Reciprocal times Derivative of Gamma Function

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Theorem

Let $\Gamma$ denote the Gamma function.


Then:

$\displaystyle \dfrac {\Gamma\,' \left({z}\right)} {\Gamma \left({z}\right)} = -\gamma + \sum_{n \mathop = 1}^\infty \left({\frac 1 n - \frac 1 {z + n - 1} }\right)$

where:

$\Gamma\,' \left({z}\right)$ denotes the derivative of the Gamma function
$\gamma$ denotes the Euler-Mascheroni constant.


Proof

From the Weierstrass form of the Gamma function:

$\displaystyle \frac 1 {\Gamma \left({z}\right)} = z e^{\gamma z} \prod_{n \mathop = 1}^\infty \left({\left({1 + \frac z n}\right) e^{-z / n} }\right)$

Taking the reciprocal of both sides:

$\displaystyle \Gamma \left({z}\right) = \frac {e^{-\gamma z}} z \prod_{n \mathop = 1}^\infty \frac {e^{z/n}} {1 + \frac z n}$

Taking the derivative of both sides:

\(\displaystyle \Gamma\,' \left({z}\right)\) \(=\) \(\displaystyle -\frac {e^{-\gamma z} \left({1 + \gamma z}\right)} {z^2} \prod_{n \mathop = 1}^\infty \left({\frac {e^{z / n} } {\left({1 + \frac z n}\right)} }\right) + \frac {e^{-\gamma z} } z \sum_{n \mathop = 1}^\infty \left({\frac z {n \left({z + n}\right)} \prod_{i \mathop = 1}^\infty \frac {e^{z / i} } {1 + \frac z i} }\right)\) General Product Rule for Derivatives
\(\displaystyle \) \(=\) \(\displaystyle -\frac {e^{-\gamma z} \left({1 + \gamma z}\right)} {z^2} \frac z {e^{-\gamma z} } \Gamma \left({z}\right) + \frac {e^{-\gamma z} } z \sum_{n \mathop = 1}^\infty \left({\frac z {n \left({z + n}\right)} \frac z {e^{-\gamma z} } \Gamma \left({z} \right) }\right)\) simplifying the product notation
\(\displaystyle \) \(=\) \(\displaystyle -\frac{1 + \gamma z} z \Gamma \left({z}\right) + \sum_{n \mathop = 1}^\infty \left({\frac {z \Gamma \left({z} \right)} {n \left({z + n}\right)} }\right)\) simplifying


Dividing both sides by $\Gamma \left({z}\right)$:

\(\displaystyle \frac {\Gamma\,' \left({z}\right)} {\Gamma \left({z} \right)}\) \(=\) \(\displaystyle -\frac {1 + \gamma z} z + \sum_{n \mathop = 1}^\infty \left({ \frac z {n \left({z + n}\right)} }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle -\gamma - \frac 1 z + \sum_{n \mathop = 1}^\infty \left({ \frac 1 n - \frac 1 {z + n} }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle -\gamma + \sum_{n \mathop = 1}^\infty \left({ \frac 1 n - \frac 1 {z + n - 1} }\right)\) rearranging the series

$\blacksquare$


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