Improper integral involving sinh

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!