Integral from $0$ to $\infty$ of $\ln(x)/e^x$

Question

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$$\int_0^\infty \frac{\ln(x)}{e^x} = -\gamma$$ (gamma is Euler-Mascheroni constant).

Can anyone please prove this result?

Also

$$ \int_0^\infty \frac{\left( \ln(x) \right)^2}{e^x}\mathrm dx. $$

Answer 1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}

Answer 2

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 $-\gamma$.

Answer 3

$$\int_{0}^{+\infty}e^{-x}\log(x)\,dx = \Gamma'(1) = \Gamma(1)\,\psi(1) = -\gamma $$ 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: $$ \int_{0}^{+\infty}e^{-x}\log^3(x)\,dx = -\gamma^3-\frac{\gamma \pi^2}{2}-2\,\zeta(3),\tag{3}$$ $$ \int_{0}^{+\infty}e^{-x}\log^4(x)\,dx = \gamma^4+\gamma^2 \pi ^2+\frac{3 \pi^4}{20}+8\gamma\zeta(3).\tag{4}$$