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Looking for the general form $f_n$ a solution for an integral with $n$ strictly positive parameters $\beta_i$, $\beta_1,\beta_2,\ldots, \beta_n$ .

The integral is as follows $$ f_n=\int_0^\infty \left(1-\prod _{i=1}^n \left(1-e^{-\frac{y}{\beta _i}}\right)\right) \mathrm{d} y$$ We get for $n=1,2,3$ the following solutions:

$$f_1=\beta_1$$ $$f_2= \frac{\beta _1^2+\beta _2^2+\beta _1 \beta _2}{\beta _1+\beta _2}$$ $$f_3=\left(\frac{1}{\beta _1+\beta _3}-\frac{\beta _2^2}{\left(\beta _1+\beta _2\right) \left(\beta _2 \beta _3+\beta _1 \left(\beta _2+\beta _3\right)\right)}\right) \beta _1^2+\beta _3+\frac{\beta _2^2}{\beta _2+\beta _3}$$

For $\beta=\beta_1= \dots=\beta_n$,

$$f_n= \beta \, H_n$$ where

$$H_n= 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n} =\sum_{k=1}^n \frac{1}{k}.$$

(I have been tinkering around the binomial coefficient and more general forms with no avail: for instance $ \frac{\left(\sum _{i=1}^n \beta _i\right){}^n-\prod _{i=1}^n \beta _i}{\left(\sum _{i=1}^n \beta _i\right){}^{n-1}}$ fails except for $n=2$, so perhaps a more general form along this tinkering.)

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  • $\begingroup$ Are you sure of your integrand ? $f_1$ is not defined. Please clarify. $\endgroup$ Jun 26, 2014 at 12:58
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    $\begingroup$ It reminds me of the coupon collector's problem, where the integral for expectation is $$\int_0^\infty \left(1-\prod_{i=1}^n(1-e^{-p_it})\right)dt$$ Did you mean this instead? $\endgroup$
    – gar
    Jun 26, 2014 at 13:20
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    $\begingroup$ Thanks gar corrected typo. Indeed. $\endgroup$
    – Nero
    Jun 26, 2014 at 13:22
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    $\begingroup$ Okay. I'd suggest you to read the wikipedia entry for coupon collector's problem, which has a link to Flajolet's paper where it's discussed in more detail. $\endgroup$
    – gar
    Jun 26, 2014 at 13:24
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    $\begingroup$ @Nero. I prefer this one ! Cheers :) $\endgroup$ Jun 26, 2014 at 13:36

2 Answers 2

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Following P. Flajolet's paper http://algo.inria.fr/flajolet/Publications/FlGaTh92.pdf (corollary 4.2)

$$f_n=E\{C_n\}=\sum_{q=0}^{n-1} (-1)^{n-1-q}\sum_{|J|=q}\frac{1}{1-P_J}$$

Where $$P_J=\sum_{j\in J}p_j=\sum_{j\in J}\frac{1}{\beta_j}$$

For $n=3$

$$f_3=1-(\frac{1}{1-\beta_1^{-1}}+\frac{1}{1-\beta_2^{-1}}+\frac{1}{1-\beta_3^{-1}})+(\frac{1}{1-\beta_1^{-1}-\beta_2^{-1}}+\frac{1}{1-\beta_2^{-1}-\beta_3^{-1}}+\frac{1}{1-\beta_1^{-1}-\beta_3^{-1}})$$

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I have a fairly simple solution of the form 1/Sum[1/beta[j], {j,{subset of 1..n}}], which is the integral of the products of a given subset of the exponential terms, the individual elements of the expansions of the (1- exp) product. Unfortunately, there are, of course, 2^(n-1) such subsets. I can send you a Mathematica notebook. For modest sized n it's entirely workable; say 4 seconds for n = 20. If its computations for such sized n I can be of service. If it's deeper insights, I must confess utter failure!

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