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

  • 2
    $\begingroup$ 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 $f(a,b)$. $\endgroup$ – Ted Shifrin Sep 22 '17 at 1:13
  • $\begingroup$ 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 $\endgroup$ – Brendan S Sep 22 '17 at 2:24

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