Reciprocal times Derivative of Gamma Function
Definition
The derivative of the logarithm of the Gamma function is called the digamma function.
$$ \psi_{0} (z) := \dfrac{d}{dz} \ln \Gamma (z) = \dfrac{\Gamma^{\prime}(z)}{\Gamma (z)} $$
Theorem
For the Gamma function $\Gamma$ and the Euler-Mascheroni constant $\gamma$, the following holds: $$ {{ \Gamma ' (z) } \over { \Gamma (z) }} = - \gamma + \sum_{n=1}^{\infty} \left( {{ 1 } \over { n }} - {{ 1 } \over { z + n - 1 }} \right) $$
Proof 1
Weierstrass’s product representation for the Gamma function: For the Gamma function $\Gamma : (0, \infty) \to \mathbb{R}$, the following holds: $$ {1 \over \Gamma (x)} = x e^{\gamma x } \lim_{n \to \infty} \prod_{k=1}^{n} \left( 1 + {x \over k} \right) e^{- {x \over k} } $$
Taking the reciprocal of the Weierstrass product representation gives: $$ \Gamma (z) = {{ e^{-\gamma z} } \over { z }} \prod_{n=1}^{n} {{ e^{z/n} } \over { 1 + z/n }} $$ According to the product rule of differentiation, $$ \begin{align*} & \Gamma ' (z) \\ =& - {{ e^{-\gamma z} \left( 1 + \gamma z \right) } \over { z^{2} }} \prod_{n=1}^{n} {{ e^{z/n} } \over { 1 + z/n }} + {{ e^{-\gamma z} } \over { z }} \sum_{n=1}^{\infty} \left[ {{ z } \over { n \left( z + n \right) }} \prod_{k=1}^{\infty} {{ e^{z/k} } \over { 1 + z/k }} \right] \\ =& - {{ e^{-\gamma z} \left( 1 + \gamma z \right) } \over { z^{2} }} {{ z } \over { e^{-\gamma z} }} \Gamma (z) + {{ e^{-\gamma z} } \over { z }} \sum_{n=1}^{\infty} \left[ {{ z } \over { n \left( z + n \right) }} {{ z } \over { e^{-\gamma z} }} \Gamma (z) \right] \\ =& - {{ 1 + \gamma z } \over { z }} \Gamma (z) + \Gamma (z) \sum_{n=1}^{\infty} \left[ {{ z } \over { n \left( z + n \right) }} \right] \end{align*} $$ and dividing both sides by $\Gamma (z)$ gives: $$ \begin{align*} {{ \Gamma ' (z) } \over { \Gamma (z) }} =& - {{ 1 + \gamma z } \over { z }} + \sum_{n=1}^{\infty} \left[ {{ z } \over { n \left( z + n \right) }} \right] \\ =& - \gamma - {{ 1 } \over { z }} + \sum_{n=1}^{\infty} \left[ {{ 1 } \over { n }} - {{ 1 } \over { z + n }} \right] \\ =& - \gamma + \sum_{n=1}^{\infty} \left[ {{ 1 } \over { n }} - {{ 1 } \over { z + n - 1 }} \right] \end{align*} $$
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