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

• I
In summary, the conversation discusses the use of an improper integral in finding the electric field at a point inside a volume charge distribution. The integral is in Cartesian coordinates and is necessary because of the integrand singularity. The steps to solve for the x-component of the electric field involve creating a cavity and taking the Riemann integral over it, repeating this for different values of epsilon, and finding the function that relates the integral and epsilon. However, in practice, this method is not used and instead, a suitable sum is made to avoid infinities. The singularity at r=r' is introduced in finding the green's function for the Poisson equation and is not a singularity that will cause the integral to diverge.

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.

Delta2
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.

mfb said:
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?

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.

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.

## 1. What is a three dimensional improper integral?

A three dimensional improper integral is a mathematical concept used in multivariable calculus to calculate the volume under a surface in three-dimensional space. It is similar to a regular integral, but instead of integrating over a one-dimensional curve, it integrates over a three-dimensional region.

## 2. How do you interpret a three dimensional improper integral?

The interpretation of a three dimensional improper integral is the volume under a surface in three-dimensional space. This means that the result of the integral represents the amount of space that is enclosed by the surface.

## 3. What makes an integral improper?

An integral is considered improper if one or more of the following conditions are met: the limits of integration are infinite, the integrand is undefined at a point within the limits of integration, or the integrand has an infinite discontinuity within the limits of integration.

## 4. How do you know if your interpretation of a three dimensional improper integral is correct?

To ensure the accuracy of your interpretation of a three dimensional improper integral, you should first check that the integral is set up correctly and the limits of integration are appropriate for the given problem. Then, you can use mathematical techniques to evaluate the integral and compare your result to known solutions or use visualization methods to verify the calculated volume.

## 5. What are some real-world applications of three dimensional improper integrals?

Three dimensional improper integrals have numerous applications in physics, engineering, and other fields. They can be used to calculate the volume of irregularly shaped objects, the center of mass of a three-dimensional object, and the probability of events occurring in three-dimensional space. Some examples of real-world applications include calculating the volume of a water tank, finding the center of mass of a car, and determining the probability of a particle's position in quantum mechanics.