📂Functions

Derivative of Gamma Function at 1

Theorem

For the Gamma function $\Gamma$ and the Euler-Mascheroni constant $\gamma$, the following holds: $$\Gamma ' (1) = - \gamma$$

Proof ¹

The derivative of the Gamma function times its reciprocal: $${{ \Gamma ' (z) } \over { \Gamma (z) }} = - \gamma + \sum_{n=1}^{\infty} \left( {{ 1 } \over { n }} - {{ 1 } \over { z + n - 1 }} \right)$$

Substituting $z = 1$ into the product of the derivative and the reciprocal of the Gamma function gives $$\begin{align*} {{ \Gamma ' (1) } \over { \Gamma (1) }} =& - \gamma + \sum_{n=1}^{\infty} \left( {{ 1 } \over { n }} - {{ 1 } \over { 1 + n - 1 }} \right) \\ =& - \gamma + \sum_{n=1}^{\infty} \left( {{ 1 } \over { n }} - {{ 1 } \over { n }} \right) \\ =& - \gamma + 0 \end{align*}$$ and since $\Gamma (1) = 0! = 1$, we obtain $\Gamma ' (1) = - \gamma$.

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