Derivation of an Integral Expression of the Euler-Mascheroni Constant – Part 2

I derived the expression
\begin{equation}
\int\limits_{0}^{\infty} \mathrm{e}^{-x} \mathrm{ln}(x) \mathrm{d} x = -\gamma
\label{eq:1608161}
\tag{1}
\end{equation}
for the Euler-Mascheroni constant here. However, there is a far easier method that was fully derived in Advanced Integration Techniques by Zaid Alyafeai. I recommend this book to readers of this blog. It is free and contains many useful and interesting results.

We start with
\begin{equation}
\int\limits_{0}^{\infty} \mathrm{e}^{-x} x^t \mathrm{d} x = \Gamma(t+1)
\label{eq:1608162}
\tag{2}
\end{equation}
Differentiate with respect to \(t\)
\begin{equation}
\int\limits_{0}^{\infty} \mathrm{e}^{-x} x^t \mathrm{ln}(x) \mathrm{d} x = \frac{d\Gamma(t+1)}{dt} = \Gamma(t+1) \psi^{(0)}(t+1)
\label{eq:1608163}
\tag{3}
\end{equation}

Taking the limit of equation \eqref{eq:1608163}, \(t \to 0\) yields
\begin{equation}
\int\limits_{0}^{\infty} \mathrm{e}^{-x} \mathrm{ln}(x) \mathrm{d} x = \Gamma(1) \psi^{(0)}(1) = -\gamma
\label{eq:1608164}
\tag{4}
\end{equation}