How to prove this multivariable integral identity? 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.
 A: 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.
A: 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
$$ 2 \int_0^\infty  f(\delta)d\delta
$$
