How to compute $\int_0^{\infty} x^{t-1} e^{-x}\ln(x)\,dx$?

Question

I have hit the following integral (in the process of trying to derive a finite-sample correction for the Maximum Likelihood fitting of the Generalized Extreme Value distribution...):

$$\int_0^{\infty} x^{t-1} e^{-x}\ln(x)\,dx$$

It's a sort of cross between a definition of the $\Gamma$ function and an Euler-Mascheroni integral. Help would be greatly appreciated!

Thanks in advance.

Answer 1

By definition, we know that $$ \Gamma(t)= \int_0^\infty x^{t-1}e^{-x}\,dx $$ then our problem is simply the first derivative of gamma function w.r.t. variable $t$ $$ \int_0^\infty x^{t-1}e^{-x}\ln x\,dx=\frac{d}{dt} \Gamma(t)=\Gamma(t)\cdot\psi(t) $$ where $\psi(t)$ is the digamma function.

Answer 2

Expanding Daniel Fischer's comment, $$\int_{0}^{+\infty}x^{t-1}e^{-x}\log x\,dx = \frac{d}{dt}\int_{0}^{+\infty}x^{t-1}e^{-x}\,dx = \frac{d}{dt}\Gamma(t) = \psi(t)\cdot\Gamma(t) $$ where $\psi(t)$ is the digamma function: $$\psi(t)=-\gamma+\sum_{n=1}^{+\infty}\frac{t-1}{n(n+t-1)}.$$