Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ) ?.

share|improve this question
    
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

1 Answer 1

up vote 3 down vote accepted

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 $\Gamma(2/\alpha)$.

share|improve this answer
    
..... 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
    
would it be possible to approximate it in a simpler form? –  ACAC Sep 9 '11 at 1:27

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.