Suppose $N$ integral, $N\geq |u|$, prove that $$ \lim_{N\rightarrow\infty}\int_{|z|=N+\frac{1}{2}}\frac{\pi\cot\pi z}{(u+z)^2}dz =0. $$
I have got $$ \left|\int_{|z|=N+\frac{1}{2}}f(z)dz\right|\leq\frac{2\pi^2(N+\frac{1}{2})}{{(N+\frac{1}{2}-|u|})^2}\sup_{|z|=N+\frac{1}{2}}|\cot z|. $$
If we can prove that $\sup_{|z|=N+\frac{1}{2}}|\cot z|=o(N) (N\rightarrow \infty)$, then we complete the proof. But I stuck here. And I want to use the result and the residue formula to prove an indentity, so the formula is of no use here. Appreciate any help!
Hint: Consider the square closed contour $$ Q:\quad (\nu,i\nu)\to(-\nu,i\nu)\to(-\nu,-i\nu)\to(\nu,-i\nu)\to (\nu,i\nu) $$ where $\nu=N+1/2$.
Since the function $f(z)$ is analytic inside the area enclosed by $Q$ and $C:\ |z|=(N+1/2)$ the integrals along both contours are equal.
Now recall that $$ \cot(x+iy)=\frac{\sin 2x-i\sinh2y}{\cosh2y-\cos2x}, $$ so that $$ |\cot(x+iy)|^2=\frac{\sin^22x+\sinh^22y}{(\cosh2y-\cos2x)^2} =\frac{\cosh^22y-\cos^22x}{(\cosh2y-\cos2x)^2}= \frac{\cosh2y+\cos2x}{\cosh2y-\cos2x}. $$