## Cauchy distribution instead of Coulomb law?

A recent question by alexv - and his comment that the answer will eventually be used in Gravity modeling - has triggered the following in my mind. It's about Electric modeling instead of Gravity modeling, but the inverse square law is similar.

The energy density in the electric field of a point charge $q$ is given by $\frac{1}{2} \epsilon_0 E^2$ where the electric field $E$ at a distance $r$ is: $$E = \frac{q}{4 \pi \epsilon_0 r^2}$$ The total energy in the field is thus given by the integral: $$U = \int_0^\infty\frac{1}{2}\epsilon_0 \left(\frac{q}{4\pi\epsilon_0 r^2} \right)^2 4 \pi r^2 dr \ = \frac{q^2}{8 \pi \epsilon_0} \int_0^\infty \frac{dr}{r^2} = \frac{q^2}{8 \pi \epsilon_0} \left[ \frac{1}{r} \right]_0^\infty = \infty$$ Hence there is an infinite outcome for de self energy of the electron. As is well known, this is quite a serious problem in classical electrodynamics.
It is shown below how this problem can be resolved by sort of renormalization, as understood by this author. Replace $r^2$ by $(r^2+\sigma^2)$ where $\sigma$ is interpreted as the "size" of the electron. Such a Cauchy distribution is in agreement with Coulomb's law at reasonable distances from the origin.
At the same time the singularity at the origin is effectively removed, because we have the following extremely simple result: $$U = \frac{q^2}{8 \pi \epsilon_0} \int_0^\infty \frac{r^2\,dr}{(r^2+\sigma^2)^2} = \frac{q^2}{8 \pi \epsilon_0} \frac{1}{4}\frac{\pi}{\sigma} = \frac{q^2}{32\,\epsilon_0\sigma}$$ The outcome must be equal to $m_0 c^2$, hence we calculate for the electron radius: $$\sigma = \frac{q^2}{32\,\epsilon_0 m_0 c^2} = \frac{\pi}{8} r_e$$ Where $r_e$ is the classical electron radius: Classical electron radius
I'm wondering if an approach as shown here been tried before. And if not, why not?