Derivative of Gamma Function at 1

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Theorem

Let $\Gamma$ denote the Gamma function.


Then:

$\map {\Gamma'} 1 = -\gamma$

where:

$\map {\Gamma'} 1$ denotes the derivative of the Gamma function evaluated at $1$
$\gamma$ denotes the Euler-Mascheroni constant.


Proof 1

From Reciprocal times Derivative of Gamma Function:

$\ds \dfrac {\map {\Gamma'} z} {\map \Gamma z} = -\gamma + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {z + n - 1} }$

Setting $z = 1$:

\(\ds \frac {\map {\Gamma'} 1} {\map \Gamma 1}\) \(=\) \(\ds -\gamma + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {1 + n - 1} }\)
\(\ds \) \(=\) \(\ds -\gamma + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 n}\)
\(\ds \) \(=\) \(\ds -\gamma + 0\)
\(\ds \) \(=\) \(\ds -\gamma\)

Using Gamma Function Extends Factorial:

$\map \Gamma 1 = \paren {1 - 1}! = 1$

Hence:

$\map {\Gamma'} 1 = -\gamma \map \Gamma 1 = -\gamma$

$\blacksquare$


Proof 2

Derivative of Gamma Function at 1/Proof 2

Sources