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

  • $\begingroup$ Is it allowed to use any integral identity for $\gamma$, or what do you know about $\gamma$? $\endgroup$ – mickep Sep 23 '15 at 16:22
  • $\begingroup$ @mickep I just know the basic definition of the euler mascheroni constant and some integrals that are equivalent. $\endgroup$ – Kunal Gupta Sep 23 '15 at 16:25
  • $\begingroup$ 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. $\endgroup$ – rVitale Sep 23 '15 at 16:29
  • $\begingroup$ @rVitale that'll work but from where did that integral come from?? $\endgroup$ – Kunal Gupta Sep 23 '15 at 16:34
  • $\begingroup$ @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. $\endgroup$ – rVitale Sep 23 '15 at 16:42

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: $$ \psi(z)+\psi\left(z+\frac{1}{2}\right) = -2\log 2+2\,\psi(2z).\tag{3} $$

  • $\begingroup$ The way i would go, very nice. On the other side, thinking hard for some time i don't find another one. would be very interested $\endgroup$ – tired Sep 23 '15 at 17:21
  • $\begingroup$ Looks like I will have to learn about digamma functions! $\endgroup$ – Kunal Gupta Sep 24 '15 at 4:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.