Theorem: Mittag-Leffler Expansion for Cotangent Function

:$\ds \pi \cot \pi z = \frac 1 z + 2 \sum_{n \mathop = 1}^\infty \frac z {z^2 - n^2}$
where:
:$z \in \C$ is not an integer
:$\cot$ is the cotangent function.

Proof:
Following the method of Herglotz, define $/map f x = \pi \cot \pi x$. 
This $f$ is seen to obey four important properties: 
:$1$) It is continuous on its domain $\R\setminus \Z$, 
:$2$) It is periodic with period $1$, 
:$3$) It obeys the functional equation $f(\frac x 2) + f(\frac{x+1} 2) = 2f(x)$, and 
:$4$) It holds that $0 = \lim_{x\to 0} \map f x - \frac 1 x$. 
Now we let $g:\R\setminus \Z \to \R$ represent the desired partial fraction decomposition as below.
{{begin-eqn}}
{{eqn | l = \map g x
      | r = \frac 1 x + 2x \sum_{n=1}^{\infty} \frac 1 {x^2 - n^2}
      | c = 
}}
{{eqn | r = \frac 1 x + \sum_{n=1}^{\infty} \paren {\frac 1 {x - n} + \frac 1 {x + n} }
      | c = 
}}
{{eqn | r = \frac 1 x + \sum_{n=1}^{\infty} \paren {\frac 1 {x - n} + \frac 1 {x - \paren {- n} } }
      | c = 
}}
{{eqn | r = \lim_{N\to \infty} \sum_{n=-N}^N \frac 1 {x - n}
      | c = 
}}
{{end-eqn}}
We show that $g$ obeys those same four properties, which then forces $f = g$. 
Using the first representation above, apply the Comparison Test and Convergence of P-Series to find that $g$ converges. 
Moreover $g$ converges uniformly over any compact domain, so $g$ is continuous by the Uniform Limit Theorem. 
Using the second representation above, we find $\lim_{x\to 0} \map g x - \frac 1 x = 0$. 
Next, we see $g$ is periodic.
{{begin-eqn}}
{{eqn | l = \map g {x + 1}
      | r = \lim_{N\to \infty} \sum_{n=-N}^N \frac 1 {\paren {x + 1} - n}
      | c =
}}
{{eqn | r = \lim_{N\to \infty} \sum_{n=-N - 1}^{N - 1} \frac 1 {\paren {x + 1} - \paren {n + 1} }
      | c = 
}}
{{eqn | r = \lim_{N\to \infty} \paren {\frac 1 {x - (-N - 1) } - \frac 1 {x - N} + \sum_{n=-N}^N \frac 1 {x - n} }
      | c = 
}}
{{eqn | r = \map g x
      | c = $\lim_{N\to \infty} \dfrac 1 N$ equals Zero 
}}
{{end-eqn}}
Finally, we observe that $g$ obeys the required functional equation.
{{begin-eqn}}
{{eqn | l = \map g {\frac x 2} + \map g {\frac {x + 1} 2}
      | r = \lim_{N\to \infty} \paren {\sum_{n=-N}^N \frac 1 {\frac{x}{2} - n} + \sum_{n=-N}^N \frac 1 {\frac { x + 1} 2 -n} }
      | c = 
}}
{{eqn | r = \lim_{N\to \infty} \paren {\sum_{n=-N}^N \frac 2 {x - 2n} + \sum_{n=-N}^N \frac 2 {x - \paren {2 n - 1} } }
      | c = 
}}
{{eqn | r = \lim_{N\to \infty} 2 \sum_{n=-2N - 1}^{2 N} \frac 1 {x - n}
      | c = 
}}
{{eqn | r = 2 \map g x
      | c = 
}}
{{end-eqn}}
Now define $\map h x = \map f x - \map g x$, using $\map h n = 0$ for all $n \in \Z$. 
By proxy of the established properties of $f$ and $g$, it holds that $h$ is continuous on $\R\setminus \Z$, periodic with period $1$, and obeys the same functional equation. 
Moreover $h$ is continuous at $0$, and therefore on all $\Z$ due to periodicity, as shown below.
{{begin-eqn}}
{{eqn | l = \lim_{x\to 0} \map h x
      | r = \lim_{x\to 0} \map f x - \map g x
      | c = 
}}
{{eqn | r = \lim_{x\to 0} \paren {\map f x - \frac 1 x} - \paren {\map g x - \frac 1 x}
      | c = 
}}
{{eqn | r = 0
      | c = 
}}
{{end-eqn}}
By the Weierstrass Extreme Value Theorem, we find $x_0 \in \closedint 0 1$ such that all $x \in \closedint 0 1$ obey $\size {\map h x} \leq \size {\map h {x_0} }$. 
By periodicity, this extends to all $x \in \R$. 
Letting $M = \size {\map h {x_0} }$, the functional equation gives $|h(\frac {x_0} 2) + h(\frac {x_0 + 1} 2)| = 2M$. 
By the Triangle Inequality for Real numbers and the maximality of $M$, we infer:
:$\ds 2M \leq \size {\map h {\dfrac {x_0} 2 } } + \size {\map h {\dfrac {x_0 + 1} 2 } } \leq \size {\map h {\dfrac {x_0} 2 } } + M \leq 2M$
Therefore $\size {\map h {\dfrac {x_0} 2 } } + M = 2M$ so that $\size {\map h {\dfrac {x_0} 2 } } = M$. 
By induction on $n$, infer $\size {\map h {\dfrac {x_0} {2^n} } } = M$ for all $n\in \N$. 
Finally, we take a limit.
:$\ds M = \lim_{n\to\infty} \left|h\left(\frac{x_0}{2^n}\right)\right| = \size {\map h 0} = 0$
It follows that $M = 0$, and since $|h(x)| \leq M$ for all $x$ then $\map h x = 0$, therefore $f = g$.
{{qed}}