# Mittag-Leffler Expansion for Cotangent Function

 It has been suggested that this page or section be merged into Partial Fractions Expansion of Cotangent. (Discuss)
 It has been suggested that this page or section be merged into Series Expansion for Pi Cotangent of Pi Lambda. (Discuss)

## 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.

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

## Proof 1

Let $\mathcal L$ denote the logarithmic derivative.

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

 $\ds \pi \cot \pi z$ $=$ $\ds \mathcal L \left({\sin \left({\pi z}\right)}\right)$ Primitive of Cotangent Function, or a complex version thereof $\ds$ $=$ $\ds \mathcal L \left({\pi z \prod_{n \mathop = 1}^\infty \left({1 - \frac {z^2} {n^2} }\right)}\right)$ Euler Formula for Sine Function $\ds$ $=$ $\ds \mathcal L \left({\pi z}\right) + \sum_{n \mathop = 1}^\infty \mathcal L \left({1 - \frac {z^2} {n^2} }\right)$ Logarithmic Derivative of Infinite Product of Analytic Functions $\ds$ $=$ $\ds \frac \pi {\pi z} + \sum_{n \mathop = 1}^\infty \frac 1 {1 - \frac {z^2} {n^2} } \cdot \frac {\mathrm d} {\mathrm d z} \left({1 - \frac {z^2} {n^2} }\right)$ 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 $\displaystyle \map g z = \sum_{n \mathop = 1}^\infty z^n \map \zeta {2 n}$ 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:

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

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.