 2^{1-s}(1-2^{1-s})\Gamma(s+1)\zeta(s)$$

