I am trying to calculate the following improper integral:

$$ \int_{-\infty }^{\infty }\frac{x}{x-\sinh \left( \frac{\pi x}{2}\right) }ds $$

I try to calculate it using the residues theorem. Therefore I extend the integrand away from the real axis as:

$$ \frac{z}{z-\sinh \left( \frac{\pi z}{2}\right) } $$

The integrand has only poles on $ z=\pm i $, because at $z=0$ the limit is $ -\frac{2}{\pi-2}\ $. Then I consider a contour integral that comprises the integral along the real axis from −R to +R together with the integral along the semi-circular arc on the upper half-plane. When R→∞ the contribution from the straight line part approaches the required integral because the curved section vanishes in the limit.

According to this, the integral should be equal to: $$ 2 \pi i Res(\frac{z}{z-\sinh \left( \frac{\pi z}{2}\right) },z=i)= 2 \pi i (i)=-2\pi=-6.283185308 $$

But when I numerically compute the above integral has a value of $ -3.838888154$.

What I am doing wrong?

Thanks a lot!


There are further poles, as you can see e.g. in these plots. I doubt that these poles can be expressed in closed form.

  • 1
    $\begingroup$ Totally True!! In view of this, I think that it is not possible to obtain a closed form solution for this integral. Thanks a lot for your help and time! $\endgroup$ – jesusvaleo Jun 7 '18 at 11:16

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