Past exam of the algebra and analysis II for mathematics and philosophy course of the University of Oxford / 2013 / 5 / d / Solution

Periodicity reduces all the indicated vertical segments to the compact segment $1/2+i[-1,1]$, which misses every pole. Continuity therefore gives the first uniform bound. In fact an explicit identity gives a stronger estimate for every real $y$:Thus we can take $D_1=\pi$.

Write $L=m+1/2$. On the vertical sides of the square the same bound holds, since both real coordinates are half-integers. On either horizontal side $|\mathrm{Im}z|=L\ge 1/2$. For $\mathrm{Im}z=y>0$, put $w=e^{2\pi iz}$, so $|w|=e^{-2\pi y}$. Usinggives $|g(z)|\le \pi (1+e^{-2\pi y})/(1-e^{-2\pi y})\le \pi \coth (\pi /2)$. Conjugation gives the same estimate below the real axis. Hence an explicit bound valid on every square isFor $L>|b|$, the contour has length $8L$ and every point on it has $|z|\ge L$. The reverse triangle inequality yields $|z^2-b^2|\ge L^2-b^2$, so the ML inequality givesOrient the square positively. Its enclosed poles at the integers $-m,\dots ,m$ have residues $1/(n^2-b^2)$. Since $b$ is nonintegral, the additional poles $b$ and $-b$ are distinct from those integers and from one another. Their residues are $g(b)/(2b)$ and $g(-b)/(-2b)$, both equal to $g(b)/(2b)$ because $g$ is odd. Their combined contribution is therefore $\pi \cot (\pi b)/b$.

The residue theorem givesThe series is absolutely convergent, since its terms are $O(n^{-2})$. Taking the limit, then including the $n=0$ term, yieldsThis is cotangent contour summation. Keeping the $n=0$ term separate until the last step determines the sign of the $1/(2b^2)$ contribution.