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

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.

• Do you have any more information about your function $f$? or are these the only hypotheses? – Kayle of the Creeks Nov 21 '15 at 13:05
• @KayleoftheCreeks: Yeah, in the case I am considering \begin{align} f(u) = p(u) U(u) \rho U^(u)^{\dagger}, \end{align} where $p(u)$ is a probability distribution, $\rho$ is a density matrix, and $U(u)$ is a unitary operator parameterized by $u$. Having said that, I think whatever method used to evaluate the limit when $f$ is just a functions from the reals to the reals, should carry over fairly easily to the case I'm considering. – e4alex Nov 21 '15 at 13:11

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 $F(2x)-F(-2x) \to \int_{-\infty}^\infty f.$

• Thank you for your answer! I think it's right, but I don't see how you went from the second last equality to the last one. Starting with the second last equality, and making the change of variables $y=x-t$: \begin{align} \int_{-x}^x \Big( F(x-t) - F(-x-t) \Big) dt &= \int_0^{2x} \Big( F(y) - F(y-2x) \Big) dy, \end{align} which differs from what you have above. – e4alex Nov 21 '15 at 21:17
• Applying L'Hopital rule to the above yields \begin{align} \lim_{x\to \infty} \frac{1}{2x} \int_0^{2x} \Big( F(y) - F(y-2x) \Big) dt &= \frac{1}{2} \lim_{x\to \infty} \frac{d}{dx} \int_0^{2x} \Big( F(y) - F(y-2x) \Big) dt \\ &= \frac{1}{2} \lim_{x\to \infty}\left[ 2 \Big( F(2x) - F(0) \Big) + \int_0^{2x} 2 F'(y-2x) \Big) dy \right] \\ &= \lim_{x\to \infty} \left[ F(2x) - F(0) + F(0) - F(-2x) \right] \\ &= \lim_{x\to \infty} \left[ F(2x) - F(-2x) \right]\\ &= \int^{\infty}_\infty f(u) du, \end{align} Would you agree with the above? Thanks again :-) – e4alex Nov 21 '15 at 21:18
• It's much easier in $\int_{-x}^x F(-x-t) \,dt$ to just let $t= t'-x$ to get $\int_{0}^{2x} F(-t') \,dt'.$ – zhw. Nov 21 '15 at 23:16

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

• Perhaps I'm missing something, but I don't see how you can swap the order of integration. The bounds on the inner integral depend on the integration variable of the outer integral. Do you mind explaining that part? – e4alex Nov 21 '15 at 13:24
• My first goes were wrong. Draw the parallelogram in $(u,v)$ space, which I think has vertices (0,-x),(2x,-x),(0,x),(-2x,x). That is two isosceles right triangles. for each triangle, work out the limits with $u$ the outer variable, then the limits for $v$. – Empy2 Nov 21 '15 at 13:36