Unusual integral I got a clock as a gift recently. It has a very novel face in that the hour positions are given by a complex formula. For the most part, I have been able to verify the calculations presented as accurate, but the two o'clock identity has me stumped.
$$
\frac{\gamma}{\displaystyle{\int_0^\infty\frac{e^{-x^2}-e^{-x}}{x}\,dx}}
$$
I know that $\gamma$ is the Euler-Mascheroni constant. And WolframAlpha tells me that $\int_0^\infty\frac{e^{-x^2}-e^{-x}}{x}\,dx=\frac{\gamma}{2}$ which  makes sense because
$$
\frac{\gamma}{\displaystyle{\int_0^\infty\frac{e^{-x^2}-e^{-x}}{x}\,dx}}=\frac{\gamma}{\displaystyle{\frac{\gamma}{2}}}=2
$$
It is not clear how I would show $\int_0^\infty\frac{e^{-x^2}-e^{-x}}{x}\,dx=\frac{\gamma}{2}$. Could anyone shed some light on this or point me to a source (book, article, etc...) where I can read up on this. The usual internet (listed above) resources have not been helpful to me.
 A: $$\frac{e^{-x^2}-e^{-x}}{x} = \int_x^1 dt \, e^{-x t} = \int_{x^2}^x dy \, e^{-y}$$
Therefore, let's integrate and reverse the order of integration as follows.
$$\begin{align} \int_0^{\infty} dx\, \frac{e^{-x^2}-e^{-x}}{x} &= \int_0^{\infty} \frac{dx}{x} \,  \int_{x^2}^x dy \, e^{-y} \\ &= \int_0^{\infty} dy \, e^{-y} \int_y^{\sqrt{y}} \frac{dx}{x} \\ &= \int_0^{\infty} dy \, e^{-y} \left [\log{\sqrt{y}}-\log{y} \right ] \\ &=-\frac12 \int_0^{\infty} dy \, e^{-y} \, \log{y} \\ &= \frac{\gamma}{2} \end{align}$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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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}
