TITLE: Integral $\int_0^\infty x^2\,e^{-x^2}\operatorname{erf}(x)\,\log(x)\,dx$
QUESTION [26 upvotes]: I need to evaluate this integral:
$$I=\int_0^\infty x^2\,e^{-x^2}\operatorname{erf}(x)\,\log(x)\,dx\tag1$$
I tried to do this in Mathematica and it returned a result of the form
$$I=\frac{(\pi+2)\,(1-\gamma)}{16\,\sqrt\pi}+\frac1{2\,\sqrt\pi}\left.\frac{d}{d\xi}\Bigg({_2F_1}\left(\tfrac12,\xi;\tfrac32;-1\right)\Bigg)\right|_{\xi=2}\tag2$$
I tried to find a closed form for the derivative using an integral representation of the hypergeometric function, but this way returned me back to my original integral. 

Is it possible to represent $I$ in a closed form that does not contain unevaluated integrals or derivatives?

REPLY [19 votes]: $\newcommand{\+}{^{\dagger}}%
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$\ds{I \equiv \int_{0}^{\infty}x^{2}\expo{-x^{2}}{\rm erf}\pars{x}\ln\pars{x}
     \,\dd x}$. Let's
  $\ds{{\cal I}\pars{\mu}
     \equiv \int_{0}^{\infty}x^{\mu}\expo{-x^{2}}{\rm erf}\pars{x}\,\dd x}$ such that $\ds{I = \lim_{\mu \to 2}\totald{{\cal I}\pars{\mu}}{\mu}}$

Since
$\ds{{\rm erf}\pars{x}
     \stackrel{{\rm def.}}{=}{2 \over \root{\pi}}\int_{0}^{x}\expo{-y^{2}}\,\dd y}$:
\begin{align}
{\cal I}\pars{\mu}&=\int_{0}^{\infty}x^{\mu}\expo{-x^{2}}
{2 \over \root{\pi}}\int_{0}^{\infty}\Theta\pars{x - y}\expo{-y^{2}}\,\dd y\,\dd x
\\[3mm]&=
{2 \over \root{\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}
\\[3mm]&={1 \over \root{\pi}}\,\Gamma\pars{1 + {\mu \over 2}}
\int_{0}^{\pi/4}\dd\theta\,\cos^{\mu}\pars{\theta}
\end{align}

\begin{align}
I&=\lim_{\mu \to 2}\totald{{\cal I}\pars{\mu}}{\mu}=
{1 \over 2\root{\pi}}\,\overbrace{\Psi\pars{2}}^{\ds{1 - \gamma}}\
\overbrace{\int_{0}^{\pi/4}\dd\theta\,\cos^{2}\pars{\theta}}^{\ds{\pars{\pi + 2}/8}}
\\[3mm]&+
{1 \over \root{\pi}}\,\overbrace{\Gamma\pars{2}}^{\ds{1}}
\overbrace{%
\int_{0}^{\pi/4}\dd\theta\,\cos^{2}\pars{\theta}\ln\pars{\cos\pars{\theta}}}
^{\ds{\braces{4G + \pi - 2\bracks{1 + \ln\pars{2}} - \pi\ln\pars{4}}/16}}
\end{align}
$\Gamma\pars{z}$ and $\Psi\pars{z}$ are the $\it Gamma$ and $\it Digamma$ functions, respectively. $\gamma$ and $G$ are the $\it Euler-Mascheroni$ and Catalan constants, respectively.

$$
\begin{array}{|l|}\hline \mbox{}\\
\quad{\displaystyle\int_{0}^{\infty}x^{2}\expo{-x^{2}}
\,\mathrm{erf}\pars{x}\ln\pars{x}\,\dd x}\quad
\\[2mm] =
\quad{{\displaystyle\quad\pars{\pi + 2}\pars{1 - \gamma} + 4G + \pi -
2\bracks{1 + \ln\pars{2}} - \pi\ln\pars{4}\quad} \over
{\displaystyle 16\root{\pi}}}\quad
\\ \mbox{}\\ \hline
\end{array}
\approx 0.0436462
$$