Integral $\int_0^{\infty} \log(x) e^{-x^2} \mathrm{d}x = -\frac{1}{4}\sqrt{\pi} (\gamma + \log(4)).$

Question

While trying to compute the expected value $E[\log(X)]$ for a normally distributed variable $X$ I found the following integral $$\int_{0}^{\infty}\log\left(x\right) {\rm e}^{-x^{2}}\,{\rm d}x =-\,{1 \over 4}\,\,\sqrt{\,\pi\,}\,\left[\,\gamma + \log\left(4\right)\,\right] $$. Can anybody explain this?

Answer 1

Consider the integral

$$I(a) = \int_0^{\infty} dx \, x^a \, e^{-x^2}$$

Sub $x=y^{1/2}$ and get

$$I(a) = \frac12 \int_0^{\infty} dy \, y^{(a-1)/2} \, e^{-y} = \frac12 \Gamma\left ( \frac{a+1}{2}\right)$$

The integral in question is

$$I'(0) =\frac12 \left [\frac{d}{da} \Gamma\left ( \frac{a+1}{2}\right)\right ]_{a=0} = \frac14 \Gamma\left (\frac12\right) \psi\left ( \frac12 \right)$$

where

$$\Gamma\left (\frac12\right)=\sqrt{\pi}$$ $$\psi\left ( \frac12 \right) = -(\log{4}+\gamma)$$

Answer 2

Your result is correct for the integral. I uppose that your question is : how to arrive there ? In fact, concerning the antiderivative, integrate by parts [u = Log(x)] and you arrive to
Sqrt[Pi] Log[x] Erf[x] / 2 - Sqrt[Pi] /2 Integral[Erf(x) / x dx]
The result of the last integration involves the hypergeometric FPQ function.

I really apologize for this awful format but I am almost blind and I do not "see" what I am typing.