Mittag-Leffler Expansion for Cotangent Function/Proof 1

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.

Outline of proof

Informally, we can say:

$\pi \cot \pi z = \map {\dfrac \d {\d z} } {\ln \map \sin {\pi z} }$.

We then use the Euler Formula for Sine Function to write $\map \sin {\pi z}$ as an infinite product and differentiate its logarithm.

Formally, we work with logarithmic derivatives and use Logarithmic Derivative of Infinite Product of Analytic Functions.

Proof

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

\(\ds \)

\(=\)

\(\ds \frac 1 z + 2 \sum_{n \mathop = 1}^\infty \frac z {z^2 - n^2}\)

$\blacksquare$