Integral from $0$ to $\infty$ of $\ln(x)/e^x$ Show
$$\int_0^\infty \frac{\ln(x)}{e^x} = -\gamma$$
(gamma is Euler-Mascheroni constant).
Can anyone please prove this result?
Also
$$
\int_0^\infty \frac{\left( \ln(x) \right)^2}{e^x}\mathrm dx.
$$
 A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
\color{#00f}{\int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x}&=
\lim_{\mu \to 0}\totald{}{\mu}\int_{0}^{\infty}x^{\mu}\expo{-x}\,\dd x
\\[5mm] & =
\lim_{\mu \to 0}\totald{\Gamma\pars{\mu + 1}}{\mu}
\\[5mm] & =\lim_{\mu \to 0}\bracks{\Gamma\pars{\mu + 1}\Psi\pars{\mu + 1}}
\\[3mm]&=\Gamma\pars{1}\Psi\pars{1}=
\color{#00f}{\large -\gamma}
\end{align}
A: For the first integral, consider the substitution
$x = \ln(1/u)$ i.e. $u = e^{-x}$. 
This gives you
\begin{align*}
\int_0^\infty \ln(x) e^{-x} \; dx
&= \int_1^0 \ln \left(\ln \frac{1}{u}\right)  (-du) \\
&= \int_0^1 \ln \left(\ln \frac{1}{u}\right) \; du
\end{align*}
This answer now shows the answer is $-\gamma$.
A: $$\int_{0}^{+\infty}e^{-x}\log(x)\,dx = \Gamma'(1) = \Gamma(1)\,\psi(1) = -\gamma $$
follows from differentiation under the integral sign, as shown by Felix Marin. In a similar way:
$$ \int_{0}^{+\infty}e^{-x}\log^2(x)\,dx = \Gamma''(1).\tag{1}$$
We may consider that $\Gamma'(x)=\Gamma(x)\psi(x)$ implies $\Gamma''(x)=\Gamma(x)\psi(x)^2+\Gamma(x)\psi'(x)$.
Since $\psi'(1)=\zeta(2)=\frac{\pi^2}{6}$,
$$ \int_{0}^{+\infty}e^{-x}\log^2(x)\,dx = \color{blue}{\gamma^2+\frac{\pi^2}{6}}.\tag{2} $$
In a similar way:
$$ \int_{0}^{+\infty}e^{-x}\log^3(x)\,dx = -\gamma^3-\frac{\gamma \pi^2}{2}-2\,\zeta(3),\tag{3}$$
$$ \int_{0}^{+\infty}e^{-x}\log^4(x)\,dx = \gamma^4+\gamma^2 \pi ^2+\frac{3 \pi^4}{20}+8\gamma\zeta(3).\tag{4}$$
