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Johan Wästlund
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The integral essentially asks for the probability that, for $n$ independent "events" uniformly distributed in $[0,1]$, at least one happens after $c_n$, at least two happen after $c_{n-1}$, etc (thinking of the unit interval as time). Let $P_n(c_1,\dots,c_n) = n!\cdot J$ denote this probability (the integral $J$ also requires the $n$ events to occur in a specified order).

If we condition on the number $k$ of events in the interval $[c_n,1]$, then the remaining events will be uniformly distributed in $[0,c_n]$, and we can write $P_n$ as $$P_n(c_1,\dots,c_n) = Pr(k=1)\cdot P_{n-1}\left(\frac{c_1}{c_n},\dots,\frac{c_{n-1}}{c_n}\right) + Pr(k=2)\cdot P_{n-2}\left(\frac{c_1}{c_n},\dots,\frac{c_{n-2}}{c_n}\right)+\dots.$$

To complete a recursive computation, we only need to compute the $O(n^2)$ integrals of the form $$P_i\left(\frac{c_1}{c_j},\dots,\frac{c_i}{c_j}\right)$$ for $1\leq i < j\leq n$. I'm not an expert on the complexity of rational arithmetic, but it seems to me that this should be doable in polynomial time.

Johan Wästlund
  • 5.5k
  • 3
  • 28
  • 35