# Integrating:$\displaystyle \int_0 ^ {\infty}e^{-x^2}\ln(x)dx$

$$\displaystyle \int_0 ^ {\infty}e^{-x^2}\ln(x)dx = -\frac{1}{4}(\gamma +2\ln(2))\sqrt{\pi}$$ 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

• Is it allowed to use any integral identity for $\gamma$, or what do you know about $\gamma$? – mickep Sep 23 '15 at 16:22
• @mickep I just know the basic definition of the euler mascheroni constant and some integrals that are equivalent. – Kunal Gupta Sep 23 '15 at 16:25
• Maybe you can use the integral equivalence $\gamma = -4 \int_{0}^{\infty} e^{-x^2}x \ln(x) dx$. Then replace $\gamma$ in your equation and combine the integrals, then integrate. – rVitale Sep 23 '15 at 16:29
• @rVitale that'll work but from where did that integral come from?? – Kunal Gupta Sep 23 '15 at 16:34
• @Kunal $\gamma$ can be defined by a convergent integral, and then it is equal to a bunch of other things, and appears in integral expressions. – rVitale Sep 23 '15 at 16:42

$$I(\alpha) = \int_{0}^{+\infty}x^{\alpha}e^{-x^2}\,dx = \frac{1}{2}\int_{0}^{+\infty}x^{\frac{\alpha-1}{2}}e^{-x}\,dx = \frac{1}{2}\,\Gamma\left(\frac{\alpha+1}{2}\right).\tag{1}$$ 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: $$\psi(z)+\psi\left(z+\frac{1}{2}\right) = -2\log 2+2\,\psi(2z).\tag{3}$$