Derivative of the Gamma function How do you prove that
$$
\Gamma'(1)=-\gamma,
$$
where $\gamma$ is the Euler-Mascheroni constant?
 A: 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}
A: 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:
$$ \psi(1) = \Gamma'(1) = -\gamma.\tag{3}$$
A: 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 $$-\gamma_{\,_\mathrm{EM}}$$
