# Electrostatic potential resolving singularity

I'm trying to determine the electrostatic potential caused by a specified charge density function: $$\rho_c(\vec{r}) =\begin{cases} 1 & \vec{r} \in V\\ 0 & \text{otherwise} \end{cases}$$

The solution using Green's function is then: $$\phi(\vec{r}) = \iiint_{V} \frac{\rho_c(\vec{r}')}{4 \pi \epsilon_0 \|\vec{r}-\vec{r}'\|} d \vec{r}'$$

I can understand how this integral can be easily evaluated so long as $$\vec{r} \not\in V$$ since there is no singularity in the integral, however I don't know how to resolve the integral if $$\vec{r} \in V$$, since this appears to cause the integral to become infinite/undefined.

Suppose for a concrete example that $$V$$ is an axis-aligned rectangular prism with opposing corners at $$(x_0,y_0,z_0)$$ and $$(x_1,y_1,z_1)$$. How do I find $$\phi(\vec{r})$$ inside of this box, and how do I generalize that procedure to any arbitrary $$V$$?

• The singularity is integrable... Aug 31 at 11:46
• I'd imagine it is since I can see one alternative solution to be to use the Green's function method to produce boundary conditions outside of $V$ for a Poisson PDE solver, however I don't know what mathematical tools I need to use to evaluate this integral "directly". Suppose I took the 1D case with $\rho_c(x) = 1$ for $x \in [-1, 1]$. I think this produces the integral $\phi(0) = \int_{-1}^1 \frac{1}{4\pi \epsilon_0 |x|} dx$, which doesn't converge, but using the "Poisson with BC's" approach gives $\phi(0) = \frac{6 \pi + \log(3)}{4 \pi \epsilon_0}$. Aug 31 at 12:38
• In 1D there is no Green function, but in 2D there is and it is proportional to $\ln \sqrt{x^2+ y^2}$ which is locally integrable as well in $dxdy$ Aug 31 at 13:17

Here's a very simple example: what is the improper integral $$\int_0^1\frac{dt}{\sqrt{t}}$$? It is $$\frac{1}{2}$$ even though $$\frac{1}{\sqrt{t}}\to \infty$$ as $$t\to 0^+$$. The reason is that although the function is unbounded, it is not "bad enough" for the purposes off integration. This is what @Valter Moretti means by "the singularity is integrable". It goes off to $$\infty$$ slowly enough that it can be integrated over to yield a finite result. On the other hand, $$\int_0^1\frac{dt}{t}=\infty$$.
So, for integration purposes, mere unboundedness of the function alone is not enough to deduce anything about the finiteness of the integral. In $$3$$ dimensions, \begin{align} \int_{\|\mathbf{r}'\|\leq 1}\frac{dV'}{\|\mathbf{r}'\|}=\int_0^1\frac{1}{r'}4\pi r'^2\,dr'=4\pi\int_0^1r'\,dr' \end{align} is certainly finite. And in general, in $$\Bbb{R}^n$$, \begin{align} \int_{\|\xi\|\leq 1}\frac{1}{\|\xi\|^p}\,d^n\xi&=\int_0^1\frac{1}{r^p}A_{n-1}r^{n-1}\,dr=A_{n-1}\int_0^1\frac{dr}{r^{p+1-n}} \end{align} so this is finite if and only if $$p+1-n<1$$, if and only if $$p (here $$A_{n-1}=\frac{2\pi^{n/2}}{\Gamma(n/2)}$$ is the surface area of the unit sphere $$S^{n-1}\subset\Bbb{R}^n$$).
• hmm, so the 1D case is still unbounded? Or am I miss-understanding the second criteria since I can still directly solve the 1D Poisson's equation for $\rho_c = 1$ for $x \in [-1,1]$? Aug 31 at 12:48
• @helloworld922 in the 1-D case it is not integrable as I mentioned: $\int_0^1\frac{dt}{t}=\infty$. In 3-dimensions there's no issues. I even gave the condition in $n$-dimensions for when such inverse-powers of distance yield finite integrals. (though for $1D$ solving Poisson's equation is trivial, you just integrate $u''(x)=\rho(x)$ twice (I'm ignoring the various constants)) Aug 31 at 12:49
• I suppose a follow-up question is why can I solve Poisson's equation "directly" by applying appropriate BC's at $x=\pm 2$ and find $\phi(0)$? Aug 31 at 12:51
• $\iint \frac{d^2 \phi(x)}{dx^2} dx dx = \phi(x)$, $\iint h(1-x) h(1+x) dx dx = \frac{1}{2} (2 c_0 x + (x - 1)^2 h(x-1) - (x+1)^2 h(x+1))+c_1$, then I just need to apply appropriate boundary conditions to find the constants of integration? Aug 31 at 13:03