Definite Integral to Infinity of Exponential of -x^2 by Logarithm of x

Theorem

 * $\displaystyle \int_0^\infty e^{-x^2} \ln x \rd x = -\frac {\sqrt \pi} 4 \paren {\gamma + 2 \ln 2}$

where $\gamma$ denotes the Euler-Mascheroni constant.

Proof
Consider the integral:


 * $\displaystyle \int_0^\infty x^t e^{-x^2} \rd x$

for positive real parameter $t$.

Using Definite Integral to Infinity of $x^m e^{-a x^2}$, we have:


 * $\displaystyle \int_0^\infty x^t e^{-x^2} \rd x = \frac 1 2 \map \Gamma {\frac {1 + t} 2}$

Differentiating the with respect to $t$ we have:

Differentiating the using the Chain Rule for Derivatives gives:


 * $\displaystyle \int_0^\infty x^t e^{-x^2} \ln x \rd x = \frac 1 4 \map {\Gamma'} {\frac {1 + t} 2}$

We therefore have: