Cotangent residues sum a rational function over the integers

Let $F(z)=\pi \cot (\pi z)f(z)$. By [L1], $\sin (\pi z)$ has a simple zero at [L1, L2, algebra] each integer $n$, so applied to $\pi \cos (\pi z)f(z)\mathrm{/}\sin (\pi z)$ gives $\mathrm{Res}(F,n)=\frac{\pi \cos (\pi n)f(n)}{\pi \cos (\pi n)}=f(n)\mathrm{.}$

Integrate $F$ around the rectangle with vertices $N+\frac{1}{2}\pm iN$ and $-N-\frac{1}{2}\pm iN$. On the vertical sides one has $\cot (\pi (x+iy))=\mp i\tanh (\pi y)$ because $x=\pm (N+\frac{1}{2})$, so $\mathrm{\mid }\cot \mathrm{\mid }$ is uniformly bounded there. On the horizontal sides $\cot (\pi (x\pm iN))$ tends uniformly to $\mp i$ as $N\to \mathrm{\infty }$. Since $f(z)=O(z^{-2})$, the integrand is $O(z^{-2})$ on every side, and the boundary integral tends to $0$.

By [L3], the sum of all residues of $F$ inside the rectangle is therefore $0$. Those residues are the integer residues from step 1.1 together with the nonintegral poles of $f$. Letting $N\to \mathrm{\infty }$ yields ${\sum }_{n\in \mathbb{Z}}f(n)+{\sum }_{a\mathrm{\notin }\mathbb{Z}}\mathrm{Res}(\pi \cot (\pi z)f(z),a)=0,$ which is the stated identity.