How do I find the closed form of this integral?

Question

I was trying to calculate the expected value of $\log(Y)$ where $Y$ has gamma distribution and I got to something like this:

$$\int_0^\infty \log(z)z^{\phi-1}e^{-z} \, dz,$$ wich based on other results should give me $\Gamma'(\phi)$ but I have no idea how to solve the integral, what are the steps to get the answer?

Answer 1

It looks like you already know the answer. Since: $$ \Gamma(\alpha)=\int_{0}^{+\infty}x^{\alpha-1}e^{-x}\,dx \tag{1}$$ by differentiating (with respect to $\alpha$) under the integral sign we get: $$ \frac{d}{d\alpha}\Gamma(\alpha) = \int_{0}^{+\infty} x^{\alpha-1}e^{-x}\log x\,dx\tag{2}$$ and since $$\frac{\Gamma'(\alpha)}{\Gamma(\alpha)}=\frac{d}{d\alpha}\log\Gamma(\alpha) = \psi(\alpha) = H_{\alpha-1}-\gamma, \tag{3}$$

where $H_n$ stands for the $n$-th harmonic number, $\psi(z)$ for the digamma function and $\gamma$ for the Euler-Mascheroni constant, we have:

$$ \int_{0}^{+\infty}x^{\alpha-1}e^{-x}\log x\,dx = \Gamma'(\alpha) = \left(H_{\alpha-1}-\gamma\right)\Gamma(\alpha).\tag{4}$$