I'm reading a physics paper that uses the following identity:
$$ \int_{0}^{\tau} dt_1\int_{0}^{\tau} dt_2 = \int_{0}^{\tau} dt_1\int_{0}^{t_1} dt_2 \; + \; \int_{0}^{\tau} dt_2 \int_{0}^{t_2} dt_1. $$
I'm not sure where this idenity comes from? Does it come from:
$$ \int_{0}^{\tau} dt_1 = \int_{0}^{b} dt_1 + \int_{b}^{\tau} dt_1. $$
If someone can "prove" the identity, that'd be great.
I'll call $t_1 :=x$ and $t_2 := y$ just to have a picture of what happens in the $2D$ plane.
The proof : $$ (A)\ \int_0^{\tau} dx \int_0^{x} dy = \int_0^{\tau} dx (x-0) = \frac{x^2}{2}|^{\tau}_0 = \tau^2/2 \\ (B)\ \int_0^{\tau} dy \int_0^{y} dx = \int_0^{\tau} dy (y-0) = \frac{y^2}{2}|^{\tau}_0 = \tau^2/2 $$ So the RHS $ = \tau^2$ and the LHS clearly $= \tau^2$ since it is just the integral of $1\ dxdy$ over the square in the plane $[0,\ \tau]^2$
For the intuition let's look at the integral $(A)$ and think of the $2D$ plane. We first fix an $x$ coordinate (the outer integral is done over $x$) which represents a vertical line in the plane. We then integrate over this vertical line from the bottom of the square (the $x$-axis) to the fixed $x$ value. Then varying our fixed $x$ will cover the lower rectangular triangle that appears when you cut the square domain $[0,\ \tau]^2$ by the curve $y = x$ (a simple diagonal). This represents half of the domain to integrate. The $(B)$ integral is exactly the same, but this time you first fix $y$ and the integral will cover the upper rectangular triangle.
So it more or less comes from the fact you quoted in the sense that you can cut the integral domain in several parts, integrate over them, and sum the results. However the fact you proposed represents this idea in one dimension, here we work in $2D$.