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I asked this question on MSE, but received no answer.

Recently, reading this problem, I found out that

$$ \lim_{n\to \infty} \int_{0}^{1} \dotsi \int_{0}^{1} \frac{x_1^q + \dotsb + x_n^q}{x_1^p + \dotsb + x_n^p} \, \mathrm{d}x_1 \dotsm \mathrm{d}x_n =\frac{p+1}{q+1}. $$

Is a general formula known for that multiple integral when we fix $ n$, $p $ and $q$ as positive integers? Otherwise is it possible to have a general formula if $q=2$, $p=1$ and $n$ is a fixed positive integer?

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    $\begingroup$ the case $q=2$, $p=1$ is answered at mathoverflow.net/q/288085/11260 $\endgroup$
    – Carlo Beenakker
    Commented Apr 28 at 14:54
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    $\begingroup$ I think for the case (q,1) there could be a formula similar to the case (2,1); for general (q,p) it may be far more complicated $\endgroup$
    – Pietro Majer
    Commented Apr 28 at 16:52
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    $\begingroup$ @PietroMajer thank you Professor, I was one of your students at University of Pisa long time ago! $\endgroup$
    – user967210
    Commented Apr 28 at 19:37
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    $\begingroup$ Hi! I'm sorry, I'm not sure I remember of you now... I'm getting old... But welcome to MO! $\endgroup$
    – Pietro Majer
    Commented Apr 29 at 9:54

2 Answers 2

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Using the substitution $u_i=x_i^p$ and letting $r:=q/p$ and $a:=1-1/p\in[0,1)$, we see that the integral in question is \begin{align} I&=np^{-n}\int_0^1\cdots\int_0^1\frac{du_1\cdots du_n}{u_1^a\cdots u_n^a} u_1^r\int_0^\infty dt\,e^{-t(u_1+\cdots+u_n)} \\ &=np^{-n}\int_0^\infty dt\,\int_0^1\frac{du_1}{u_1^{a-r}}\,e^{-tu_1} \Big(\int_0^1\frac{du_2}{u_2^a}\,e^{-tu_2}\Big)^{n-1} \\ &=np^{-n}\int_0^\infty \frac{dt}{t^{r+n(1-a)}}\gamma(1+r-a,t)\, \gamma(1-a,t)^{n-1}, \end{align} where $\gamma$ is the lower incomplete gamma function.

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  • $\begingroup$ Great answer, thank you. Is it possible to further simplify, as in the case $p=1, q=2$ mathoverflow.net/questions/288085/… ? $\endgroup$
    – user967210
    Commented Apr 30 at 6:07
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    $\begingroup$ @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. $\endgroup$
    – Iosif Pinelis
    Commented Apr 30 at 14:39
  • $\begingroup$ 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. $\endgroup$
    – esg
    Commented May 2 at 20:05
  • $\begingroup$ @esg : Indeed, the case $p=1$ seems much easier, because then $\gamma(1-a,t)=1-e^{-t}$ for $t>0$. $\endgroup$
    – Iosif Pinelis
    Commented May 3 at 18:01
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Analogously to the answer for $p=1,q=2$ at https://mathoverflow.net/a/289068/11260 , one has for $p=1$ and general $q$ the integral expression $$\int_{0}^{1} \cdots \int_{0}^{1} \frac{x_1^q + \cdots + x_n^q}{x_1 + \cdots + x_n} \, \mathrm{d}x_1 \cdots \mathrm{d}x_n=$$ $$\qquad =n\int_0^1 du\int_0^\infty dt\, t^{1-n}u^q e^{-tu}(1-e^{-t})^{n-1}$$ $$\qquad=nq!\int_0^\infty dt\,(1-e^{-t})^{n-1}\frac{1}{t^{n+q}}\left(1-e^{-t}\sum_{k=0}^q\frac{t^{k}}{k!} \right).$$

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