Electric Potential of an off axis charge (Legendre Generating Function) An insulated disk, uniform surface charge density $\sigma$, of radius R is laid on the xy plane. Deduce the electric potential $V(z)$ along the z-axis. Next consider an off axis point $p'$, with distance $\rho$ from the center, Making an angle $\theta$ with the z-axis. Expand the potential at $p'$ in terms of Legendre polynomials $P_l(\cos\theta)$ for $\rho < R$ and $\rho > R$
for the point on the z-axis, this is pretty easy. The differential Voltage from a differential ring of charge with radius $r$ is:
$$dV = \frac{1}{4 \pi \epsilon_o} \frac{dq}{ \mathscr{R}}$$
$$dq = \sigma dA = \sigma 2 \pi r dr$$ 
$$\mathscr{R} = \sqrt{r^2 + z^2}$$
$$ \Delta V(z) = \frac{ \sigma}{2 \epsilon_o}\int_0^R \frac{ r dr}{\sqrt{r^2 + z^2}} = \frac{ \sigma}{2 \epsilon_o} \left( \sqrt{R^2 + z^2} - |z| \right)$$
Which is obtained by using a U substitution.
As for the second part, The only thing that changes is the distance from the differential of charge and the point of interest so I have:
$$dV = \frac{ \sigma}{2 \epsilon_o} \frac{r dr}{ \mathscr{R}}$$
But now using the law of cosines, I use the angle between r and $\mathscr{R}$, Note: this is not the angle recommended in the problem.
$\mathscr{R} = (r^2 + p^2 - 2rp\cos \phi)^{1/2} = r(1 - 2 \frac{p}{r}cos \phi + \frac{p^2}{r^2})^{1/2}$
Using Spherical Polar coordinates,
where $p =$ distance from origin to point of interrest p'
This is the Generating function of the Legendre polynomials
$$\therefore \frac{1}{\mathscr{R}} = \frac1r G( \frac{p}{r}, \cos \phi)$$
$$dV = \frac{ \sigma}{2 \epsilon_o} G( \frac{p}{r}, \cos \phi) dr =  \frac{ \sigma}{2 \epsilon_o} \sum_{l = 0} ^{\infty} p_l(\cos \phi) \left( \frac{p}{r} \right)^l dr$$
Okay, so my question is this, assuming all of this is correct (which I believe is not) How would possibly integrate this? Is it as simple as 
$\int_0^R \left( \frac{p}{r} \right)^l dr$? This creates an infinity. Any help would save me so very much.
 A: There are three variables involved, and it's important not to mix them up, or you'll go astray.  There's the distance from the origin to the field point, call this $r$.  There's the distance from the point on the surface of the disc being integrated to the field point, call this $\mathscr{R}$.  And there's the distance from the origin on the disc to the point being integrated, call this $r'$.  Assuming $\sigma$ is not a function of $r'$ the last equation will then look like:
$\frac{ \sigma}{4\pi \epsilon_o}\frac{1}{r} \sum_{l = 0} ^{\infty} p_l(\cos \phi) \left( \frac{r'}{r} \right)^l dt$, 
for a general surface or volume element $dt$. You are integrating with respect to $r'$, so the $r$ comes outside the integral and you get (in polar coordinates):
$\frac{ \sigma}{4\pi \epsilon_o} \sum_{l = 0} ^{\infty} \frac{1}{r^{1+l}}\int p_l(\cos \phi) \left( r' \right)^l r'dr'd\phi.$ 
It's then just a matter of "pulling out" as many terms as you like, like:
$\frac{ \sigma}{4\pi \epsilon_or}\int r'dr' + \frac{ \sigma}{4\pi \epsilon_o r^2}\int r'^2\cos(\phi)dr'd\phi...$
to get an approximation for the potential to any accuracy you desire.
