Hi guys I'm solving this integral : $$\int_{+\partial D}\frac{\sin(2zj)}{z(z^{2}+\frac{\pi^{2}}{4})^{2}}dz\,,$$ where $D=\{z\in\mathbb{C}:|z|<\pi \}$
I have found that for $z=0$ the residue is $0$ for $z=\frac{-\pi}{2}$ and $z=\frac{\pi}{2}$ ( that are simple poles) the residue are $\frac{4}{\pi^{3}}$ and $\frac{-4}{\pi^{3}}$ so summing every residue the sum is 0 so the integral is $0$. Unfortunately I don't have the solution, and I can't know if I'm doing it right.