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I am trying to obtain a closed form solution of this definite integral, or in a form at least which simplify its numerical treatment. $$\int_{x_1=0}^1...\int_{x_N=0}^1 \prod_r \left( \frac {x_r f_r} {\sum_g x_g f_g}\right) ^{n_r} dx_1...dx_N $$ where $n_r$ are positive integers, and $f_r\ge0$ constants.

I tried to introduce a contour transformation, by $s=x_1 f_1 + ... + x_n f_n, y_2=x_2 f_2,..., y_N=x_N f_N$, |det Jacobian|=constant, which "reduce" to $$\int_{s=0}^{\sum_g f_g}ds\int_{y_2=0}^{?_2}dy_2...\int_{y_N=0}^{?_N}dy_N \prod_r \left( \frac {y_r} s\right) ^{n_r}$$

where $y_r$ are constrained (integral upper limits $?_r$) to total $s$ and be no greater than $f_r$, which seems no advance at all to me .

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Your multiple integral $I$ can be converted into a single integral (to start with - more simplifications are perhaps possible). Use the identity $$ x^{-s}=\frac{1}{\Gamma(s)}\int_0^\infty d\xi\ \xi^{s-1}e^{-\xi x} $$ with $s=\sum_r n_r$ to 'exponentiate' the denominator and obtain $$ I = \left(\prod_r f_r^{n_r}\right)\frac{1}{\Gamma(\sum_r n_r)}\int_0^\infty d\xi\ \xi^{\sum_r n_r-1}\prod_{r=1}^N \left[\int_0^1 dx\ x^{n_r}e^{-\xi f_r x}\right] $$ $$ = \left(\prod_r f_r^{n_r}\right)\frac{1}{\Gamma(\sum_r n_r)}\int_0^\infty d\xi\ \xi^{\sum_r n_r-1}\prod_{r=1}^N \left[(f_r \xi )^{-n_r-1}\gamma(n_r+1,\xi f_r)\right]\ $$ $$ =\left(\prod_r\frac{1}{f_r}\right)\frac{1}{\Gamma(\sum_r n_r)}\int_0^\infty d\xi \frac{\prod_{r=1}^N \gamma(n_r+1,\xi f_r)}{\xi^{N+1}}\ , $$ where $\gamma(a,x)=\int_0^x dt\ t^{a-1}e^{-t}$ is the lower incomplete Gamma function http://mathworld.wolfram.com/IncompleteGammaFunction.html

Further simplifications are perhaps possible, noting that the lower incomplete Gamma function whose first argument is an integer (as in this case) has an elementary representation as $(1-\text{a finite sum})$ (see the link above).

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    $\begingroup$ that seems to be more than what I expected. $\endgroup$ – jss Feb 12 '16 at 18:33
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    $\begingroup$ @jss done, thanks...I have checked the final formula numerically. It should be much more stable and easier to evaluate numerically than your initial beast. I'll also try to see if the integral can be somehow written in closed form. $\endgroup$ – Pierpaolo Vivo Feb 13 '16 at 7:55
  • $\begingroup$ @jss What prevents you from using the same trick? $\endgroup$ – Pierpaolo Vivo Feb 17 '16 at 15:45
  • $\begingroup$ @jss better keep this answer clean (in this current question, there is no reference to higher moments). I can write another answer to your second question there (when I get time). But if you got the trick, you should be able to find it yourself....why don't you detail (in your OTHER question) what you have tried in full, so I might just point out where the mistake is? $\endgroup$ – Pierpaolo Vivo Feb 17 '16 at 16:57
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    $\begingroup$ I posted the question considering moments here $\endgroup$ – jss Feb 17 '16 at 18:40

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