# Calculating higher order moments of a product of weigthed average parameters

I am trying to find a closed form or a transformation which simplify the numerical treatment of this multiple integral:
$$\int_{x_1=0}^1…\int_{x_N=0}^1 \prod_r x_r^{m_r}\left(\frac {x_r f_r} {\sum_g x_g f_g}\right) ^{n_r} dx_1…dx_N$$
where $m_r$ are nonegative integers, $n_r$ are positive integers, and $f_r\ge0$ constants.

There is an answer for moment $m=0$ here. Following the same exponentiation procedure suggested there by @Pierpaolo

$$\left(\prod_r {f_r^{n_r}}\right)\int_{x_1=0}^1…\int_{x_N=0}^1 \left(\sum_g{x_gf_g}\right)^{-\sum_r {n_r}} \prod_r x_r^{n_r+m_r} dx_1…dx_N$$

$$=\left(\prod_r f_r^{n_r}\right)\frac{1}{\Gamma(\sum_r n_r)}\int_0^\infty d\xi\ \xi^{\sum_r n_r-1}\prod_{r=1}^N \left[\int_0^1 dx\ x^{n_r + m_r}e^{-\xi f_r x}\right]$$

$$=\left(\prod_r{f_r^{-(m_r+1)}}\right)\frac{1}{\Gamma(\sum_r n_r)}\int_0^\infty d\xi {\frac 1 {\xi^{N+1+\sum_r {m_r}}}}{\prod_{r=1}^N \gamma(n_r+m_r+1,\xi f_r)}$$

Is that correct? There exist a closer form solution?