Reciprocal times Derivative of Gamma Function/Proof 2

Theorem

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

Then:

$\ds \dfrac {\map {\Gamma'} z} {\map \Gamma z} = -\gamma + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {z + n - 1} }$

where:

$\map \Gamma z$ denotes the Gamma function

$\map {\Gamma'} z$ denotes the derivative of the Gamma function

$\gamma$ denotes the Euler-Mascheroni constant.

Proof

\(\ds \frac 1 {\map \Gamma z}\)

\(=\)

\(\ds z e^{\gamma z} \prod_{n \mathop = 1}^\infty \paren {\paren {1 + \frac z n} e^{-z / n} }\)

Weierstrass Form of Gamma Function

\(\ds \leadsto \ \ \)

\(\ds \map \Gamma z\)

\(=\)

\(\ds \frac {e^{-\gamma z} } z \prod_{n \mathop = 1}^\infty \frac {e^{z/n} } {1 + \frac z n}\)

reciprocal of both sides

\(\ds \leadsto \ \ \)

\(\ds \map \ln {\map {\Gamma} z}\)

\(=\)

\(\ds \map \ln {\frac {e^{-\gamma z} } z \prod_{n \mathop = 1}^\infty \frac {e^{z/n} } {1 + \frac z n} }\)

logarithm of both sides

\(\ds \)

\(=\)

\(\ds \map \ln {e^{-\gamma z} } - \ln z + \sum_{n \mathop = 1}^\infty \paren { \map \ln {e^{z/n} } - \map \ln {1 + \frac z n} }\)

Sum of Logarithms and Difference of Logarithms

\(\ds \)

\(=\)

\(\ds -\gamma z \ln e - \ln z + \sum_{n \mathop = 1}^\infty \paren {\frac z n \ln e - \map \ln {1 + \frac z n} }\)

Logarithm of Power

\(\ds \)

\(=\)

\(\ds -\gamma z - \ln z + \sum_{n \mathop = 1}^\infty \paren {\frac z n - \map \ln {1 + \frac z n} }\)

Natural Logarithm of e is 1

\(\ds \leadsto \ \ \)

\(\ds \map {\dfrac \d {\d z} } {\map \ln {\map {\Gamma} z} }\)

\(=\)

\(\ds \map {\dfrac \d {\d z} } {-\gamma z - \ln z + \sum_{n \mathop = 1}^\infty \paren {\frac z n - \map \ln {1 + \frac z n} } }\)

differentiating with respect to $z$

\(\ds \leadsto \ \ \)

\(\ds \frac {\map {\Gamma'} z} {\map \Gamma z}\)

\(=\)

\(\ds -\gamma - \frac 1 z + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac {\frac 1 n} {1 + \frac z n} }\)

Derivative of Composite Function and Derivative of Natural Logarithm Function

\(\ds \)

\(=\)

\(\ds -\gamma - \frac 1 z + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {z + n} }\)

\(\ds \)

\(=\)

\(\ds -\gamma - \frac 1 z + \paren {\paren {\frac 1 1 - \frac 1 {z + 1} } + \paren {\frac 1 2 - \frac 1 {z + 2} } + \paren {\frac 1 3 - \frac 1 {z + 3} } + \cdots}\)

\(\ds \)

\(=\)

\(\ds -\gamma + \paren {\paren {\frac 1 1 - \frac 1 z } + \paren {\frac 1 2 - \frac 1 {z + 1} } + \paren {\frac 1 3 - \frac 1 {z + 2} } + \cdots}\)

shifting the terms with $z$ one to the right

\(\ds \)

\(=\)

\(\ds -\gamma + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {z + n - 1} }\)

$\blacksquare$