Reciprocal times Derivative of Gamma Function

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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.


Corollary 1

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

Then:

$\ds \dfrac {\map {\Gamma'} z} {\map \Gamma z} = -\gamma + \int_{\mathop \to 0}^{\mathop \to 1} \paren {\dfrac {1 - t^{z - 1} } {1 - t} } \rd t$

where:

$\map \Gamma z$ denotes the Gamma function
$\map {\Gamma'} z$ denotes the derivative of the Gamma function
$\gamma$ denotes the Euler-Mascheroni constant.


Corollary 2

Let $a$ and $b \in \C$ such that $\paren {\dfrac a {2 b} + 1} \in \C \setminus \Z_{\le 0}$ and $\paren {\dfrac a {2 b} + \dfrac 1 2 } \in \C \setminus \Z_{\le 0}$

Then:

$\ds \map \psi {\dfrac a {2 b} + 1} - \map \psi {\dfrac a {2 b} + \dfrac 1 2} = 2 b \sum_{k \mathop = 1}^\infty \dfrac {\paren {-1}^{k + 1} } {a + b k} $

where:

$\psi$ is the digamma function


Corollary 3

$\ds \map \psi {z + 1} = -\gamma + \harm 1 z$


Proof 1

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

$\blacksquare$


Proof 2

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

$\blacksquare$


Examples

Example: $\ds \sum_{k \mathop = 1}^\infty \dfrac {\paren {-1}^{k + 1} } {1 + k}$

$\ds \sum_{k \mathop = 1}^\infty \dfrac {\paren {-1}^{k + 1} } {1 + k} = 1 - \ln 2$


Example: $\ds \sum_{k \mathop = 1}^\infty \dfrac {\paren {-1}^{k + 1} } {1 + 2 k}$

$\ds \sum_{k \mathop = 1}^\infty \dfrac {\paren {-1}^{k + 1} } {1 + 2 k} = 1 - \dfrac \pi 4$


Example: $\ds \sum_{k \mathop = 1}^\infty \dfrac {\paren {-1}^{k + 1} } {1 + 3 k}$

$\ds \sum_{k \mathop = 1}^\infty \dfrac {\paren {-1}^{k + 1} } {1 + 3 k} = 1 - \dfrac 1 3 \ln 2 - \dfrac {\pi \sqrt 3} 9$


Example: $\map {H^{\paren 1} } {\dfrac 1 2}$

$\harm 1 {\dfrac 1 2} = 2 - 2 \ln 2$


Example: $\map {H^{\paren 1} } {-\dfrac 1 2}$

$\harm 1 {-\dfrac 1 2} = -2 \ln 2$


Also see


Sources

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