Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2016 / S1 / 1

The simple poles occur at $z=n-\alpha$, $n\in \mathbb{Z}$, since the derivative of the sine there is $\pi (-1{)}^n$. ThusThere is also a double pole at zero, not included in that list because $0<\alpha <1$. Its residue, needed for the next contour calculation, is the derivative of $\csc \pi (z+\alpha )$ at zero, namely $-\pi \csc (\pi \alpha )\cot (\pi \alpha )$.

On the square, $|z|\ge N$. The identity $|\sin \pi (x+\alpha +iy){|}^2={\sin }^2\pi (x+\alpha )+{\sinh }^2\pi y$ bounds the sine away from zero on the vertical sides because $x=\pm N$; on horizontal sides it grows exponentially. Therefore the whole contour integral is $O(N^{-1})$ or smaller and vanishes. The residue theorem then giveswhich proves $\boxed{\sum _{n\in \mathbb{Z}}(-1{)}^n/(n-\alpha {)}^2={\pi }^2\csc \pi \alpha \cot \pi \alpha }$. Absolute convergence justifies the limiting residue sum.

Separate the $n=0$ term and pair $n$ with $-n$. For small $\alpha$,Compare the constant and quadratic coefficients with the supplied expansion after subtracting ${\alpha }^{-2}$. It follows thatThe paired expansion is uniformly summable on any sufficiently small closed interval of $\alpha$, so coefficient comparison is legitimate.