TITLE: How to prove the result of this definite integral?
QUESTION [15 upvotes]: 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?

REPLY [4 votes]: 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 $12-8\gamma$.