Solving an $n$-dimensional integral Let $p> - 1$. Find the integral
$$ \int\limits_{(0,1)^n} \left( \frac{\min(x_1,...,x_n)}{\max(x_1,...,x_n)} \right)^p dx_1...dx_n. $$
 A: Consider the integral on subset of $(0,1)^n$ where $x_1=\min(x_1,\dots,x_n)$ and $x_n=\max(x_1,\dots,x_n)$,
$$I:=\int_{x_1=0}^1\int_{x_n=x_1}^1\int_{x_1\leq x_2,\dots,x_{n-1}\le x_n} \left(\frac{x_1}{x_n}\right)^p dx_1\dots dx_n\\
=\int_{x_1=0}^1\int_{x_n=x_1}^1 \left(\frac{x_1}{x_n}\right)^p(x_n-x_1)^{n-2} dx_n dx_1\\
=\int_{x_n=0}^1x_n^{n-1}\left(\int_{t=0}^1 t^p(1-t)^{n-2} dt\right)
 dx_n\\
=\int_{x_n=0}^1x_n^{n-1}B(p+1,n-1) dx_n=\frac{B(p+1,n-1)}{n}
$$
where $B$ is the Beta function.
The integral on $(0,1)^n$ should be $n(n-1)I$, because we have $n(n-1)$ ways to choose an ordered couple of distinct variables among $x_1,\dots,x_n$, that is $\min(x_1,\dots,x_n)$ and $\max(x_1,\dots,x_n)$.
Hence
$$\int\limits_{(0,1)^n} \left( \frac{\min(x_1,...,x_n)}{\max(x_1,...,x_n)} \right)^p dx_1...dx_n=(n-1)B(p+1,n-1)=\frac{(n-1)!}{\prod_{i=1}^{n-1} (p+i)}=\frac{1}{\binom{p+n-1}{n-1}}.$$
A: There are $n(n-1)$ pairs $j,k$ so that $x_j=\min\limits_ix_i$ and $x_k=\max\limits_ix_i$. On each of these identical regions, the integral is
$$
\int_0^1\int_0^b\left(\frac ab\right)^p(b-a)^{n-2}\,\mathrm{d}a\,\mathrm{d}b
$$
So the sum over the $n(n-1)$ identical regions would be
$$
\begin{align}
&n(n-1)\int_0^1\int_0^b\left(\frac ab\right)^p(b-a)^{n-2}\,\mathrm{d}a\,\mathrm{d}b\\
&=n(n-1)\int_0^1b^{n-1}\int_0^1a^p(1-a)^{n-2}\,\mathrm{d}a\,\mathrm{d}b\\
&=n(n-1)\int_0^1b^{n-1}\frac{\Gamma(p+1)\Gamma(n-1)}{\Gamma(p+n)}\,\mathrm{d}b\\
&=(n-1)!\frac{\Gamma(p+1)}{\Gamma(p+n)}\\
&=\frac{(n-1)!}{(p+1)(p+2)\cdots(p+n-1)}\\
&=\frac1{\binom{p+n-1}{n-1}}
\end{align}
$$
