Prove that $\int_0^\infty e^{-x} \ln x d x = - \gamma $ I can see it is right by using some knowledge of the Gamma function. We have
$$ \Gamma(\alpha ) = \int_0^\infty e^{-x} x^{\alpha - 1 } dx  . $$
Differentiating with respect to $\alpha$, we get 
$$ \frac{d\Gamma}{d \alpha} = \int_0^\infty e^{-x} x^{\alpha - 1 }\ln x dx  .  $$
Setting $\alpha = 1 $, we get 
$$ \int_0^\infty e^{-x} \ln x dx = \frac{d\Gamma}{d\alpha }\bigg|_{\alpha = 1 }= - \gamma.   $$
But the knowledge $d \Gamma/d \alpha |_{\alpha = 1}=-\gamma$ is a mystery to me. 
Can anyone find an elementary proof? 
 A: 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 $-\gamma$.
A: 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}
A: Your method seems like a reasonable way to derive this result, and you are almost there.  Now use the Taylor expansion of the digamma function, which is:
$$\psi(z+1) = -\gamma - \sum_{k=1}^\infty \zeta(k+1) \cdot (-z)^k.$$
Substituting $z=0$ gives:
$$\psi(1) = -\gamma - \sum_{k=1}^\infty \zeta(k+1) \cdot 0^k = -\gamma.$$
So you have:
$$\int_0^\infty e^{-x} \ln x dx = \Gamma'(1) = \psi(1) \cdot \Gamma(1) = -\gamma \cdot 1 = -\gamma.$$
