Let $U_i$ be $[0,1]$ i.i.d. uniform random variables, for $i=1,\ldots,n$. As an example, let $n=3$. Now pick an ordering, say $x_1>x_2<x_3$. and consider the order statistics integral
$$3!\int\cdots\int_{x_1>x_2<x_3;\ \ 1>x_i>0} dx_1\,dx_2\,dx_3=2. $$
We get that this integral equals the number of permutations $\pi=(\pi_1,\pi_2,\pi_3)$ in $S_3$ with $\pi_1>\pi_2<\pi_3$. The only ones are $(3,1,2)$ and $(2,1,3)$ for a total of 2 as expected. In general, we have for a given fixed ordering $x_1?x_2\cdots?x_n$, where the question-marks correspond to $<$ or $>$:
$$|\{\pi: \pi_1?\cdots?\pi_n\}|=n!\int\cdots\int_{x_1?x_2\cdots?x_n;\ \ 1>x_i>0} dx_1\,dx_2\,dx_3.$$
Question: is there a sensible meaning for the integral:
$$n!\int_{x_1?x_2\cdots?x_n;\ \ 1>x_i>0} \,x_1\,dx_1\,dx_2\,dx_3\cdots dx_n.$$
I want to conclude that it's (related to) the expected value of the first element $\pi_1$ of a uniformly random permutation drawn from the set $\{\pi: \pi_1?\cdots?\pi_n\}$. Unfortunately, this does not seem to be the case. Is there a way to remedy this?