It holds that
$$
\int_0^\infty \int_0^\infty f(x_1,x_2) dx_1 dx_2 = 2 \int_0^\infty \int_0^{x_2} f(x_1,x_2) dx_1 dx_2
$$
when $f$ is symmetric, or, in other words, it is invariant under permuting the variables.
Does it hold for some $C$ that
$$
\int_0^\infty \cdots \int_0^{\infty}f(x_1, \cdots, x_n) dx_1 \cdots dx_n = C \int_0^\infty\int_0^{x_n}\cdots\int_0^{x_n}f(x_1, \cdots, x_n) dx_1 \cdots dx_n
$$
where $n$ is the number of integrals, again for $f$ being symmetric. If so, how does $C$ depend on $n$?
Best Answer
In your exact formulation of the question, there is no such constant $C$. This can be seen by trying the pair of functions
$$f_1(x,y,z)=\mathrm{e}^{-x^2-y^2-z^2} \\ f_2(x,y,z)=\mathrm{e}^{-(x-1)^2-y^2-z^2}$$ The reasoning behind the first equation $$ \int_0^\infty \int_0^\infty f(x_1,x_2) dx_1 dx_2 = 2 \int_0^\infty \int_0^{x_2} f(x_1,x_2) dx_1 dx_2 $$
is that the RHS is equal to the sum
$$\iint_{x_1 \leq x_2} f + \iint_{x_1 \geq x_2} f .$$
In the general case with $n$ variables, there are $n!$ permutations of the variables, with $x_1 \leq x_2 \leq \cdots \leq x_n$ being one of them. Thus, the true analogue of your 2-d observation is
$$\int_{\mathbf{R}^n} f(x_1,x_2,\dots,x_n) \mathrm{d} \mathbf{x} = n! \int_0^\infty \int_0^{x_n} \cdots \int_{0}^{x_2} f(x_1,x_2,\dots, x_n)\mathrm{d}x_1 \mathrm{d} x_2 \dots \mathrm{d} x_n. $$