# Double integral of the product of Dirac delta functions

I'm curious how one would proceed in integrating a function of the form -

$$\iint dx \hspace{1mm} dx' \hspace{1mm} \delta(x - x') \delta(x' - x) f(x, x')$$

My intuition (and the solution to the problem, which is given) tells me that the answer is -

$$f(x', x)$$

but I'm having trouble convincing myself. Thanks

• That makes absolutely no sense. Both $x$ and $x'$ are dummy variables. If you had $\iint dx\,dx'\,\delta(x-a)\delta(x'-b)f(x,x')$, then you would get $f(a,b)$. – Ted Shifrin Sep 22 '17 at 1:13
• Sorry, I was being pretty loose with notation (I've learned most of my math from physicists) and it made a lot more sense written down in front of me. It should have looked like something closer to $\iint dy \hspace{1mm} dz \hspace{1mm} \delta(x - y) \delta(z - w) f(y, z)$ in which case yeah, you're right. In any case, I worked out an easier solution that only involves one integral and everything comes out looking like it should. Thanks – Brendan S Sep 22 '17 at 2:24