Mittag-Leffler Expansion for Cotangent Function/Proof 2

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.

Proof

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 $\size z < 1$:

\(\ds \pi \map \cot {\pi z}\)

\(=\)

\(\ds \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n \pi^{2 n} 2^{2 n} B_{2 n} \, z^{2 n - 1} } {\paren {2 n}!}\)

\(\ds \leadsto \ \ \)

\(\ds \pi z \map \cot {\pi z}\)

\(=\)

\(\ds 1 + \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n \pi^{2 n} 2^{2 n} B_{2 n} \, z^{2 n} } {\paren {2 n}!}\)

\(\ds \leadsto \ \ \)

\(\ds \frac {\pi z \map \cot {\pi z} - 1} {-2}\)

\(=\)

\(\ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} \pi^{2 n} 2^{2 n - 1} B_{2 n} \, z^{2 n} } {\paren {2 n}!}\)

By Riemann Zeta Function at Even Integers:

$\map \zeta {2 n} = \dfrac {\paren {-1}^{n + 1} \pi^{2 n} 2^{2 n - 1} B_{2 n} } {\paren {2 n}!}$

Thus:

\(\ds \dfrac {\pi z \map \cot {\pi z} - 1} {-2}\)

\(=\)

\(\ds \sum_{n \mathop = 1}^\infty \map \zeta {2 n} z^{2 n}\)

\(\ds \)

\(=\)

\(\ds \map g {z^2}\)

By Analytic Continuation of Generating Function of Dirichlet Series and Uniqueness of Analytic Continuation:

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

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

Thus:

\(\ds \pi z \map \cot {\pi z} - 1\)

\(=\)

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

\(\ds \leadsto \ \ \)

\(\ds \pi z \map \cot {\pi z}\)

\(=\)

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

\(\ds \leadsto \ \ \)

\(\ds \pi \map \cot {\pi z}\)

\(=\)

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

$\blacksquare$