# Mittag-Leffler Expansion for Cotangent Function

## Theorem

$\ds \pi \cot \pi z = \frac 1 z + 2 \sum_{n \mathop = 1}^\infty \frac z {z^2 - n^2}$

where:

$z \in \C$ is not an integer
$\cot$ is the cotangent function.

### Real Domain

Let $\alpha \in \R$ be a real number which is specifically not an integer.

$\ds \dfrac 1 \alpha + \sum_{n \mathop \ge 1} \dfrac {2 \alpha} {\alpha^2 - n^2} = \pi \cot \pi \alpha$

## Proof 1

Let $\LL$ denote the logarithmic derivative.

On the open set $\C \setminus \Z$ we have:

 $\ds \pi \cot \pi z$ $=$ $\ds \map \LL {\map \sin {\pi z} }$ Primitive of Cotangent Function, or a complex version thereof $\ds$ $=$ $\ds \map \LL {\pi z \prod_{n \mathop = 1}^\infty \paren {1 - \frac {z^2} {n^2} } }$ Euler Formula for Sine Function $\ds$ $=$ $\ds \map \LL {\pi z} + \sum_{n \mathop = 1}^\infty \map \LL {1 - \frac {z^2} {n^2} }$ Logarithmic Derivative of Infinite Product of Analytic Functions $\ds$ $=$ $\ds \frac \pi {\pi z} + \sum_{n \mathop = 1}^\infty \frac 1 {1 - \frac {z^2} {n^2} } \cdot \map {\frac \d {\d z} } {1 - \frac {z^2} {n^2} }$ Definition of Logarithmic Derivative of Meromorphic Function $\ds$ $=$ $\ds \frac 1 z - 2 \sum_{n \mathop = 1}^\infty \frac z {n^2 - z^2}$ Derivative of Power $\ds$ $=$ $\ds \frac 1 z + 2 \sum_{n \mathop = 1}^\infty \frac z {z^2 - n^2}$

$\blacksquare$

## Proof 2

Let $\map \zeta s$ be the Riemann zeta function.

Let $\ds \map g z = \sum_{n \mathop = 1}^\infty z^n \map \zeta {2 n}$ be the generating function of $\map \zeta {2 n}$

By Power Series Expansion for Cotangent Function, for $\size z < 1$:

 $\ds \pi \map \cot {\pi z}$ $=$ $\ds \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n \pi^{2 n} 2^{2 n} B_{2 n} \, z^{2 n - 1} } {\paren {2 n}!}$ $\ds \leadsto \ \$ $\ds \pi z \map \cot {\pi z}$ $=$ $\ds 1 + \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n \pi^{2 n} 2^{2 n} B_{2 n} \, z^{2 n} } {\paren {2 n}!}$ $\ds \leadsto \ \$ $\ds \frac {\pi z \map \cot {\pi z} - 1} {-2}$ $=$ $\ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} \pi^{2 n} 2^{2 n - 1} B_{2 n} \, z^{2 n} } {\paren {2 n}!}$
$\map \zeta {2 n} = \dfrac {\paren {-1}^{n + 1} \pi^{2 n} 2^{2 n - 1} B_{2 n} } {\paren {2 n}!}$

Thus:

 $\ds \dfrac {\pi z \map \cot {\pi z} - 1} {-2}$ $=$ $\ds \sum_{n \mathop = 1}^\infty \map \zeta {2 n} z^{2 n}$ $\ds$ $=$ $\ds \map g {z^2}$
$\ds \dfrac {\pi z \map \cot {\pi z} - 1} {-2} = \sum_{n \mathop = 1}^\infty \dfrac {z^2} {n^2 - z^2}$

for all of $\C$, as this is the overlap of their domains.

Thus:

 $\ds \pi z \map \cot {\pi z} - 1$ $=$ $\ds 2 \sum_{n \mathop = 1}^\infty \frac {z^2} {z^2 - n^2}$ $\ds \leadsto \ \$ $\ds \pi z \map \cot {\pi z}$ $=$ $\ds 1 + 2 \sum_{n \mathop = 1}^\infty \frac {z^2} {z^2 - n^2}$ $\ds \leadsto \ \$ $\ds \pi \map \cot {\pi z}$ $=$ $\ds \frac 1 z + 2 \sum_{n \mathop = 1}^\infty \frac z {z^2 - n^2}$

$\blacksquare$

## Source of Name

This entry was named for Magnus Gustaf Mittag-Leffler.