Reciprocal times Derivative of Gamma Function/Proof 1

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

\(\ds \leadsto \ \ \)

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

\(=\)

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

differentiating with respect to $z$

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds -\frac {e^{-\gamma z} \paren {1 + \gamma z} } {z^2} \prod_{n \mathop = 1}^\infty \paren {\frac {e^{z / n} } {\paren {1 + \frac z n} } } + \frac {e^{-\gamma z} } z \sum_{n \mathop = 1}^\infty \paren {\frac z {n \paren {z + n} } \prod_{i \mathop = 1}^\infty \frac {e^{z / i} } {1 + \frac z i} }\)

Product Rule for Derivatives

\(\ds \)

\(=\)

\(\ds -\frac {e^{-\gamma z} \paren {1 + \gamma z} } {z^2} \frac z {e^{-\gamma z} } \map \Gamma z + \frac {e^{-\gamma z} } z \sum_{n \mathop = 1}^\infty \paren {\frac z {n \paren {z + n} } \frac z {e^{-\gamma z} } \map \Gamma z}\)

simplifying the continued product

\(\ds \)

\(=\)

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

simplifying

\(\ds \leadsto \ \ \)

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

\(=\)

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

dividing both sides by $\map \Gamma z$

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

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

rearranging the series

$\blacksquare$