I encountered a double integral property for the Dyson series in Quantum mechanics. The derivation seemed to make use of this property:
$$ \int_0^y \int_0^{x_1}f(x_1)f(x_2) dx_2 dx_1 = \frac{1}{2} \int_0^y \int_0^y f(x_1)f(x_2) dx_2 dx_1$$
I dont see how to prove this, but after trying it a few functions on Mathematica, it seems to hold up. Can it also be generalized to an arbitrary number of integration varables?
$$ \int_0^y \int_0^{x_1} … \int_0^{x_{n-1}} f(x_1)…f(x_n) dx_n … dx_2 dx_1 = \frac{1}{n!} \int_0^y \int_0^{y} … \int_0^{y} f(x_1)…f(x_n) dx_n … dx_2 dx_1$$
Best Answer
Hint: the domain of the second integral is the square $[0,y]\times[0,y]$, while in the first integral is a triangle (draw it), subset of the square. Finally, consider the symmetry of the integrand $f(x_1)f(x_2)$.