Consider a volume charge distribution $\rho({\bf r'})$. The electric field at ${\bf r}$ is
$${\bf E}({\bf r})=\iiint\frac{1}{4\pi\epsilon_0}\frac{\rho({\bf r'})}{R^2}\hat{\bf R}\, \mathrm d^3{\bf r'}$$
where ${\bf R}={\bf r}-{\bf r'}$.
My question is at positions ${\bf r}$ where $\rho \ne 0$, the integrand blows up when ${\bf R}={\bf 0}$. But I know from EM textbooks that ${\bf E}({\bf r})$ at where $\rho({\bf r})\ne 0$ is well-defined.
Can one explain this mathematically?
Edit: Here is own thought, but I am not sure whether it's correct.
Consider this as a purely mathematics question first.
Then I think the question is similar to whether
$$\int_0^1\frac{1}{\sqrt{x}}dx$$
is undefined or equals $2$.
In my opinion, strictly speaking, the above integral is undefined, because the integrand is undefined at $x=0$.
However, since
$$\lim_{\delta\rightarrow0^+}\int_\delta^1\frac{1}{\sqrt{x}}dx=2$$
we usually just say
$$\int_0^1\frac{1}{\sqrt{x}}dx=2$$
So in my opinion, when we write
$${\bf E}({\bf r})=k\iiint\frac{\rho({\bf r'})}{|{\bf r}-{\bf r'}|^2}\hat{\bf R}d^3{\bf r'}$$
we are actually saying
$${\bf E}({\bf r})=\lim_{\delta\rightarrow0^+}k\iiint_{D_{\delta}}\frac{\rho({\bf r'})}{|{\bf r}-{\bf r'}|^2}\hat{\bf R}d^3{\bf r'}$$
where $D_\delta$ is the space minus a ball with radius $\delta$ centered at ${\bf r}$.
It can be shown that if $\rho({\bf r})$ is finite or does not go to infinity $\textit{too fast}$, then the limit exists.
If the above is true, then the physics question one should ask is whether this gives the correct answer in physics.
In classical EM, the $\textit{macroscopic}$ field ${\bf E}$ is in fact the average of the $\textit{microscopic}$ field ${\bf e}$. (Landau and Lifshitz, $\textit{Electrodynamics of Continuous Media}$, p.1.)
So you can consider ${\bf E}({\bf r})$ as the average of ${\bf e}$ over a ball with radius $\delta$ centered at ${\bf r}$.
Now, it can be shown that
(1) the average field over the ball due to charges outside the ball is the same as the total field due to all charges outside the ball at the center of the ball, and
(2) the average field over the ball due to charges inside the ball is
$$-k\frac{{\bf p}}{\delta^3}$$
where ${\bf p}$ is the total dipole moment inside the ball.
(Griffith, $\textit{Introduction to Electrodynamics}$, p. 156-157, problem 3.41).
So when $\delta$ is small macroscopically, ${\bf p}\rightarrow{\bf 0}$, and we have
$${\bf E}({\bf r})=k\iiint_{D_{\delta}}\frac{\rho({\bf r'})}{|{\bf r}-{\bf r'}|^2}\hat{\bf R}d^3{\bf r'}$$
Best Answer
I actually agree with Ruslan's comment. You cannot say that the integral blows up when $\textbf{r} = \textbf{r}'$, with $\textbf{r}'$ spanning the integration domain where $\rho \neq 0$.
The reason is simply that this is a triple integral and that the volume form in the integral may compensate the diverging behaviour of the Green function.
To see this you can make a change of variable $\textbf{r}' \rightarrow \textbf{R}$ and swap to spherical coordinates $(R, \theta, \phi)$ (with $\theta$ and $\phi$ defined with phase shift to account for the minus sign arising with the change of variable), you will see that your formula for the electric field becomes:
\begin{equation} \textbf{E(r)} = \iiint \frac{\rho(\textbf{r}-\textbf{R})}{R^2} \hat{\textbf{R}} \: R^2 \sin \theta \: d\theta \: d\phi \:dR = \iiint \rho(\textbf{r}-\textbf{R})\hat{\textbf{R}} \: \sin \theta \: d\theta \: d\phi \:dR \end{equation}Of course the price to pay to write it like that in general is that the domain of integration becomes very non trivial and all the difficulty is encapsulated in expressing the function $\rho$ as a function of $R, \: \theta$ and $\phi$.