Digamma Additive Formula

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Theorem

Let $n \in \N_{>0}$ where $\N_{>0}$ denotes the non-zero natural numbers.

Let $z \in \C \cap z \notin \set {-\dfrac m n: m \in \N}$


Then:

$\ds \map \psi {n z} = \frac 1 n \sum_{k \mathop = 0}^{n - 1} \map \psi {z + \frac k n} + \ln n$

where:

$\psi$ is the digamma function
$\ln$ is the complex natural logarithm.


Corollary

Let $n \in \N_{>0}$ where $\N_{>0}$ denotes the non-zero natural numbers.

Then:

$\ds \sum_{k \mathop = 1}^{n - 1} \map \psi {\frac k n} = -\paren {n - 1} \gamma - n \ln n$

where:

$\psi$ is the digamma function
$\ln$ is the complex natural logarithm.
$\gamma$ denotes the Euler-Mascheroni constant.


Proof

\(\ds \paren {2 \pi}^{\paren {n - 1} / 2} n^{1/2 - n z} \map \Gamma {n z}\) \(=\) \(\ds \prod_{k \mathop = 0}^{n - 1} \map \Gamma {z + \frac k n}\) Gauss Multiplication Formula
\(\ds \leadsto \ \ \) \(\ds \paren {\frac {n - 1} 2 } \map \ln {2 \pi} + \paren {\frac 1 2 - n z} \map \ln n + \map \ln {\map \Gamma {n z} }\) \(=\) \(\ds \sum_{k \mathop = 0}^{n - 1} \map \ln {\map \Gamma {z + \frac k n} }\) taking the complex natural logarithm of both sides, Sum of Logarithms
Work In Progress
In particular: There still needs to be a complex form of Sum of Logarithms
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\(\ds \leadsto \ \ \) \(\ds -n \map \ln n + n \map \psi {n z}\) \(=\) \(\ds \sum_{k \mathop = 0}^{n - 1} \map \psi {z + \frac k n}\) taking the derivative of both sides, Definition of Digamma Function
\(\ds \leadsto \ \ \) \(\ds \map \psi {n z}\) \(=\) \(\ds \frac 1 n \sum_{k \mathop = 0}^{n - 1} \map \psi {z + \frac k n} + \ln n\) dividing by $n$ and moving $\ln n$ to right hand side

$\blacksquare$


Also see


Sources

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