How to prove the result of this definite integral? During my work I came up with this integral:
$$\mathcal{J} = \int_0^{+\infty} \frac{\sqrt{x}\ln(x)}{e^{\sqrt{x}}}\ \text{d}x$$
Mathematica has a very elegant and simple numerical result for this, which is
$$\mathcal{J} = 12 - 8\gamma$$
where $\gamma$ is the Euler-Mascheroni constant.
I tried to make some substitutions, but I failed. Any hint to proceed?
 A: Start from the well-known integral
$$-\gamma=\int_0^\infty\exp(-x)\log x\,\mathrm dx$$
A round of integration by parts yields
$$\begin{align*}
-\gamma&=\int_0^\infty\exp(-x)\log x\,\mathrm dx\\
&=\int_0^\infty x\exp(-x)\log x\,\mathrm dx-\int_0^\infty x\exp(-x)\,\mathrm dx\\
1-\gamma&=\int_0^\infty x\exp(-x)\log x\,\mathrm dx
\end{align*}$$
A second round gives
$$\begin{align*}
1-\gamma&=\int_0^\infty x\exp(-x)\log x\,\mathrm dx\\
&=\int_0^\infty x\exp(-x)(x-1)(\log x-1)\,\mathrm dx\\
&=\int_0^\infty x\exp(-x)\,\mathrm dx-\int_0^\infty x^2\exp(-x)\,\mathrm dx-\int_0^\infty x\exp(-x)\log x\,\mathrm dx+\int_0^\infty x^2\exp(-x)\log x\,\mathrm dx\\
3-2\gamma&=\int_0^\infty x^2\exp(-x)\log x\,\mathrm dx
\end{align*}$$
where the last integral is the one obtained by Dr. MV after an appropriate substitution. Thus, the original integral is equal to $12-8\gamma$.
A: Enforce the substitution $x\to x^2$.  Then, we have
$$\begin{align}
\mathcal{I}&=4\int_0^\infty x^2\log(x)e^{-x}\,dx\\\\
&=4\left.\left(\frac{d}{da}\int_0^\infty x^{2+a}e^{-x}\,dx\right)\right|_{a=0}\\\\
&=4\left.\left(\frac{d}{da}\Gamma(a+3)\right)\right|_{a=0}\\\\
&=4\Gamma(3)\psi(3)\\\\
&=4(2!)(3/2-\gamma)\\\\
&=12-8\gamma
\end{align}$$
as was to be shown!
Note the we used (i) the integral representation of the Gamma function
$$\Gamma(x)=\int_0^\infty t^{x-1}e^{-t}\,dt$$
(ii) the relationship between the digamma and Gamma functions
$$\psi(x)=\frac{\Gamma'(x)}{\Gamma(x)}$$
and (iii) the recurrence relationship 
$$\psi(x+1)=\psi(x)+\frac1x$$
In addition, we used the special values
$$\Gamma(3)=2!$$
and
$$\psi(1)=-\gamma$$
A: Hint. One may recall that
$$
\int_0^\infty u^{s}e^{-u}\:du=\Gamma(s+1), \quad s>0,\tag1
$$ giving, by differentiating under the integral sign,
$$
\begin{align}
\int_0^\infty u^{2}\ln u \:e^{-u}\:du&=\left.\left(\Gamma(s+1)\right)'\right|_{s=2}\\\\
&=\left.\left(s(s-1)\Gamma(s-1)\right)'\right|_{s=2}\\\\
&=3+2\Gamma'(1)\\\\
&=3-2\gamma,\tag2
\end{align}
$$ where we have used $\Gamma'(1)=-\gamma$.
Then, one may rewrite the initial integral as
$$
2\int_0^\infty \sqrt{x}\:(\ln \sqrt{x}) \:e^{-\sqrt{x}}\:dx,
$$ then perform the change of variable $x=u^2$, $dx=2udu$, obtaining
$$
\begin{align}
\int_0^\infty \sqrt{x}\ln x \:e^{-\sqrt{x}}\:dx&=4\int_0^\infty u^{2}\ln u \:e^{-u}\:du.\tag3
\end{align}
$$ Considering $(2)$ and $(3)$ gives the announced result.
