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

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}
