Integral Form of Polygamma Function

Theorem

Let $z$ be a complex number with a positive real part.

Then:

$\ds \map {\psi_n} z = \paren {-1}^{n + 1} \int_0^\infty \frac {t^n e^{-z t} } {1 - e^{-t} } \rd t$

where $\map {\psi_n} z$ denotes the $n$th polygamma function.

Corollary

Let $z$ be a complex number with a positive real part.

Then:

$\ds \map {\psi_n} z= -\int_0^1 \frac {u^{z - 1} \paren {\ln u}^n } {1 - u} \rd u$

where $\map {\psi_n} z$ denotes the $n$th polygamma function.

Proof

From Gauss's Integral Form of Digamma Function, we have:

$\ds \map \psi z = \int_0^\infty \paren {\frac {e^{-t} } t - \frac {e^{-z t} } {1 - e^{-t} } } \rd t$

Therefore:

\(\ds \map \psi z\)

\(=\)

\(\ds \int_0^\infty \paren {\frac {e^{-t} } t - \frac {e^{-z t} } {1 - e^{-t} } } \rd t\)

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds \dfrac {\d^n} {\d z^n} \paren {\int_0^\infty \paren {\frac {e^{-t} } t - \frac {e^{-z t} } {1 - e^{-t} } } \rd t }\)

taking $n$th derivative

\(\ds \)

\(=\)

\(\ds \paren {-1}^{n + 1} \int_0^\infty \frac {t^n e^{-z t} } {1 - e^{-t} } \rd t\)

Derivative of Exponential Function: Corollary $1$

$\blacksquare$

Also see

• Gauss's Integral Form of Digamma Function