# Derivative of the Gamma function

How do you prove that $$\Gamma'(1)=-\gamma,$$ where $\gamma$ is the Euler-Mascheroni constant?

• What definition the the gamma function are you using? Apr 22, 2015 at 15:22
• And what definition of $\gamma$? Apr 22, 2015 at 15:32
• Would this answer your question ? Apr 22, 2015 at 16:01

Consider the integral form of the Gamma function, \begin{align} \Gamma(x) = \int_{0}^{\infty} e^{-t} \, t^{x-1} \, dt \end{align} taking the derivative with respect to $x$ yields \begin{align} \Gamma'(x) = \int_{0}^{\infty} e^{-t} \, t^{x-1} \, \ln(t) \, dt. \end{align} Setting $x=1$ leads to \begin{align} \Gamma'(1) = \int_{0}^{\infty} e^{-t} \, \ln(t) \, dt. \end{align} This is one of the many definitions of the Euler-Mascheroni constant. Hence, \begin{align} \Gamma'(1) = - \gamma = \int_{0}^{\infty} e^{-t} \, \ln(t) \, dt. \end{align}
• How to take derivative with respect to x of$\int_{0}^{\infty} e^{-t} \, t^{x-1} \, dt$? Oct 7, 2022 at 2:36
The Weierstrass product for the $\Gamma$ function gives: $$\Gamma(z+1)=e^{-\gamma z}\cdot\prod_{n\geq 1}\left(1+\frac{z}{n}\right)^{-1}e^{z/n}\tag{1}$$ hence by considering $\frac{d}{dz}\log(\cdot)$ of both terms we get: $$\psi(z+1)=\frac{\Gamma'(z+1)}{\Gamma(z+1)}=-\gamma+\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+z}\right) \tag{2}$$ and by evaluating the previous identity in $z=0$ it follows that: $$\psi(1) = \Gamma'(1) = -\gamma.\tag{3}$$
I was wrong I cannot delete my post because I having trouble singing in sorry for my lapse in judgement and failed math skills I will try to be better the solutions above work just fine. $$\Gamma^{\prime}(z) = \frac{d}{dz} \int^{\infty}_{0} e^{-t}t^{z-1}dt = \int^{\infty}_{0} \frac{d}{dz} e^{-t}t^{z-1}dt = \int^{\infty}_{0} e^{-t} \frac{d}{dz} t^{z-1} dt = \int^{\infty}_{0} e^{-t} ln(t) t^{z-1} dt$$ $$\Gamma^{\prime}(1) = \int^{\infty}_{0} e^{-t} ln(t) t^{1-1} dt = \int^{\infty}_{0} e^{-t} ln(t) dt$$ this integral can be solved numerically to show that it comes out to $$-\gamma_{\,_\mathrm{EM}}$$