Following this question, I have the PDF of a gamma-exponential compound distribution as
$$f(y) = \frac{\alpha\beta^{\alpha}} {(y+\beta)^{(\alpha+1)}} $$
For my application I need the entropy of this distribution. So far I've evaluated this numerically using quadrature, but it's too slow. Is there a closed-form expression for the entropy of this distribution? I've attempted to derive it myself by evaluating
$$\int_{-\infty}^{\infty} f(y) \log(f(y)) dy$$
as a sum of two improper integrals $$\int_{-\infty}^{0} ... dy + \int_{0}^{\infty} ... dy$$ but I get lost when trying to evaluate the infinite limits:
$$\int f(y)\log(f(y)) dy = \frac{ \beta^{\alpha} (\beta+y)^{-\alpha} (1+\alpha-\alpha\cdot{}\log(\alpha \beta^{\alpha} (\beta+y)^{-1-\alpha})) }{ \alpha }$$ (according to Wolfram)
$$\int_{-\infty}^{\infty} f(y) \log(f(y)) dy = \lim_{m\to\infty}[ \frac{ \beta^{\alpha} (\beta+m)^{-\alpha} (1+\alpha-\alpha\cdot{}\log(\alpha \beta^{\alpha} (\beta+m)^{-1-\alpha})) }{ \alpha } ] - \lim_{n\to-\infty}[\frac{ \beta^{\alpha} (\beta+n)^{-\alpha} (1+\alpha-\alpha\cdot{}\log(\alpha \beta^{\alpha} (\beta+n)^{-1-\alpha})) }{ \alpha }] $$
(Skipped a step or two there.)
But here I get stuck because I'm not sure how to evaluate the limits when the logarithms are part of the function, as I haven't found (here or here) a clear limits of logarithms rule that I can apply here.
Sorry my typesetting is ugly. I'd love to know if there's a known entropy for this distribution or if somebody can help me past where I'm stuck in the derivation of it. Thanks
As commented in the original question, this is a Pareto distribution with parameter $\alpha$ (it's shifted, but that's irrelevant - you can consider instead the variable $z=y+\beta$). It's entropy is then
$$H(y)=\log\left(\frac{\beta}{\alpha}\right)+1 + 1/\alpha$$
as computed here