Timeline for Closed form for $ \int_{0}^{1} \dotsi \int_{0}^{1} \frac{x_1^q + \dotsb + x_n^q}{x_1^p + \dotsb + x_n^p} \, \mathrm{d}x_1 \dotsm \mathrm{d}x_n $

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May 3 at 18:01	comment	added	Iosif Pinelis		@esg : Indeed, the case $p=1$ seems much easier, because then $\gamma(1-a,t)=1-e^{-t}$ for $t>0$.
May 2 at 20:05	comment	added	esg		For $p=1$ and general $q$ formulas similar to the case $p=1,q=2$ exist, as in the linked answer one can show that $$\mathbb{E}\bigg(\frac{X_1^q+\ldots + X_{n+1}^q}{X_1+\ldots+X_{n+1}}\bigg)=\frac{n(n+1)}{n!} \sum_{i=0}^n {n \choose i} (-1)^{n-i} \int_0^1 u^q(u+i)^{n-1}\log(u+i)\,du$$, and one proceeds from here simarly as there.
Apr 30 at 14:39	comment	added	Iosif Pinelis		@user967210 : Thank you for your appreciation. Apparently, in general this expression cannot be further simplified. Even for $n=3$ (and arbitrary natural $p$ and $q$), Mathematica can only reduce the triple integral to a double integral involving a hypergeometric function.
Apr 30 at 6:07	comment	added	user967210		Great answer, thank you. Is it possible to further simplify, as in the case $p=1, q=2$ mathoverflow.net/questions/288085/… ?
Apr 29 at 16:25	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 2 characters in body
Apr 28 at 19:33	history	edited	Iosif Pinelis	CC BY-SA 4.0	deleted 2 characters in body
Apr 28 at 19:22	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 7 characters in body
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Apr 28 at 17:55	history	edited	Iosif Pinelis	CC BY-SA 4.0	deleted 34 characters in body
Apr 28 at 17:50	history	answered	Iosif Pinelis	CC BY-SA 4.0