Unusual integral

Solution 1:

$$\frac{e^{-x^2}-e^{-x}}{x} = \int_x^1 dt \, e^{-x t} = \int_{x^2}^x dy \, e^{-y}$$

Therefore, let's integrate and reverse the order of integration as follows.

$$\begin{align} \int_0^{\infty} dx\, \frac{e^{-x^2}-e^{-x}}{x} &= \int_0^{\infty} \frac{dx}{x} \, \int_{x^2}^x dy \, e^{-y} \\ &= \int_0^{\infty} dy \, e^{-y} \int_y^{\sqrt{y}} \frac{dx}{x} \\ &= \int_0^{\infty} dy \, e^{-y} \left [\log{\sqrt{y}}-\log{y} \right ] \\ &=-\frac12 \int_0^{\infty} dy \, e^{-y} \, \log{y} \\ &= \frac{\gamma}{2} \end{align}$$

Solution 2:

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ The first step is 'integration by parts': \begin{align}&\color{#66f}{\large\int_{0}^{\infty}\frac{\expo{-x^{2}} - \expo{-x}}{x}\,\dd x} =-\ \overbrace{\int_{0}^{\infty}\ln\pars{x}\expo{-x^{2}}\pars{-2x}\,\dd x} ^{\ds{\dsc{x}\ \mapsto\ \dsc{x^{1/2}}}}\ +\ \int_{0}^{\infty}\ln\pars{x}\expo{-x}\pars{-1}\,\dd x \\[5mm]&=-\,\half\int_{0}^{\infty}\ln\pars{x}\expo{-x}\pars{-1}\,\dd x =-\,\half\,\lim_{\mu \to 0}\,\totald{}{\mu} \int_{0}^{\infty}x^{\mu}\expo{-x}\,\dd x =-\,\half\,\lim_{\mu \to 0}\,\totald{\Gamma\pars{\mu + 1}}{\mu} \\[5mm]&=-\,\half\,\Gamma'\pars{1} =-\,\half\,\Gamma\pars{1}\Psi\pars{1}=\color{#66f}{\large\half\,\gamma} \end{align}