Integral $\int_0^\infty x^2\,e^{-x^2}\operatorname{erf}(x)\,\log(x)\,dx$ 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?

 A: $\newcommand{\+}{^{\dagger}}%
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$\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.

$$
\begin{array}{|l|}\hline \mbox{}\\
\quad{\displaystyle\int_{0}^{\infty}x^{2}\expo{-x^{2}}
\,\mathrm{erf}\pars{x}\ln\pars{x}\,\dd x}\quad
\\[2mm] =
\quad{{\displaystyle\quad\pars{\pi + 2}\pars{1 - \gamma} + 4G + \pi -
2\bracks{1 + \ln\pars{2}} - \pi\ln\pars{4}\quad} \over
{\displaystyle 16\root{\pi}}}\quad
\\ \mbox{}\\ \hline
\end{array}
\approx 0.0436462
$$
A: Below is almost a copypaste of my answer to equivalent question - which was asked one month earlier than this one.

We start with the standard integral representation of the hypergeometric function:
\begin{align}
_2F_1\left(\frac12,a,\frac32,-1\right)=\frac12\int_0^1\frac{dt}{\sqrt{t}\left(1+t\right)^{a}}.
\end{align}
Differentiating it w.r.t. $a$, one finds
\begin{align}
S:=\left[\frac{d}{da} {}_2F_1\left(\frac12,a,\frac32,-1\right)\right]_{a=2}=
-\frac12\int_0^1\frac{\ln\left(1+t\right)dt}{\sqrt{t}\left(1+t\right)^{2}}.
\end{align}
After the change of variables $t=s^2$ the last integral can be expressed in terms of dilogarithms:
\begin{align}
S&=-\int_0^1\frac{\ln\left(1+s^2\right)ds}{\left(1+s^2\right)^2}=\\&=
\frac{\pi}{8}\left[1-3\ln 2 +\ln\left(2+\sqrt{2}\right)\right]-\frac{1+\ln 2}{4}+\Im\left(\operatorname{Li}_2\left(-e^{i\pi/4}\right)-\operatorname{Li}_2\left(1-e^{i\pi/4}\right)\right)=\\
&=\frac{\pi\left(1-2\ln 2\right)}{8}-\frac{1+\ln 2}{4}+\frac12 \Im\operatorname{Li}_2\left(i\right)=\\
&=\frac{G}{2}+\frac{\pi\left(1-2\ln 2\right)}{8}-\frac{1+\ln 2}{4}.
\end{align}
where $G$ denotes the Catalan's constant. Note that  $S$ is precisely what we need to compute.
Explanation: 
At the first step, the only nontrivial integrals (producing dilogarithms) are $\displaystyle \int\frac{\ln(1\pm i s)}{1\mp is}ds$. The other integrals are elementary.
At the last step we used the formula $\operatorname{Li}_2(i)=iG-\frac{\pi^2}{48}$ from here. 
A: $$I=\frac{2-\ln2}{16}\sqrt\pi-\frac{\gamma+\ln2}{16\,\sqrt\pi}(\pi+2)+\frac{G}{4\,\sqrt\pi},$$
where $\gamma$ is the Euler-Mascheroni constant:
$$\gamma=-\int_0^1\ln(-\ln x)\,dx$$
and $G$ is the Catalan constant:
$$G=-\int_0^1\frac{\ln x}{x^2+1}dx$$
