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An electrodynamics lecture asks me to prove that $$ \nabla \times \left( \frac{\vec{M} \times \vec{x}}{ |\vec{x}|^3} \right) = \frac{8 \pi}{3} \vec{M} \delta^3(\vec{x})- \frac{\vec{M}}{|\vec{x}|^3}+ \frac{3\vec{M} \cdot \vec{x}}{|\vec{x}|^5}\vec{x} $$ where $\vec{M}$ is some constant vector. I have already shown that for $\vec{x} \neq 0$ $$ \nabla \times \left( \frac{\vec{M} \times \vec{x}}{ |\vec{x}|^3} \right) = - \frac{\vec{M}}{|\vec{x}|^3}+ \frac{3\vec{M} \cdot \vec{x}}{|\vec{x}|^5}\vec{x} $$ To determine the value at the singularity I was thinking to consider the integral over any volume V

$$ \int \int \int_V \nabla \times \left( \frac{\vec{M} \times \vec{x}}{ |\vec{x}|^3} \right) d^3x =- \int \int _{dV} \left( \frac{\vec{M} \times \vec{x}}{ |\vec{x}|^3} \right) \times \vec{da} \ \ \ \ \ \ \ \ (*) $$ where the equality is due to a special case of the Gaussian theorem. I think I have to show $$ \int \int \int_V \nabla \times \left( \frac{\vec{M} \times \vec{x}}{ |\vec{x}|^3} \right) d^3x = \frac{8\pi}{3} \vec{M} + \int \int \int_V \left( {- \frac{\vec{M}}{|\vec{x}|^3}+ \frac{3\vec{M} \cdot \vec{x}}{|\vec{x}|^5}\vec{x} } \right )d^3x $$ for all volumes that include the origin.

Is this approach correct? I am stuck at evaluating the integral (*). Can someone give me some pointers? I did not know what this method of integrating singularities by using the delta function is called and I couldn't find anything helpful in my literature or on the internet. Thanks!

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Just to give a few hints:

The strategy is usually to use a theorem that says that two integrals must be equal. In your case the classical Gauss' and Stokes' theorems don't seem to work but some hope I see with $$\tag{1} \int_V\nabla\times \boldsymbol{F}\,dV=\oint_S\boldsymbol{n}\times \boldsymbol{F}\,dS\,. $$ For a simple proof see here. Lets take as closed surface $S$ the sphere with radius $r$ around the origin. Its unit normal is $$ \boldsymbol{n}=\frac{\boldsymbol{x}}{|\boldsymbol{x}|}\,. $$ In your case $$ \boldsymbol{F}=\frac{\boldsymbol{M}\times\boldsymbol{x}}{|\boldsymbol{x}|^3} $$ and by the Grassmann identity, $$ \boldsymbol{n}\times\boldsymbol{F}=\frac{\boldsymbol{M}}{|\boldsymbol{x}|^2}- \frac{(\boldsymbol{M}\cdot\boldsymbol{x})\,\boldsymbol{x}}{|\boldsymbol{x}|^4}\,. $$ You should now be able to calculate the RHS of (1). I believe it is independent of $|\boldsymbol{x}|=r$ so that in the limit the integrand on the LHS must be a delta function whose factor can be determined.

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By expanding the triple cross product one gets

$$ r>0 : M \nabla \cdot \frac{\vec{x}} {|\vec{x}|^3} - \frac{\vec{x}} {|\vec{x}|^3} \nabla \cdot M = M r^{-2}\partial_r(r^2 * r^{-2}) =0 $$

Per general definition or Gauss theorem, the divergence at $r=0$ is the limit of the radial field components flow integrated over the sphere giving

$$M \nabla \cdot \frac{\vec{x}} {|\vec{x}|^3} = 4 \pi r^2 r^{-2} = 4 \pi $$

independent of r.

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