Is my interpretation of this three dimensional improper integral correct?

[Beelzedad]

TL;DR Summary

Please help me understand three dimensional improper integral in electrostatics in the correct way.

In Physics/Electrostatics textbook, I am in a situation where we have to find the electric field at a point inside the volume charge distribution. In Cartesian coordinates, we can't do it the usual way because of the integrand singularity. So we use the three dimensional improper integral.

$\displaystyle \mathbf{E}=\lim\limits_{\epsilon\to 0} \int_{V'-\delta_{\epsilon}} \rho'\ \dfrac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} dV' \tag1$

where:

$\mathbf{r'}=(x',y',z')$ is coordinates of source points
$\mathbf{r}=(x,y,z)$ is coordinates of field points
$V'$ is the volume occupied by the charge
$\delta_{\epsilon}$ is an arbitrary volume contained in $V'$ around the singular point $\mathbf{r}=\mathbf{r'}$ with $\epsilon$ being its greatest chord.
$\rho'$ is the charge density and is continuous throughout the volume $V'-\delta$

While taking the limit the shape of $\delta_{\epsilon}$ is kept unaltered

From equation $(1)$, we can get the $x$-component of $\mathbf{E}$:

$\displaystyle E_x=\lim\limits_{\epsilon\to 0} \int_{V'-\delta_{\epsilon}} \rho'\ \dfrac{x-x'}{|\mathbf{r}-\mathbf{r'}|^3} dV' \tag2$

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I view the steps of solving $E_x$ as follows:

1. Make a $\delta_{\epsilon}$ cavity (with $\epsilon=a$) contained in $V'$ around the singular point $\mathbf{r}=\mathbf{r'}$. Then take the Riemann integral over $V'-\delta_{\epsilon}$

2. Repeat above procedure for all $\epsilon$ in the interval $(0,a]$ keeping the shape of $\delta_{\epsilon}$ unaltered.

3. Find the function which relates "Riemann integral over $V'-\delta_{\epsilon}$" and "$\epsilon$". Then make a graph of "$\epsilon$" ($x$-axis) and "Riemann integral over $V'-\delta_{\epsilon}$" ($y$-axis) over the interval $(0,a]$

4. Find $l$ $\ni$

$\forall \varepsilon > 0, \exists \delta \ni \text{when} |x-0|<\delta, |y-l|< \varepsilon$

Thus $l$ is the solution for $E_x$ in equation $(2)$

Am I correct? Is this the correct interpretation of equation $(2)$?

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If yes:

I know steps $(1)$ and $(4)$ can be done. But is there a way to execute step $(2)$ and $(3)$? Please explain.

 

[mfb]

In principle: Yes. In practice no one will do that. As long as the charge density is continuous there is no problem at r'=r and you can just integrate over all space, ignoring what happens at this special point.

 

[Beelzedad]

In principle: Yes. In practice no one will do that. As long as the charge density is continuous there is no problem at r'=r and you can just integrate over all space, ignoring what happens at this special point.

I guess you are saying we have to use spherical coordinates with origin at $\mathbf{r'}=\mathbf{r}$. But in some situations, (like in my situation while considering derivative of $E_x$ w.r.t. $x$) you need to take two field points, both inside the charge distribution. But in a coordinate system you cannot have two origins. So your method of using spherical coordinates with origin at $\mathbf{r'}=\mathbf{r}$ will not work. Wouldn't then we have to go for the method in my original post?
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If yes:

How shall we execute steps $(2)$ and $(3)$ in my original post?

If no:

In Cartesian coordinate, how shall we actually interpret equation $(2)$ in my original post?

 

[mfb]

Spherical coordinates are a typical choice but not the only option.
You can choose two different coordinate systems for the two calculations.

In Cartesian coordinate, how shall we actually interpret equation $(2)$ in my original post?

You never do step 2. You make some suitable sum to avoid infinities.

 

[Delta2]

I don't think you should be puzzled by the singularity at $r=r'$. This singularity is introduced when we find green's function for the Poisson equation and is needed in order for the green's function to satisfy the equation $\nabla^2G(r,r')=\delta(r-r')$ so its a singularity coming from the dirac delta function in a way. It is not a singularity that will make the integral to diverge, it is a singularity that does the trick for the green's function and to solve Poisson equation.