This seems very basic, however, it's been bugging me all weekend. A TA for one of my engineering courses wrote this on the board the other day:
$$\int_{-a}^{a} \mathrm{d}x_1\int_{-a}^{a} \mathrm{d}x_2\operatorname{sinc}(x_1-x_2)=2a\int_{-2a}^{2a}\mathrm{d}x\left(1-\frac{|x|}{2a}\right)\operatorname{sinc}(x),$$
where $\operatorname{sinc}(x)=\frac{\sin(x)}{x}$.
This is very nice as it reduces the number of integrals. I also recall that he mentioned that this works for any even function, not just $\operatorname{sinc}$.
I am stumped as to how one derives this. I am familiar with variable substitution, but I can't figure out how one removes one of the integrals. Any help?
Let's think of what we are integrating, so that you can get an intuitive idea of how this works. We are integrating with respect to two variables around a square of side length $2a$. We are only interested in the positive difference between the two variables, since we have an even function. So what we can do is take a weighted average of all the differences between the two variables in the square, and then use that to integrate across all the x values.
The weight for each x value will be proportional to the length of that diagonal on the square which represents $x-y=c$, where c is some constant between $-2a$ and $2a$. Thus the weight will be $2a-|x|$, and so we have our equality: For an even function $f(x)$, we have $$\int_{-a}^{a}\int_{-a}^{a}f(x_1-x_2)dx_1dx_2=\int_{-2a}^{2a}(2a-|x|)f(x)dx$$