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.


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 $n\in\mathbb{Z}$

  • $\begingroup$ Essentialy you have to calculate the sum of different residues in the uhp. $\endgroup$ – tired Feb 10 '16 at 13:04
  • 2
    $\begingroup$ please replace $sech(z)$ by $2/(e^z + e^{-z})$ or $1/\cos(iz)$. your integrand is a meromorphic function so your contour is the upper semicircle, no need to define its argument (the function doesn't have multiple branches as $\sqrt{z}$ so $f(z) = f(z) e^{2i\pi}$). and it is clear that $|f(z)| \to 0$ very fast on the upper half circle whose radius $\to \infty$. so you can now apply the residue theorem, where is the problem ? (assuming $a,b$ don't make the function having poles on the real axis) $\endgroup$ – reuns Feb 10 '16 at 13:11
  • $\begingroup$ When I apply the residue theorem with $z=i(2n+1)\pi$ poles, I get a Matsubara summation [en.wikipedia.org/wiki/Matsubara_frequency] and I'm not sure how to deal with that. $\endgroup$ – Medulla Oblongata Feb 10 '16 at 13:17
  • $\begingroup$ ((above I meant $f(z) = f(z e^{2i \pi})$ i.e. it has no branches)) your infinite number of residues have to be summable if the integrand respects what I wrote and the integral on the real axis is convergent (to prove that, start with a finite radius $R$, which contains only a finite number of poles, and let $R \to \infty$) so I guess you made some mistakes calculating the residues : write them $\endgroup$ – reuns Feb 10 '16 at 13:26
  • $\begingroup$ what is $z(n)?$ i assume $i(2n+1)\pi$? $\endgroup$ – tired Feb 10 '16 at 13:51

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