Derivative of Gamma Function at 1

Theorem

Let $\Gamma$ denote the Gamma function.

Then:

$\map {\Gamma'} 1 = \ds \int_0^\infty e^{-x} \ln x \rd x = -\gamma$

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 $z = 1$:

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

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 16$: Derivatives of the Gamma Function: $16.13$
• 1983: François Le Lionnais and Jean Brette: Les Nombres Remarquables ... (previous) ... (next): $0,57721 56649 \ldots$
• 2009: Murray R. Spiegel, Seymour Lipschutz and John Liu: Mathematical Handbook of Formulas and Tables (3rd ed.) ... (previous) ... (next): $\S 25$: The Gamma Function: Derivatives of the Gamma Function: $25.13.$