Question 3.6: The potential V0(θ) is specified on the surface of a hollow ...

In this case, B_{l} = 0 for all l—otherwise the potential would blow up at the origin.
Thus,

V(r, θ) =\sum\limits_{l=0}^{\infty }{A_{l}r^{l}P_{l}(\cos{\theta})}      (3.66)

At r = R this must match the specified function V_{0}(\theta):

V(R, θ) =\sum\limits_{l=0}^{\infty }{A_{l}Rr^{l}P_{l}(\cos{\theta})}=V_{0}(\theta)         (3.67)

Can this equation be satisfied, for an appropriate choice of coefficients Al? Yes: The Legendre polynomials (like the sines) constitute a complete set of functions, on the interval −1 ≤ x ≤ 1 (0 ≤ θ ≤ π). How do we determine the constants? Again, by Fourier’s trick, for the Legendre polynomials (like the sines) are orthogonal functions:^{14}

\int_{-1}^{1}{P_{l} (x)P_{\acute{l}} (x) dx}=\int_{0}^{\pi}{P_{l} (\cos{\theta})P_{\acute{l}} (\cos{\theta}) \sin{\theta} d{\theta}}  \\                                                           =\begin{cases} 0,      if         \acute{l}   \neq     l, \\ \frac{2}{2l+1} ,           if     \acute{l}  =    l\end{cases}                   (3.68)

Thus, multiplying Eq. 3.67 by P_{\acute{l}} (\cos \theta) \sin \theta and integrating, we have

or
A_{l}=\frac{2l+1}{2R^{l}} \int_{0}^{\pi}{V_{0}(\theta)P_{l} (\cos{\theta}) \sin{\theta} d{\theta}}.            (3.69)

Equation 3.66 is the solution to our problem, with the coefficients given by Eq. 3.69.
It can be difficult to evaluate integrals of the form 3.69 analytically, and in practice it is often easier to solve Eq. 3.67 “by eyeball.”^{15} For instance, suppose we are told that the potential on the sphere is

V_{0}(\theta) = k \sin^{2}(\theta/2),                 (3.70)

where k is a constant. Using the half-angle formula, we rewrite this as

Putting this into Eq. 3.67, we read off immediately that A_{0} = k/2, A_{1} = k/(2R),and all other A_{l} ’s vanish. Therefore,

V(r,0) =\frac{ k}{2}\left[r^{0}P_{0}(\cos\theta)-\frac{r^{1}}{R}P_{1}(\cos{\theta}) \right]=\frac{k}{2}(1-\frac{r}{R}\cos\theta ) .     (3.71)