Prove that $\int_0^\infty e^{-x} \ln x d x = - \gamma $

You can prove that $$\lim I_n=\int_0^\infty e^{-x}\ln x\,dx$$ using the dominated convergence theorem, where $$I_n=\int_0^n\left(1-\frac xn\right)^n\ln x\,dx.$$ Now substitute $y=1-x/n$. Then \begin{align} I_n&=n\int_0^1y^n\ln(n(1-y))\,dy\\ &=n\ln n\int_0^1 y^n\,dy+n\int_0^1y^n\ln(1-y)\,dy\\ &=\frac n{n+1}\left(\ln n-\int_0^1\frac{1-y^{n+1}}{1-y}\,dy\right)\\ &=\frac n{n+1}\left(\ln n-\sum_{k=1}^{n+1}\frac1k\right) \end{align} where we have integrated by parts along the way. This tends to $-\gamma$.

Here we will address your integral: \begin{equation} I = \int_0^\infty e^{-x} \ln(x)\:dx \nonumber \end{equation}

To do so we use the fact that: \begin{equation} \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} x^a = \ln(x) \end{equation}

As such (and by Leibniz's Integral Rule): \begin{align} I &= \int_0^\infty e^{-x} \ln(x)\:dx = \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a}\int_0^\infty e^{-x} x^a\:dx = \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} \Gamma(a + 1) \nonumber \\ &= \lim_{a \rightarrow 0^+} \Gamma'(a + 1) = \lim_{a \rightarrow 0^+} \Gamma(a + 1)\:\psi^{(0)}(a + 1)\nonumber \\ &= \Gamma(1)\:\psi^{(0)}(0) = \psi^{(0)}(0) = -\gamma \end{align}