and why is it equal to div( div (1/|x - x'|) ) ?

I calculated div( div (1/|x - x'|) ) = 0

what is this function?

The Dirac delta function is the distribution, [delta](x) that to each function f assigns the value f(0).
It is sometimes defined as a "function" such that [delta](x)= 0 if x is not 0 and is infinite AT x=0 in such a way that [delta](x)dx= 1 as long as the interval of integration includes 0. Of course, that is not really a function.

Another way to look at it (in 1-dimension) is as the limit of functions of the kind d_i(x)= 0 if |x|>1/i and d_i(x)= k/2 if |x|<= 1/i.
Notice that, for each i, d_i(x)dx= (k/2)(2k)= 1 but d_i(x)= 0 outside a very small region.
(For two or three dimension, us a disk or sphere)
If you take "limit of functions" literally, then the limit function is 0 if x is not 0 and undefined if x is 0. Of course, since the Dirac delta function is NOT a true function, you don't take it literally!

One very nice property of distributions is that we can think of distributions as being derivatives of functions that don't have normally have derivatives.
The Dirac delta function (again in one dimension) can be thought of as the derivative of a STEP function. If f(x)= 0 for x< 0, f(x)= 1 for x>=0, then its derivative is 0 as long as x is not 0, for x=0 (well, actually the derivative does not exist for x=0- but thinking of it that way gives us the Dirac delta function). A variation of this is to define f_i to be 0 for x< -1/i, 1 for x> 1/i, and the straight line connecting (-1/i, 0) and (1/i,1). Each f_i has derivative 0 for x<-1/i, i/2 for x between -1/i and 1/i and 0 for x> 1/i. The Dirac delta function is the limit as i-> of those derivatives. Of course, that will be 0 as long as x is not 0 and doesn't exist for x= 0.

Now perhaps you see why you got 0 as an answer! To get the Dirac delta function, you have to look closely at what happens at the origin (and think creatively,don't just say it doesn't exist!).

{ 0 for t > 1/ or t < 0
{ for 0 < t < 1/

and taking the limit as 1/ -> 0

It is an idealized, normalized step. One nice thing about it in the class I'm taking now is that it's Laplace transform is 1.

enigma

grad( div (1/|x-x'|) )

It is confusing that grad and div (and curl!) all involve the same symbol, but there is sense behind it.

Hurkyl

I would have used the del symol but there isn't one at this site.

Two things you need to be careful of:

• You can't change the argument inside the delta at will, even if the new argument has the same zeros. eg for const a: delta(a*x) = delta(x)/|a|
• In multidimensional spaces, the delta function in non-Cartesian coordinates is defined so that delta^3(r)dV=1. eg we don't expand out the r^2*sin() in spherical coordinates.

∇
[nabla]

What?!

div(grad(1/|x-x'|))=-4delta(x-x')