I know of the identity $\int dy \delta(x-y)\delta(y-x') = \delta(x-x')$ but what about if I have an integral like $$\int dy \delta(x-y)\delta(y-x')f(y)$$Can the above identity be used in any way? Can say something like this?$$\int dx\int dy\delta(x-y)\delta(y-x')f(y) = \int dx \delta(x-x')f(x)$$
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Intuitively that should equal $f(x)\delta(x-x')$. – Giuseppe Negro Feb 26 '23 at 12:36
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Yes. You can think of integrals with delta functions as this: the integral will only not be zero at x=x‘. You can plug the function g(x)=1 into the identity you listet and the result is g(x‘)* delta(x-x‘)=delta(x-x‘). Applying this to the second integral gives the result f(x‘)*delta(x-x‘)
David
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1For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Feb 26 '23 at 12:43
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How is this an answer to the question? I don't follow at all. Where do you "plug $g=1$"? And how does this prove the formula for an arbitrary $f$? – Giuseppe Negro Feb 27 '23 at 12:11