1

So according to Proving $\int_{0}^{\infty} \mathrm{e}^{-x^2} dx = \dfrac{\sqrt \pi}{2}$, $$\int_0^\infty e^{-x^2}dx=\frac{\sqrt{\pi}}{2}$$ I want to solve for this. $$\int_0^\infty e^{-x^2}\ln(x)dx$$ My first thought was integration by parts. Since we know what $\int_0^\infty e^{-x^2}dx$ is. But I am open to any other methods you all can give me.

Community
  • 1
hooooü
  • 11

1 Answers1

1

First, consider the following integral: $$I(\alpha)= \int_{0}^{\infty}x^{\alpha}e^{-x^2}\,dx $$ Next, set $x^2 \rightarrow x$: $$I(\alpha)= \frac{1}{2} \int_{0}^{\infty}x^{\frac{\alpha-1}{2}}e^{-x}\,dx = \frac{1}{2}\Gamma\left(\frac{\alpha+1}{2}\right) $$ In this case, we need $\alpha=0$: $$\frac{\partial I(\alpha)}{\partial \alpha} =\int_{0}^{\infty}x^{\alpha}\ln{(x)}e^{-x^2}\,dx\implies I’(\alpha)= \frac{1}{4}\Gamma'\left(\frac{\alpha+1}{2}\right) = \frac{1}{4}\Gamma\left(\frac{\alpha+1}{2}\right)\psi\left(\frac{\alpha+1}{2}\right) $$ Note that $\psi$ is the Digamma Function and $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$: $$ I'(0) = \frac{\sqrt{\pi}}{4}\psi\left(\frac{1}{2}\right) $$

To simplify this further, note that $\psi(1/2)=-\gamma-\log(4)$. $$\therefore \int_0^\infty e^{-x^2}\log(x)\,dx=\frac{\sqrt{\pi}\left(-\gamma -\log(4)\right)}{4}$$