Reciprocal times Derivative of Gamma Function/Proof 1
Jump to navigation
Jump to search
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 |
$\blacksquare$