# An integral identity, general multidimensional case

A follow-up to this question. Which of the following equations is true? Can someone give a proof to the correct equation. Thanks.

Please note the prefactors $N$ and $N!$

$$\int_0^A dx_1 \ldots \int_0^A dx_N \quad f(x_1) \ldots f(x_N) = N \int_0^A dx_1 \ldots\int_0^{x_{N-1}}dx_N \quad f(x_1) \ldots f(x_N) \qquad \mathrm{Eq. 1}$$

$$\int_0^A dx_1 \ldots \int_0^A dx_N \quad f(x_1) \ldots f(x_N) = N! \int_0^A dx_1 \ldots\int_0^{x_{N-1}}dx_N \quad f(x_1) \ldots f(x_N) \qquad \mathrm{Eq. 2}$$

The second is correct. Consider the unit cube and try to see that one sixth of it satisfies $x\le y\le z$. More generally, there are $N!$ permutations of any given $N$-tuple of values, and exactly one of them satisfies $x_1\le\dotso\le x_N$.