Let $X \sim Gamma(\alpha,\beta)$ be gamma distributed random variable with probability distribution function
$$ f_{X}(x)=\frac{\beta^{\alpha}x^{\alpha-1}e^{-\beta x}}{\Gamma(\alpha)},\;x>0 $$
where, $\Gamma(\alpha)=\int_{0}^{\infty}x^{\alpha-1}e^{-x}dx$ is Euler's gamma function. Prove that,
$$ \begin{split} H_{g}(\alpha)& =-\int_{-\infty}^{\infty}f_{X}(x)\log[f_{X}(x)] dx \\ & =\log[\Gamma(\alpha)]+\alpha -\frac{\log(\alpha)}{2}+(1-\alpha)\psi(\alpha) \end{split} $$
where, $\psi(\alpha)=\frac{\Gamma'(\alpha)}{\Gamma(\alpha)}$.
Getting rid of the syntactic sugar, we have to compute:
$$ \int_{0}^{+\infty}x^{\alpha-1}e^{-\beta x}\log\left(x^{\alpha-1}e^{-\beta x}\right)\,dx = - \frac{\Gamma(\alpha+1)}{\beta^\alpha}+(\alpha-1)\int_{0}^{+\infty}x^{\alpha-1}e^{-\beta x}\log x\,dx$$
where:
$$J(\alpha,\beta)=\int_{0}^{+\infty}x^{\alpha-1}e^{-\beta x}\log x\,dx = \frac{d}{d\alpha}\int_{0}^{+\infty}x^{\alpha-1}e^{-\beta x}\,dx $$ is: $$ J(\alpha,\beta)=\frac{d}{d\alpha}\frac{\Gamma(\alpha)}{\beta^{\alpha}}=\frac{\Gamma(\alpha)}{\beta^{\alpha}}\cdot\frac{d}{d\alpha}\log\left(\frac{\Gamma(\alpha)}{\beta^{\alpha}}\right)=\color{red}{\frac{\Gamma(\alpha)}{\beta^{\alpha}}\left(\psi(\alpha)-\log \beta\right)}. $$