I'm trying to integrate with a closed contour on the upper-half of the complex plane.
$I = \displaystyle\int_{-\infty}^\infty \dfrac{z\,\mathrm{sech(z)}}{[(z-a)^2+b^2][(z+a)^2+b^2]} dz$
There are simple poles at $z=a\pm ib$ and $z=-a\pm ib$, plus infinitely many poles along $z=i(2n+1)\pi$ where $n\in\mathbb{Z}$.
I draw an indented semicircle closed contour on the upper half complex plane, so that the diameter lies on the real axis. The contour is parametrised as:
$z=-t\,e^{i\pi}$, $-R \leq t \leq -\epsilon$
$z=\epsilon \, e^{i(\pi-\theta)}$, $0 \leq \theta \leq \pi$
$z=x$, $\epsilon \leq x \leq R$
$z=R\, e^{i\theta}$, $0 \leq \theta \leq \pi$
I get back the integral I started with using this parametrisation. I'm not sure how to proceed from here.
Edit:
residues of simple poles are $\dfrac{(-a+ib)\mathrm{sech}(-a+ib)}{i8ab(a-ib)}$ and $\dfrac{(a+ib)\mathrm{sech}(a+ib)}{i8ab(a+ib)}$
from the poles along the imaginary axis, we have
$\displaystyle \sum_n \dfrac{z(n)}{[(z(n)-a)^2+b^2][(z(n)+a)^2+b^2]}$, $z(n)=i(2n+1)\pi$ where $n\in\mathbb{Z}$