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$\ds{I \equiv \int_{0}^{\infty}x^{2}\expo{-x^{2}}{\rm fer}\pars{x}\ln\pars{x}
\,\dd x}$. Vamos
$\ds{{\cal I}\pars{\mu}
\equiv \int_{0}^{\infty}x^{\mu}\expo{-x^{2}}{\rm fer}\pars{x}\,\dd x}$ tal que $\ds{I = \lim_{\mu \2}\totald{{\cal I}\pars{\mu}}{\mu}}$
Desde
$\ds{{\rm fer}\pars{x}
\stackrel{{\rm def.}}{=}{2 \\raíz{\pi}}\int_{0}^{x}\expo{-y^{2}}\,\dd y}$:
\begin{align}
{\cal I}\pars{\mu}&=\int_{0}^{\infty}x^{\mu}\expo{-x^{2}}
{2 \\raíz{\pi}}\int_{0}^{\infty}\Theta\pars{x - y}\expo{-y^{2}}\,\dd y\,\dd x
\\[3 mm]&=
{2 \\raíz{\pi}}\int_{0}^{\pi/2}\dd\theta\,\cos^{\mu}\pars{\theta}
\Theta\pars{\cos\pars{\theta} - \sin\pars{\theta}}
\overbrace{\int_{0}^{\infty}\dd r\,r^{\mu + 1}\expo{-r^{2}}}
^{\Gamma\pars{1 + \mu/2}/2}
\\[3 mm]&={1 \over \raíz{\pi}}\,\Gamma\pars{1 + {\mu \over 2}}
\int_{0}^{\pi/4}\dd\theta\,\cos^{\mu}\pars{\theta}
\end{align}
\begin{align}
I&=\lim_{\mu \2}\totald{{\cal I}\pars{\mu}}{\mu}=
{1 \over 2\raíz{\pi}}\,\overbrace{\Psi\pars{2}}^{\ds{1 - \gamma}}\
\overbrace{\int_{0}^{\pi/4}\dd\theta\,\cos^{2}\pars{\theta}}^{\ds{\pars{\pi + 2}/8}}
\\[3 mm]&+
{1 \over \raíz{\pi}}\,\overbrace{\Gamma\pars{2}}^{\ds{1}}
\overbrace{%
\int_{0}^{\pi/4}\dd\theta\,\cos^{2}\pars{\theta}\ln\pars{\cos\pars{\theta}}}
^{\ds{\llaves{4G + \pi - 2\bracks{1 + \ln\pars{2}} - \pi\ln\pars{4}}/16}}
\end{align}
$\Gamma\pars{z}$ y $\Psi\pars{z}$ son $\it Gamma$ y $\it Digamma$ funciones, respectivamente. $\gamma$ y $G$ son $\es de Euler-Mascheroni$ catalán y constantes, respectivamente.
\begin{align}
&\color{#0000ff}{\large\int_{0}^{\infty}x^{2}\expo{-x^{2}}{\rm fer}\pars{x}\ln\pars{x}\,\dd x}
\\[3 mm]&=
\color{#0000ff}{\large{\pars{\pi + 2}\pars{1 - \gamma} + 4G + \pi - 2\bracks{1 + \ln\pars{2}} - \pi\ln\pars{4}\más de 16\raíz{\pi}}} \approx 0.0436462
\end{align}
