If I look at an exponential function, $p(t) = e^{-\mu t}$ where the parameter $\mu$ varies over a gamma distribution given by the density function
$f(\mu) = \frac{1}{\Gamma(a)b^a}\mu^{a-1}e^{-\mu/b}$ (parameters $a, b$ and $\Gamma$ the gamma function)
Then I can talk about the "average" of all exponentials with respect to $\mu$ as the integral
$\int_{0}^{\infty}p(t)f(\mu)d\mu$
I'm reading a paper where they present, without proof, the claim that this has closed form
$(1+bt)^{-a}$
That is, the "average" of an exponential curve is a power-law curve. (I'm slightly simplifying the actual claim, so I'm hoping I didn't make a mistake)
On one hand, I'm interested to know what results they used in probability calculus to get to this closed form (it is probably pretty basic, but I just never learned this stuff). More importantly, I'm interested in whether the same technique applies to a family of power-law curves. That is if I instead write $p(t) = t^{-\mu}$, can I easily derive a closed form, for this, and if so, is it a power-law?
As a side note, if the distribution is instead uniform, the answer is (sort of) yes: the integral is trivial and the closed form quickly converges to a power-law curve, as it does in the case of an exponential.