# Is my interpretation of this three dimensional improper integral correct?

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TL;DR Summary
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.

Delta2

Mentor
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.

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?

Mentor
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.

Homework Helper
Gold Member
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.