Digamma Reflection Formula
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Theorem
Let $z \in \C \setminus \Z$.
Let $\psi$ denote the digamma function.
Then:
- $\map \psi z - \map \psi {1 - z} = -\pi \cot \pi z$
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.
$\Box$
| \(\ds \map \Gamma z \map \Gamma {1 - z}\) | \(=\) | \(\ds \dfrac \pi {\sin \pi z}\) | Euler's Reflection Formula | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map \ln {\map \Gamma z \map \Gamma {1 - z} }\) | \(=\) | \(\ds \map \ln {\dfrac \pi {\sin \pi z} }\) | applying $\ln$ on both sides | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map \ln {\map \Gamma z} + \map \ln {\map \Gamma {1 - z} }\) | \(=\) | \(\ds \ln \pi - \map \ln {\sin \pi z}\) | Sum of Logarithms and Difference of Logarithms | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \dfrac \d {\d z} \map \ln {\map \Gamma z} + \dfrac \d {\d z} \map \ln {\map \Gamma {1 - z} }\) | \(=\) | \(\ds \dfrac \d {\d z} \ln \pi - \dfrac \d {\d z} \map \ln {\sin \pi z}\) | differentiating with respect to $z$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \dfrac {\map {\Gamma'} z} {\map \Gamma z} - \dfrac {\map {\Gamma'} {1 - z} } {\map \Gamma {1 - z} }\) | \(=\) | \(\ds 0 - \pi \cot \pi z\) | Derivative of Natural Logarithm Function, Derivative of Sine Function, Chain Rule for Derivatives, Derivative of Constant | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map \psi z - \map \psi {1 - z}\) | \(=\) | \(\ds -\pi \cot \pi z\) | Definition of Digamma Function |
Finally, from the Lemma, we note that:
- $\map \psi z - \map \psi {1 - z}$
is defined on the domain $\C \setminus \Z$.
$\blacksquare$
Also see
Sources
- Weisstein, Eric W. "Digamma Function." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/DigammaFunction.html