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I want a convincing proof of this...

$\nabla^2 (1/|r-r'|)$ = $-4\pi \delta^3 (r-r')$

(where r and r' are vectors and $\delta^3 (r-r')$ represents the 3-D delta function...)

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Follows from the divergence theorem, doesn't it? en.wikipedia.org/wiki/Divergence_theorem (Recall that the Laplacian of a scalar function is the divergence of its gradient.) –  Qiaochu Yuan Feb 2 '11 at 13:45
    
That is of course the standard method. But i wish to do it differently.. –  Guanidene Feb 2 '11 at 14:03
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can you say why the standard method is not convincing enough? That may help in addressing the answer. –  Willie Wong Feb 2 '11 at 14:06
    
That is of course the standard method. But i wish to do it differently.. Suppose I integrate both sides of the equation in 3d i.e. with respect to volume; then the RHS becomes $-4\pi$ . Now my task is to show that the LHS is also $-4\pi$. In LHS - Now I replace dV by $r^2 \sin \theta d\theta d\phi dr$ .Now the $\theta$ and $\phi$ can be taken out of integration to give $4\pi$ . Now am left with the radial part. I then substitute the value of $\nabla^2$ in spherical coordinates... Now I am stuck. (I have to prove that the integral is -1. I am also concerned about the limits of integration.) –  Guanidene Feb 2 '11 at 14:10
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that approach is not completely appropriate: in spherical coordinates you easily see that $\nabla^2 1/|r|$ is 0 at every point with $r \neq 0$. But at $r = 0$ your coordinate system is degenerate; but all the "action" happens at $r=0$. –  Willie Wong Feb 2 '11 at 14:24

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