Real integral with residue method and infinite singularities

Question

$$I=\int_0^\infty{\sin(ax)\over e^{2\pi x}-1}dx $$ I extend to the complex plane, so i have: $ f(z)={e^{iaz}\over e^{2\pi z}-1}$. The singularities are in $ z= ik, k\in \mathbb{Z}$. $$ \lim_{z \to ik}{e^{iaz}\over e^{2\pi z}-1}=\infty$$ so the singularities are poles (I'm not sure if this is correct), and :$$ \lim_{z \to ik}{e^{iaz}(z-ik)\over e^{2\pi z}-1}=\frac{e^{-ak}}{2\pi} $$ This proves that they are simple poles and the residue in $ik$ is $\frac{e^{-ak}}{2\pi}$. Now I don't know what integration path should I take, since the integral is from zero to infinity.