Does anyone can give me a hint how to integrate the following:
$$\int_0^\infty{\frac{x^2 {\rm d}x}{\mathrm{cosh}^2(x)}}.$$
The answer is $\frac{\pi^2}{12}$ (taken from the book).
I've started with
$$\int_0^\infty{\frac{x^2 {\rm d}x}{\mathrm{cosh}^2x}} =$$
$\int_0^\infty{\frac{x^2}{\mathrm{cosh}~x}\frac{{\rm d}(\mathrm{sinh}~x)}{1 + \mathrm{sinh}^2x}}=\left. \arctan{(\mathrm{sinh}~x)}\frac{x^2}{\mathrm{cosh}~x}\right|_0^\infty-\int_0^\infty \arctan{(\mathrm{sinh}~x)}\frac{2x~\mathrm{cosh}~x-x^2 \mathrm{sinh}~x}{\mathrm{cosh}^2x}{\rm d}=$
$$-\frac{1}{2}\int_0^\infty {\rm d}(\arctan^2(\mathrm{sinh}~x))\left[2x-x^2 \mathrm{tanh}~x\right].$$
Definitely, $$\frac{1}{2}\int_0^\infty {\rm d}(\arctan^2(\mathrm{sinh}~x)) = \frac{\pi^2}{8},$$
and I think that's the way to the answer. But I can't imagine how to deal with the part in square brackets...
I will be grateful for any help.
$$\int_0^{\infty} dx \frac{x^2}{\mathrm{cosh}^2{x}} = 4 \int_0^{\infty} dx \: x^2 e^{-2 x} (1+e^{-2 x})^{-2} $$
$$ = 4 \int_0^{\infty} dx \: x^2 e^{-2 x} \sum_{k=0}^{\infty} (-1)^k (k+1) e^{-2 k x} $$
$$ = 4 \sum_{k=0}^{\infty} (-1)^k (k+1) \int_0^{\infty} dx \: x^2 e^{-2 (k+1) x} $$
$$ = \sum_{k=0}^{\infty} (-1)^k \frac{1}{(k+1)^2} = \frac{\pi^2}{12} $$
The expression in the second line comes from the Taylor expansion of $(1+y)^{-2}$ about $y=0$.