How to prove this multivariable integral identity?

Question

By numerical experimentation I found that $$ \lim_{\beta \rightarrow \infty} \frac 1 \beta \int_0^{\beta}dx \int_0^{\beta}dy \, f\left( |x-y| \right) = 2\int_0^{\infty}dx \, f(x) $$ if $f:\mathbb{R} \rightarrow \mathbb{R} $ goes to zero sufficiently fast etc.

How can we prove this identity (I am happy with a hint/sketch) and what is its geometrical meaning? Are there generalizations?

My own attempt: write the double integral on the left-hand side as $$ \int_0^{\beta}dx \int_0^{x}dy \, f\left( x-y \right) + \int_0^{\beta}dx \int_x^{\beta}dy \, f\left( y-x \right) \\= \int_0^{\beta}dx \int_0^{x}dy \, f\left( x-y \right) + \int_0^{\beta}dy \int_0^{y}dx \, f\left( y-x \right) \\= 2 \int_0^{\beta}dx \int_0^{x}dy \, f\left( x-y \right). $$ Not sure what do do next.

Answer 1

Let us try the change of variables $$(x, \delta) = (x,x-y)$$ so that $$ \int g(x,y) dxdy = 2 \int g(x, x-\delta) dx d\delta $$

Application:

$$ \frac1\beta\int_{[0,\beta]^2} f(|x-y| ) dxdy = \frac1\beta\int_{[0,\beta]} dx \int_{x-\beta}^x d\delta f(|\delta| )\\ = \frac1\beta\int_{[-\beta,\beta]} d\delta \int_{\max(0,\delta)}^ {\min(\beta + \delta, \beta)} dx f(|\delta| ) \\= \frac1\beta\int_{[-\beta,0]} d\delta \int_{0}^{\beta + \delta} dx f(|\delta| ) + \frac1\beta\int_{[0,\beta]} d\delta \int_{\delta}^{\beta} dx f(|\delta| )\\= \int_{-\beta}^0 \frac{\beta + \delta}\beta f(-\delta)d\delta + \int_0^\beta \frac{\beta - \delta}\beta f(\delta)d\delta\\= 2 \int_0^\beta \frac{\beta - \delta}\beta f(\delta)d\delta $$

which converges under some conditions (for example: $f\ge 0$) to $$ 2 \int_0^\infty f(\delta)d\delta $$

Answer 2

Firstly, $\int_0^{\beta}dx \int_0^{\beta}dy \, f\left( |x-y| \right)$ is the integral of $g(x,y) = f(|x-y|)$ over $[0,\beta]^2$. Since $g(x,y) = g(y,x)$, we have

$$\int_0^{\beta}dx \int_0^{\beta}dy \, f\left( |x-y| \right) = 2 \int_0^{\beta}dx \int_0^{x}dy \, f\left( x-y \right)$$

i.e. the integral of $g(x,y)$ over $[0,\beta]^2$ is equal to $2$ times is integral over the triangle $x \in [0, \beta], 0\leq y \leq x$.

Remark $\int_0^{x}dy \, f\left( x-y \right) = \int_0^{x}dy \, f\left(y \right)$, so we have

$$\int_0^{\beta}dx \int_0^{\beta}dy \, f\left( |x-y| \right) = 2 \int_0^{\beta}dx \int_0^{x}dy \, f\left(y\right) = 2 \int_0^\beta dx h(x)$$

where $h(x) = \int_0^x dy f(y)$.

It's not difficult to see $\dfrac{1}{\beta}\int_0^\beta h(x)dx \to L$ when $\beta \to +\infty$ if $h(x) \to L$ when $x\to +\infty$

Under some assumptions, we can have $h(x) \to \int_0^\infty dy f(y) $, that allows to conclude.