Potential of spherical charge distribution

[Observer Two]

I want to derive this equation:

$V(r) = \frac{1}{\epsilon_0} [\frac{1}{r} \int_0^r \! r'^2 \rho(r') \, d r' + \int_r^{\infty} \! r' \rho(r') \, d r' ]$

of a spherical charge distribution.

I can do it with the general integral definition of the electrostatic potential (which is basically the utilization of the Greens function). But isn't it also possible to derive it directly from the Poisson equation:

$\nabla^2 V(r) = -\frac{\rho(r)}{\epsilon_0}$

Using spherical coordinates:

$\frac{1}{r^2} \frac{\partial}{\partial r}(r^2 \frac{\partial V(r)}{\partial r}) = -\frac{\rho(r)}{\epsilon_0}$

$(r^2 \frac{\partial V(r)}{\partial r}) = -\int \! \frac{\rho(r)}{\epsilon_0} r^2 \, d r$

Here is the part where I'm stuck. I could maybe "split" the right hand side in the following way:

$(r^2 \frac{\partial V(r)}{\partial r}) = \frac{1}{\epsilon_0} [\int_r^{\infty} \! \rho(r') r'^2 \, d r' + \int_0^r \! \rho(r') r'^2 \, d r']$

But then I'm lost how to continue since there is another derivative on the left hand side.

 

[Meir Achuz]

Use Gauss's law to get E(r). Then integrate $\phi(r)=\int_r^\infty E(r')dr'$
by parts.

 

[Observer Two]

I'm missing something here it seems. $E(r) = \frac{Q}{4 \pi \epsilon_0 r^2}$ obviously. Integrating that doesn't give me the general expression that I want though.

Edit: Nevermind, got it! $Q = Q(r) = \int_V \! \rho(r) \, d V$ Then it works. Thanks!