# expectation of function of exponential

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 ) ?.

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Bob, The formula isn't rendered very clearly to me (subscript and superscripts). So I rewrote it slightly. Hope its ok with you. Also, check that I haven't made a mistake :) –  Srivatsan Sep 7 '11 at 22:21
Is the thing inside the square brackets, the CDF instead of the PDF of the distribution? Are you calculating the $E[X]$ for some distribution (related to exponentials)? –  Srivatsan Sep 7 '11 at 22:43
yes it's an expectation. I just fixed a typo, now it should converge. –  ACAC Sep 7 '11 at 22:55

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 $\Gamma(2/\alpha)$.
..... Anyway, the binomial theorem works for noninteger $M>0$ so long as you take the sum out to $\infty$ (and extend the binomial coefficient with gamma functions). –  anon Sep 8 '11 at 0:33
@anon: You don't even need to use $\Gamma$ to extend the binomial coefficient: $\displaystyle\binom{M}{k}=\frac{M(M-1)(M-2)...(M-k+1)}{k!}$ works for non-integer $M$. –  robjohn Sep 8 '11 at 0:39
in my case $M$ is integer, so I guess the result works as is –  ACAC Sep 8 '11 at 0:40
@Bob: I had a feeling $M\in\mathbb{Z}^+$ –  robjohn Sep 8 '11 at 0:42