Electric field on the boundary of a continuous charge distribution

Question

In Purcell and Morin's Electricity and Magnetism, 3rd Edition, the claim is made that the magnitude of the electric field on the boundary of a continuous charge distribution is finite (assuming the charge distribution is finite everywhere). I have a question similar to one asked here, but I do not think my question was fully answered.

Equation ($1.22$): $$\vec{E}(x,y,z)=\dfrac{1}{4 \pi \epsilon_0} \int \dfrac{ρ\ (x^\prime, y^\prime, z^\prime)\ \hat{r}\ dx^\prime, dy^\prime, dz^\prime}{r^2}.\tag{1.22}$$

Looking at Equation (1.22), I can see how, when using spherical coordinates, the $r^2$ in the denominator of Equation (1.22) is canceled and, consequently, the integral does not become infinite when $r=0$. However, I do not see how performing the same integral using Cartesian coordinates would not blow up. The limits of integration for this example would contain $$(x,y,z)=(0,0,0)$$ making the denominator 0. There is no $r^2$, when using Cartesian coordinates, to cancel out the $r^2$ in the denominator. To my understanding, an integral cannot be finite in spherical coordinates and infinite in Cartesian. So, how is the Cartesian integral finite?

Answer 1

If you use the Cartesian coordinate, you should recall the definition of "improper integral". Let me give you a simpler example here, consider

$\int_0^1\frac{1}{\sqrt{x}}dx$,

the integrand approaches infinity as $x\to 0$, however, the "improper integral " defines it as a limit, i.e.,

$\int_0^1\frac{1}{\sqrt{x}}dx=\lim_{\epsilon\to0}\int_{\epsilon}^1\frac{1}{\sqrt{x}}dx=\lim_{\epsilon\to0}2\sqrt{x}\big|_{\epsilon}^1=2$,

thus the integral will still give you a finite value.

Note that if the singularity is even stronger, you may consider the Cauchy principle value. But this integrand has the singularity type $1/r^2$, in 3D, this integral is just weakly singular, so probably you do not even need to use the principle value.