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How do we argue that [itex]\nabla^2\frac{1}{r}[/itex] is a three dimensional delta function? I have seen some people do it using the divergence theorem, i.e. saying that

[tex]

\int_V \nabla\cdot\nabla\frac{1}{r}dv=-\oint_S \nabla\frac{1}{r}\cdot ds=-4\pi

[/tex]

if S is a surface containing the origin, but I don't think it is legal to use the divergence theorem in this case because it requires that both the vector field and its divergence be well defined.

Instead one could make the sequence [itex]\nabla^2\frac{1}{r+\varepsilon}[/itex] which approaches the correct function as [itex]\varepsilon\rightarrow0[/itex]. Here we can calculate the laplacian and then integrate this.. What do you think?

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# 3D delta function

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