Integral Form of Polygamma Function
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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 $1$ |
$\blacksquare$