Different answers for self-energy?

Question

If we calculate the self-energy using the energy density integral then we get $$U=\int^\infty_0\frac{q^2}{8\pi\epsilon_0 r^2}dr=-\frac{q^2}{8\pi\epsilon_0 r}$$ However, since the potential of a point charge is given by $\phi=\frac{q}{4\pi\epsilon_0 r}$ we can write $$U=q\phi=+\frac{q^2}{4\pi\epsilon_0 r}$$ So why are these two answers differed by a factor of $-1/2$?

Answer 1

Point charges are not great for using the integral formulation in order to find the total internal energy of the system. If you'll remember, the total energy of a system composed of $N$ point charges, when $V(\mathbf{r}_i)$ is the potential generated at the position of $q_i$ due to all the other $N-1$ charges is: $$ U = \frac{1}{2} \sum_{i=1}^{N} q_i V(\mathbf{r}_i), $$ where we clearly have not counted the influence of the $i$-th charge over itself. This formulation may be extended to a continuum, with charge density $\rho$: $$ \frac{1}{2} \int \rho V\,\mathrm d\tau = \frac{\epsilon_0}{2} \int E^2\,\mathrm d\tau, $$

where $E$ is the electric field due to all the charges in the configuration. The integral is over all space. In this situation, $V$ is the total potential, whereas in the former, $V(\mathbf{r}_i)$ would not count the potential due to the $i$-th charge. For a continuous distribution there is no distinction, since the amount of charge right at the point $\mathbf{r}$ is so small.

Actually, if you'd calculate the internal energy of a point charge via an integral formulation, you'd get

$$ \frac{\epsilon_0}{2} \int E^2\mathrm d\tau = \frac{\epsilon_0}{2} \int_{0}^{2 \pi} \int_{0}^{\pi} \int_{0}^{\infty} \left( \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2} \right) r^2\sin\theta \,\mathrm dr\,\mathrm d\theta\,\mathrm d\phi \rightarrow \infty, $$

which is an embarrassment afflicting both quantum AND classical EM theory.