Showing that $\int \frac{ \sinh (az)}{\sinh (\pi z)} \, e^{ibz} \, dz $ vanishes along three sides of a rectangle in the upper half-plane One of several ways to evaluate $$\int_{0}^{\infty} \frac{\sinh (ax)}{\sinh (\pi x)} \, \cos (bx) \, dx \, ,  \quad  \, |a|< \pi,$$  is to sum the residues of $$ f(z) = \frac{\sinh (az)}{\sinh (\pi z)} \,e^{ibz}$$ in the upper half-plane.
But if you restrict $b$ to positive values, how do show that $\int f(z) \, dz$ vanishes along the right, left, and upper sides of a rectangle with vertices at $\pm N, \pm N + i\left(N+\frac{1}{2} \right)$ as $N \to \infty$ through the positive integers?
I think we can use the M-L inequality (in combination with the triangle and reverse triangle inequalities) to show that that integral vanishes along the vertical sides of the rectangle.
But showing that the integral vanishes along the top of the rectangle seems a bit tricky.

So that my question doesn't remain unanswered, I'm going to post an answer using the hints that Marko Riedel provided in the comments.

On the right side of the rectangle (and similiarly on the left side of the rectangle), we have 
\begin{align} \left| \int_{0}^{N + \frac{1}{2}}  \frac{\sinh\big(a(N+it)\big)}{\sinh\big(\pi(N+it)\big)} \, i e^{ib(N+it)}  \, dt \right| &\le \int_{0}^{N + \frac{1}{2}} \left| \frac{\sinh\big(a(N+it)\big)}{\sinh\big(\pi(N+it)\big)} \, i e^{ib(N+it)} \right| \, dt \\  &\le  \int_{0}^{N+\frac{1}{2}} \frac{e^{aN}+e^{-aN}}{e^{\pi N}-e^{-\pi N}} \, e^{-bt} \, dt \\ &\le \left(N+ \frac{1}{2}\right) \frac{e^{aN}+e^{-aN}}{e^{\pi N}-e^{-\pi N}} \to 0 \ \text{as} \  N \to \infty\end{align} since $|a| < \pi$.
And on the upper side of the rectangle, we have
$$ \begin{align} \left| \int_{-N}^{N} \frac{\sinh \big(a(t+i(N+\frac{1}{2}) \big)}{\sinh \big(\pi (t + i (N+\frac{1}{2}) \big)} \,  e^{ib\left(t+i(N+1/2)\right)}  \, dt \right| &\le \int_{-N}^{N} \left|\frac{\sinh \big(a(t+i(N+\frac{1}{2}) \big)}{\sinh \big(\pi (t + i (N+\frac{1}{2}) \big)} \,  e^{ib\left(t+i(N+1/2)\right)} \right| \, dt \\ &\le \frac{e^{-b (N+1/2)}}{2} \int_{-N}^{N} \frac{e^{at}+e^{-at}}{\cosh \pi t}  \, dt \\ &< e^{-b(N+1/2)} \int_{-\infty}^{\infty} \frac{\cosh(at)}{\cosh(\pi t)} \, dt \to 0 \ \text{as} \ N \to \infty \end{align} $$ since the restriction $|a| < \pi$ means that$\int_{-\infty}^{\infty} \frac{\cosh(at)}{\cosh(\pi t)} \, dt$ converges.
