Proving Gauss's Law with Coulomb's Law: Challenges and Solutions

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sadraj
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Hi guys. My question is related to proving Gauss's law by using Columb's law. Let start with a charge density [tex]\rho(\vec{r})[/tex] in [tex]R^3[/tex] . by Columb's law we have :
[tex]E(\vec{r})=\int{\frac{\rho(\vec{r'})d^3r'(\vec{r'}-\vec{r})}<br /> {|\vec{r'}-\vec{r}|^3}}[/tex]

suppose that domain of function [tex]\rho(\vec{r})[/tex] is finite in [tex]R^3[/tex]. Clearly it can be showed that above integral exists for any [tex]\vec{r}[/tex] out of the domain. & it's divergence is zero out of the domain. So for any Gauss's surface out of domain we can use Divergence theoreom to prove that
[tex]\int{E(\vec{r}).\vec{dS}}=\int{\rho(\vec{r})d^3\vec{r}}=\frac{Q}{\epsilon 0}[/tex]

But for a Gauss's surface that goes through domain of [tex]\rho(\vec{r})[/tex] we have two problems :
1. it is not clear that for all [tex]\vec{r}[/tex] the integral of electric field exist and converges.
2. Suppose that E converges for any [tex]\vec{r}[/tex]. But it is not easy to bring devergence operator in the integral. Divergence theoreom doesn't work here.

What is your idea? Please don't use delta dirac function. I don't understand it. Because I haven't studied distributions in mathematics.
Similar question can be asked when we have surface charge density on the Gauss's surface.
Sorry for bad english
 
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I think its more important that you understand the dirac delta function right now than prove gauss's law. This is a fundamental idea which can greatly simplify many of your future calculations.
 
I believe most of physicists or at least undergraduate people in physics don't understand delta dirac function (Like many things in mathematics) , They just use it. Like me !

I think similar to proving Gauss's law , clearly with divergence theorem out of charge's distribution , there is a way to show Gauss's law is true in other conditions that I interpreted.