The Mittag-Leffler expansion of pi cotangent

Fix $z\in \mathbb{C}\setminus \mathbb{Z}$. For $N$ large enough that $z,-z\in R_N^{\circ }$, let $R_N$ be the positively oriented rectangle with [L2, L3, given, algebra] vertices $\pm (N+1\mathrm{/}2)\pm i(N+1\mathrm{/}2)$ and set $F_N(w):=\frac{\pi \cot (\pi w)}{w^2-z^2}\mathrm{.}$ By [L2], the poles of $F_N$ inside $R_N$ are the integers $n$ with $\mathrm{\mid }n\mathrm{\mid }\le N$ and the points $w=\pm z$. The residue at an integer $n$ is $(n^2-z^2{)}^{-1}$, while the residues at $w=z$ and $w=-z$ sum to $\frac{\pi \cot (\pi z)}{2z}+\frac{\pi \cot (-\pi z)}{-2z}=\frac{\pi \cot (\pi z)}{z}\mathrm{.}$ Therefore [L3] gives ${\int }_{\mathrm{\partial }{R}_N}{F}_N(w)dw=2\pi i(\frac{\pi \cot (\pi z)}{z}+{\sum }_{n=-N}^N\frac{1}{n^2-z^2})\mathrm{.}$

On the vertical sides of $\mathrm{\partial }{R}_N$, write $w=\pm (N+1\mathrm{/}2)+iy$. [L1, step 1.1, algebra] $\sin (\pi w)=\pm \cosh (\pi y)$ and $\cos (\pi w)=\mp i\sinh (\pi y)$ by [L1], so $\mathrm{\mid }\cot (\pi w)\mathrm{\mid }=\mathrm{\mid }\tanh (\pi y)\mathrm{\mid }\le 1$. On the horizontal sides, $w=x\pm i(N+1\mathrm{/}2)$, and [L1] gives $\mathrm{\mid }\sin (\pi w){\mathrm{\mid }}^2={\sin }^2(\pi x)+{\sinh }^2(\pi (N+1\mathrm{/}2)),\mathrm{\mid }\cos (\pi w){\mathrm{\mid }}^2={\cos }^2(\pi x)+{\sinh }^2(\pi (N+1\mathrm{/}2)),$ so $\mathrm{\mid }\cot (\pi w)\mathrm{\mid }\le 2$. Also $\mathrm{\mid }w\mathrm{\mid }\ge N+1\mathrm{/}2$ on $\mathrm{\partial }{R}_N$, hence $\mathrm{\mid }{w}^2-z^2\mathrm{\mid }\ge \mathrm{\mid }w{\mathrm{\mid }}^2-\mathrm{\mid }z{\mathrm{\mid }}^2\ge (N+1\mathrm{/}2{)}^2-\mathrm{\mid }z{\mathrm{\mid }}^2$. Thus $\mathrm{\mid }{F}_N(w)\mathrm{\mid }\le C_z\mathrm{/}{N}^2$ on $\mathrm{\partial }{R}_N$, and since the boundary length is $8N+4$, one gets ${\int }_{\mathrm{\partial }{R}_N}{F}_N(w)dw\to 0$ as $N\to \mathrm{\infty }$.

Letting $N\to \mathrm{\infty }$ in step 1.1 and using step 2.1 yields [step 1.1, step 2.1, algebra] $\frac{\pi \cot (\pi z)}{z}=-{\sum }_{n\in \mathbb{Z}}\frac{1}{n^2-z^2}=\frac{1}{z^2}+{\sum }_{n\ge 1}\frac{2}{z^2-n^2}\mathrm{.}$ Multiplying by $z$ gives $\pi \cot (\pi z)=\frac{1}{z}+{\sum }_{n\ge 1}\frac{2z}{z^2-n^2}=\frac{1}{z}+{\sum }_{n\ge 1}(\frac{1}{z-n}+\frac{1}{z+n})\mathrm{.}$ On every compact subset of $\mathbb{C}\setminus \mathbb{Z}$, the last series is bounded termwise by $C_K\mathrm{/}{n}^2$ for all large $n$, so it converges locally uniformly there. This is exactly the claimed expansion.