Integrating:$\int\limits_0 ^ {\infty}e^{-x^2}\ln(x)dx $
$$ \displaystyle \int_0 ^ {\infty}e^{-x^2}\ln(x)dx = -\frac{1}{4}(\gamma +2\ln(2))\sqrt{\pi} $$ This is a well known integral. But I want to know how to solve it?? Also, please refrain using contour integration etc, as I don't know it.
$\gamma $ is the Euler-Mascheroni constant
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$\begingroup$Differentiation under the integral sign gives a pretty fast way. Let:
$$ I(\alpha) = \int_{0}^{+\infty}x^{\alpha}e^{-x^2}\,dx = \frac{1}{2}\int_{0}^{+\infty}x^{\frac{\alpha-1}{2}}e^{-x}\,dx = \frac{1}{2}\,\Gamma\left(\frac{\alpha+1}{2}\right).\tag{1}$$ Our integral is just $I'(0)$: since $\Gamma' = \psi\cdot\Gamma$, $$ \int_{0}^{+\infty}e^{-x^2}\log(x)\,dx = \frac{1}{4}\Gamma\left(\frac{1}{2}\right)\psi\left(\frac{1}{2}\right)=\color{red}{-\frac{\sqrt{\pi}}{4}\left(\gamma+2\log 2\right)}.\tag{2} $$ $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ is well-known and $\psi\left(\frac{1}{2}\right)$ can be computed from $\psi(1)=-\gamma$ and the duplication formula for the $\psi$ function: $$ \psi(z)+\psi\left(z+\frac{1}{2}\right) = -2\log 2+2\,\psi(2z).\tag{3} $$
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