Please help with finding the limit of the integral

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Homework Statement



I want to show that

\lim_{\epsilon\to 0}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{\epsilon\over 2\pi [(x-x')^2+(y-y')^2+\epsilon^2]^{3\over2}}f(x,y)\;dx\,dy\,\,=f(x',y')? I am not sure what conditions there is on f(x,y), though I do know that \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{z\over 2\pi [(x-x')^2+(y-y')^2+z^2]^{3\over2}}f(x,y)\;dx\,dy is well-defined for all x,y\in R and z>0.

Homework Equations



Please see above section.

The Attempt at a Solution



It is possible that we couldchange variables or sth? Or maybe show that in the limit, the green's function is the delta function? Please help!
 
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Yes, you need to show that in the limit it is a representation of a delta function. You need to i) as \epsilon->0 that that {\epsilon\over [(x-x')^2+(y-y')^2+\epsilon^2]^{3\over2}}
approaches zero unless x=x' and y=y'. And ii) that for fixed \epsilon that the integral of that is 1. Use polar coordinates around the point (x',y'). I think you actually need a bit more than that but that's a good start. To go beyond that I think you do need some conditions on f(x,y).
 
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Thanks, @Dick . So we are showing that that bit is acting like a distribution, right? What conditions are required for f? Is it enough to have f tending to 0?
 
MarkovMarakov said:
Thanks, @Dick . So we are showing that that bit is acting like a distribution, right? What conditions are required for f? Is it enough to have f tending to 0?

The main condition you need is that f is continuous at x',y'. And sure, it shouldn't blow up at infinity so fast the integral doesn't exist.
 
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