This question involves maths. I do not think it is inappropriate here.

I am reading Purcell and Morin's Electricity and Magnetism 3rd Edition.

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

On page 22, it says:

"A continuous charge distribution $ρ\ (x^\prime, y^\prime, z^\prime)$ that is nowhere infinite gives no trouble at all. Equation $(1.22)$ can be used to find the field at any point within the distribution. The integrand doesn’t blow up at $r = 0$ because the volume element in the numerator equals $r^2 \sin \phi\ d \phi\ d \theta\ dr$ in spherical coordinates, and the $r^2$ here cancels the $r^2$ in the denominator in Eq. $(1.22)$. That is to say, so long as $ρ$ remains finite, the field will remain finite everywhere, even in the interior or on the boundary of a charge distribution."

According to the above quoted paragraph, equation $(1.22)$ becomes:

$$\vec{E}(x,y,z)=\dfrac{1}{4 \pi \epsilon_0} \int ρ\ (x^\prime, y^\prime, z^\prime)\ \hat{r}\ \sin \phi\ d \phi\ d \theta\ dr$$

Here there is no particular direction for $\hat{r}$ at $r=0$. Then how can we say that in spherical coordinates the integral doesn't blow up at $r=0$.

I have more questions on this:

(2) How can we be sure that the integral doesn't blow up at $r=0$ in other coordinate systems?

(3) Are there any analogous expressions for electric field (independent of $r$) due to surface charge density and line charge density?

  1. The question wether the integral blows up or not is not a "pointwise" question, since the singleton $\{r=0\}$ has $0$ volume, we have $$ \int_{\overrightarrow r\in \mathbb R^3} f(r) \mathrm d \overrightarrow r = \int_{\overrightarrow r\in \mathbb R^3 \setminus\{0\}} f(r) \mathrm d \overrightarrow r$$ This remark is valid in any coordinate system. In you case, you have a gentle function writtent in spherical coordinates which is well defined everywhere but at $r=0$. No problem : you can choose the value of the function to be whatever you want at $\{r=0\}$ even infinite, the integral will still be well defined and has the same value.
  2. Here his a theorem for yout peace of mind : let $(X,\mu)$ be a measures set, let $f:X\rightarrow \mathbb R$ be a measurable function and let $g:Y\rightarrow X$ be change of variables, then $$\int_{X} |f\circ g|\, \mathrm d \mu< +\infty \quad\Leftrightarrow\quad \int_{Y} |f|\,g\#\mathrm d \mu < +\infty$$ This theorem is valid in a very wide generality : the change of variable $g$ do not need to be continuous or bijective and it includes all change of variable you might encounter. I wrote $\phi \# \mathrm d \mu$ for the image measure. For instance, let $X=\mathbb R^3$ and $Y= [0,2\pi]\times [0,\pi]\times \mathbb R_ +^*$, let $$ g : \begin{matrix}Y&\rightarrow &X \\ (\theta,\phi,r)&\mapsto& (r\cos(\theta)\cos(\phi),r\sin(\theta)\cos(\phi),r\sin(\phi))\end{matrix}$$ You recognize your spherical change of variable and $$\mathrm d\mu = \mathrm d x\mathrm dy \mathrm d z \quad\quad \quad g\#\mathrm d \mu = r^2\sin(\phi)\mathrm d \theta \mathrm d \phi \mathrm d r$$ here $f=\rho \hat r$.
  3. If you're interested for more precise mathematical treatement of non-smooth electrival charge distribution, you could have a look at distribution theory. It allows you to write such integrals even for "line-wise" or "surface-wise" or "point-wise" charges. In this case, you only need to understand the most basic examples of distributions which are actually measures as in Lebesgue measure theory. If you go in that direction, you rewrite $$E(\rho)= \int_{\mathbb R^3} \frac{\rho\left(\hat r\right) \hat r}{r^2} \mathrm d \hat r = \int_{\mathbb R^3} \frac{ \hat r}{r^2} \left(\rho(\hat r)\mathrm d \hat r\right) = \int_{\mathbb R^3} \frac{ \hat r}{r^2} \mathrm d \mu = E(\mu)$$ Your $\mathrm d \mu$ can now be any charge distribution you want as long as the behaviour around the origin is not too pathological.

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