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Once again, I was reading my Jackson [1], which characteristically had the statement “the […] integral can easily be shown to have the value \( 4 \pi \)”, in a discussion of electrostatic energy and self energy.

The integral is
\begin{equation}\label{eqn:selfEnergyIntegral:20}
I = \int \frac{\Brho}{\rho^3} \cdot \frac{\Brho + \Bn}{\Norm{\Brho + \Bn}^3} d^3 \rho.
\end{equation}

This is something that I once figured out once before (see [2] appendix C). However, trying to do it a second time around, I think that I found the “easy” way.

As Jackson hints, the starting point is
\begin{equation}\label{eqn:selfEnergyIntegral:40}
\frac{\Bx}{\Norm{\Bx}^3}
=
-\spacegrad \inv{\Norm{\Bx}},
\end{equation}
but we don’t have to apply it to both the vector terms, as I did in my initial attempt (which results in a Laplacian to reduce.) Inserting this and applying chain rule, we find
\begin{equation}\label{eqn:selfEnergyIntegral:60}
\begin{aligned}
I
&= -\int \frac{\Brho}{\rho^3} \cdot \spacegrad_\Brho \inv{\Norm{\Brho + \Bn}} d^3 \rho \\
&=
-\int
\spacegrad_\Brho \cdot \lr{
\frac{\Brho}{\rho^3} \cdot
\inv{\Norm{\Brho + \Bn}}
}
d^3 \rho
+
\int
\lr{
\spacegrad_\Brho \cdot
\frac{\Brho}{\rho^3}
}
\inv{\Norm{\Brho + \Bn}}
d^3 \rho
\\
\end{aligned}
\end{equation}

The first integral can be evaluated using an infinite spherical shell
\begin{equation}\label{eqn:selfEnergyIntegral:80}
\begin{aligned}
I_1
&= -\int
\spacegrad_\Brho \cdot \lr{
\frac{\Brho}{\rho^3} \cdot
\inv{\Norm{\Brho + \Bn}}
}
d^3 \rho \\
&=
\lim_{\rho \rightarrow \infty}
-\frac{\Brho}{\rho} \cdot \lr{
\frac{\Brho}{\rho^3} \cdot
\inv{\Norm{\Brho + \Bn}}
} 4 \pi \rho^2 \\
&=
\lim_{\rho \rightarrow \infty}
\frac{-4 \pi}{\Norm{\Brho + \Bn}} \\
&=
0.
\end{aligned}
\end{equation}

The divergence term in the second integral, provided \( \Bx \ne 0 \), has the form
\begin{equation}\label{eqn:selfEnergyIntegral:100}
\begin{aligned}
\spacegrad \cdot \frac{\Bx}{\Norm{\Bx}^3}
&=
\inv{\Norm{\Bx}^3} \spacegrad \cdot \Bx
+
\lr{ \Bx \cdot \spacegrad } \inv{\Norm{\Bx}^3} \\
&=
\frac{3}{\Norm{\Bx}^3}
+
2 \frac{x_k x_j}{\Norm{\Bx}^5} \lr{-\frac{3}{2}} \partial_k x_j \\
&=
\frac{3}{\Norm{\Bx}^3}
– \frac{3}{\Norm{\Bx}^3}
\end{aligned}
\end{equation}
However, in a neighbourhood of the origin, this actually has a delta function structure. We can see that from Gauss’s law, where we have
\begin{equation}\label{eqn:selfEnergyIntegral:120}
\spacegrad \cdot \BE = \frac{\rho}{\epsilon_0}.
\end{equation}
If we plug in the integral representation of \( \BE \) on the LHS, we have
\begin{equation}\label{eqn:selfEnergyIntegral:140}
\begin{aligned}
\spacegrad \cdot \BE
&=
\spacegrad \cdot \int \frac{\rho(\Bx’)}{4 \pi \epsilon_0} \frac{\Bx – \Bx’}{\Norm{\Bx – \Bx’}^3} d^3 x \\
&=
\int \frac{\rho(\Bx’)}{4 \pi \epsilon_0} \spacegrad \cdot \frac{\Bx – \Bx’}{\Norm{\Bx – \Bx’}^3} d^3 x \\
&=
-\int \frac{\rho(\Bx’)}{4 \pi \epsilon_0} \spacegrad’ \cdot \frac{\Bx – \Bx’}{\Norm{\Bx – \Bx’}^3} d^3 x.
\end{aligned}
\end{equation}
Comparing the LHS and RHS, we must have
\begin{equation}\label{eqn:selfEnergyIntegral:160}
\spacegrad’ \cdot \frac{\Bx’ – \Bx}{\Norm{\Bx’ – \Bx}^3} = 4 \pi \delta^3\lr{\Bx’ – \Bx}.
\end{equation}

We can now substitute that into the second integral to find
\begin{equation}\label{eqn:selfEnergyIntegral:n}
\begin{aligned}
I_2 &=
\int
\lr{
\spacegrad_\Brho \cdot
\frac{\Brho}{\rho^3}
}
\inv{\Norm{\Brho + \Bn}}
d^3 \rho \\
&=
\frac{4 \pi}{\Norm{\Bn}} \\
&=
4 \pi.
\end{aligned}
\end{equation}

Sure enough, the integral has a \( 4 \pi \) value. But was that easy?  I think Hitler would disagree.

References

[1] JD Jackson. Classical Electrodynamics. John Wiley and Sons, 2nd edition, 1975.

[2] Peeter Joot. Electromagnetic Theory. Kindle Direct Publishing, Toronto, 2016.