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:

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

Setting $n = 1$:

\(\displaystyle \frac {\Gamma' \left({1}\right)} {\Gamma \left({1}\right)}\) \(=\) \(\displaystyle -\gamma + \sum_{n \mathop = 1}^\infty \left({\frac 1 n - \frac 1 {1 + n - 1} }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle -\gamma + \sum_{n \mathop = 1}^\infty \left({\frac 1 n - \frac 1 n}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle -\gamma + 0\)
\(\displaystyle \) \(=\) \(\displaystyle -\gamma\)

$\blacksquare$


Proof 2

Derivative of Gamma Function at 1/Proof 2

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