Contour integral of $\int_{0}^\infty \frac{\sinh(kx)}{\sinh(x)}dx = \frac{1}{2}\tan{\frac{a}{2}}$

Question

From Calculating the Fourier transform of $\frac{\sinh(kx)}{\sinh(x)}$

In the case of zero $\omega$ and integral starts as 0, how do I prove that using contour integral

$\int_{0}^\infty \frac{\sinh(kx)}{\sinh(x)}dx = \frac{1}{2}\tan{\frac{k}{2}}$, $|k|<1$?

Thanks in advance.