How to take the limit: x goes to infinity, when x appears in the bounds of an integral?

Question

To be specific, I am wondering how to take the limit of the following expression: \begin{align} L=\lim_{x \to \infty} \frac{1}{2x} \int_{-x}^x dh \int_{-x}^x dg \ f(g-h), \end{align} where $f(g-h)$ is a well behaved function.

I can change coordinates to $u=g-h$ and $v=h$; the ‎Jacobian of this transformation is J=1, so the limit becomes \begin{align} L=\lim_{x \to \infty} \frac{1}{2x} \int_{-x}^x dv \int_{-x-v}^{x-v} du \ f(u). \end{align}

At this point I'm stuck. I believe, by looking at it, the limit is \begin{align} L = \int_{-\infty}^{\infty} du \ f(u), \end{align} but I would really like a way to see this formally. It would be nice to know of general methods to solve the above limit as well.

Answer 1

We can stick to simple one variable changes of variable. Let $F$ be an antiderivative for $f.$ Then

$$\int_{-x}^x\int_{-x}^x f(s-t)\,ds\,dt = \int_{-x}^x\int_{-x-t}^{x-t} f(s)\,ds\,dt$$ $$ = \int_{-x}^x (F(x-t)-F(-x-t)) \,dt = \int_0^{2x}(F(t)-F(-t))\, dt.$$

Thus the expression we're interested in is

$$\frac{\int_0^{2x}(F(t)-F(-t))\, dt}{2x}.$$

By L'Hopital, the limit of this equals the limit of $F(2x)-F(-2x) \to \int_{-\infty}^\infty f.$

Answer 2

Swap the order of integration, so you have $$\int_{-2x}^{0}duf(u)\int_{-x-u}^{x}dv+\int_0^{2x}udf(u)\int_{-x}^{x-u}dv$$