Evaluate $\int_{0}^{\infty}e^{-x^2}\ln(x)dx$

Question

Can a step-by-step answer be shown how to prove: $$\int_{0}^{\infty}e^{-x^2}\ln(x)dx = -\frac{{\pi^\frac{1}{2}}}{4}(\gamma+\ln(4))$$

I have a feeling differentiating under the integral sign could be done, but I'm not sure how.

Answer 1

Use $$\int_{0}^{\infty} e^{- x^{2}} \, x^{u -1} \, dx = \frac{1}{2} \, \Gamma\left(\frac{u}{2}\right)$$ and differentiate with respect to $u$ to obtain $$\int_{0}^{\infty} e^{- x^{2}} \, x^{u -1} \, \ln(x) \, dx = \frac{1}{4} \, \Gamma\left(\frac{u}{2}\right) \, \psi\left(\frac{u}{2}\right).$$ Set $u =1$ and use the appropriate value of the digamma function to obtain the desired result.