Polygamma Reflection Formula/Proof 1

Theorem

Let $z \in \C \setminus \Z$.

Let $\psi_n$ denote the $n$th polygamma function.

Then:

$\map {\psi_n} z - \paren {-1}^n \map {\psi_n} {1 - z} = -\pi \dfrac {\d^n} {\d z^n} \cot \pi z$

Proof

Lemma

The expression:

$\map \psi z - \map \psi {1 - z}$

is defined on the domain $\C \setminus \Z$.

That is, on the set of complex numbers but specifically excluding the integers.

$\Box$

By definition:

$\map {\psi_n} z = \dfrac {\d^n} {\d z^n} \map \psi z$

where:

$\psi$ denotes the digamma function

$z \in \C \setminus \Z_{\le 0}$.

Then:

\(\ds \map \psi z - \map \psi {1 - z}\)

\(=\)

\(\ds -\pi \cot \pi z\)

Digamma Reflection Formula

\(\ds \leadsto \ \ \)

\(\ds \dfrac {\d^n} {\d z^n} \map \psi z - \dfrac {\d^n} {\d z^n} \map \psi {1 - z}\)

\(=\)

\(\ds -\pi \dfrac {\d^n} {\d z^n} \cot \pi z\)

taking $n$th derivative

\(\ds \leadsto \ \ \)

\(\ds \map {\psi_n} z - \paren {-1}^n \map {\psi_n} {1 - z}\)

\(=\)

\(\ds -\pi \dfrac {\d^n} {\d z^n} \cot \pi z\)

Definition of Polygamma Function

Finally, from the Lemma, we note that:

$\map \psi z - \map \psi {1 - z}$

is defined on the domain $\C \setminus \Z$.

$\blacksquare$