How to calculate divergence of some special fields

[netheril96]

$$\[<br /> \nabla \cdot \frac{{\vec e_r }}{{r^2 }} = 4\pi \delta (\vec r)<br /> \]$$
This can be seen from$$\[<br /> \nabla \cdot \frac{{\vec e_r }}{{r^2 }} = \frac{1}{{r^2 }}\frac{\partial }{{\partial r}}(r^2 \cdot \frac{1}{{r^2 }}) = \frac{1}{{r^2 }}\frac{\partial }{{\partial r}}(1) = 0(r \ne 0)<br /> \]$$
And from Gauss' Theorem$$\[<br /> \int_V {(\nabla \cdot \frac{{\vec e_r }}{{r^2 }})dV = \oint_S {\frac{{\vec e_r }}{{r^2 }} \cdot d\vec S} } = 4\pi <br /> \]$$
But if I want to directly using the formula of divergence in spherical coordinates,I can only get$$\[<br /> \nabla \cdot \frac{{\vec e_r }}{{r^2 }} = \frac{1}{{r^2 }}\frac{\partial }{{\partial r}}(\frac{{r^2 }}{{r^2 }})<br /> \]$$
And integrating this over a volume cannot give me the result of 4π$$\[<br /> \int_V {(\nabla \cdot \frac{{\vec e_r }}{{r^2 }})dV = } \int_0^\pi {\sin \theta d\theta \int_0^{2\pi } {d\phi \int_0^R {\frac{\partial }{{\partial r}}(\frac{{r^2 }}{{r^2 }})} } } dr = 4\pi \int_0^R {\frac{\partial }{{\partial r}}(\frac{{r^2 }}{{r^2 }})} dr<br /> \]$$
(Here V is a sphere with radius of R)
So how can I connect it with Dirac Delta?
By the way,I post this here because this problem arises in the electrostatic field of a point charge and I found nothing about such thing in any book concerning δ(x).

 

[gabbagabbahey]

The problem is that $\frac{r^2}{r^2}=\infty$ at $r=0$.

 

[netheril96]

The problem is that $\frac{r^2}{r^2}=\infty$ at $r=0$.

So how can I get$$\[<br /> \int_0^R {\frac{\partial }{{\partial r}}(\frac{{r^2 }}{{r^2 }})} dr = 1<br /> \]$$
Without integration,you cannot conclude some function with a singularity is δ(x)

 

[gabbagabbahey]

Other than just using Gauss' Law, I suppose an appropriate limiting procedure can be used. I'd start with your expression for $\mathbf{\nabla}\cdot\left(\frac{\textbf{e}_r}{r^2}\right)$ and calculate the limit of it as $r\to 0$

 

[Meir Achuz]

As you have seen $$\delta({\vec r})$$ is not easily treated in spherical coordinates.
What is wrong with your first two lines? They constitute one of the definitions of the delta function, which is as 'direct' as you can get.