In a recent theoretical physics problem, I stumbled across two integrals that I thought look quite easy but it turned out that I can't solve them. I appreciate any hint on how to get to a solution :)
The two integrals are defined by the functions:
$f(z) = \frac{\left(\pi ^2-4 z^2\right) \text{sech}^2(z)}{\left(4 z^2+\pi ^2\right)^2}$
$g(z) = \frac{\text{sech}^2(z)}{4 z^2+\pi ^2}$
as
$\int_{-\infty}^{\infty} f(z) \,dz $ and $ \int_{-\infty}^{\infty} g(z) \,dz$
Plot of f(z) and g(z) within some range of z
Thanks in advance to y'all
The residue theorem is the standard technique for dealing with such integrals. We have
$$ \frac{1}{\cosh^2(x)} = -\sum_{k\geq 0}\frac{1}{\left(x-(2k+1)\frac{\pi i}{2}\right)^2}+\frac{1}{\left(x+(2k+1)\frac{\pi i}{2}\right)^2} $$
and
$$ \int_{\mathbb{R}}\frac{dx}{(x-z)^2(4x^2+\pi^2)} = \frac{2}{(2z+i\pi)^2}$$ for any $z\in\mathbb{C}$ in the upper half-plane. It follows that
$$ \int_{\mathbb{R}}\frac{dx}{(4x^2+\pi^2)\cosh^2(x)} = \sum_{k\geq 0}\frac{1}{\pi^2(k+1)^2}=\frac{1}{6} $$ and the other integral can be tackled in a similar fashion: $$ \int_{\mathbb{R}}\frac{(\pi^2-4z^2)\,dz}{(\pi^2+4z^2)^2\cosh^2(z)}=\frac{\zeta(3)}{\pi^2}. $$