TITLE: expectation of function of exponential
QUESTION [5 upvotes]: I need to compute the following integral
$$
\int_{0}^{\infty}x\,\left\{\vphantom{\LARGE A}%
1- \left[\vphantom{\Large A}1- \exp(-a\,x^{\alpha}) \right]^M
\right\}\,{\rm d}x
\qquad
\mbox{with}\quad \alpha > 0\quad\mbox{and}\quad a,M > 0. 
$$  
Is this a known integral ( if possible, without approximating the exponential function ) ?.

REPLY [3 votes]: I am assuming that $M$ is an integer. Let $\beta=1/\alpha$. Then substituting $x\to x^\beta$
$$
\begin{align}
\int_0^\infty x\left[1-(1-\exp(-ax^\alpha))^M\right]\mathrm{d}x
&=\int_0^\infty \beta x^{2\beta-1}\left[1-(1-\exp(-ax))^M\right]\mathrm{d}x\\
&=\int_0^\infty \beta x^{2\beta-1}\left[\sum_{k=1}^M(-1)^{k-1}\binom{M}{k}\exp(-akx)\right]\mathrm{d}x\\
&=\beta\;a^{-2\beta}\;\Gamma(2\beta)\;\sum_{k=1}^M(-1)^{k-1}k^{-2\beta}\binom{M}{k}
\end{align}
$$
This can be evaluated as long as you can evaluate $\Gamma(2/\alpha)$.