why does $n$ have to be an integer in order for the solutions to be bounded? $$(1-x^2)\frac{d^2}{dx^2}f(x)-2x\frac{d}{dx}f(x)+n(n+1)f(x)=0$$
this is Legendre's equation. I've stumbled into it when studying Laplace's equation in spherical coordinates.
in documents and books that I've seen they all claim that the only $n$'s for which the solutions are bounded on $[-1, 1]$ are the integers.
any references or explanations will be highly appreciated. 
 A: Applying the Method of Frobenius at $x=0$, substituting in $f(x) = \sum_{m=0}^{\infty} a_m x^{m+\sigma}$, we obtain the recurrence relation
$$ (m+\sigma+2)(m+\sigma+1)a_{m+2} = (m+\sigma-n)(m+\sigma+n+1)a_m, $$
and putting $n=-2$ gives the indicial equation $\sigma(\sigma-1)=0$. So there is always an even solution $\sum_{m=0}^{\infty} a_{2m} x^{2m}$ with coefficients satisfying
$$ a_{m+2} = \frac{(m-n)(m+n+1)}{(m+2)(m+1)} a_m $$
and an odd solution $\sum_{m=0}^{\infty} a_{2m+1} x^{2m+1}$ with coefficients satisfying
$$ a_{m+2} = \frac{(m+1-n)(m+n+2)}{(m+3)(m+2)} a_m. $$
If $n$ is an integer, one of $m-n$ or $m+1-n$ eventually vanishes, so any coefficients beyond $a_n$ or $a_{n+1}$ vanish, and there is a polynomial solution (whose degree is equal to $n$, and whose parity is the same as that of $n$).
Otherwise, the coefficients all eventually have the same sign, since every factor in the fraction is positive for $m>n$. Also, the ratio of the coefficients converges to $1$, since
$$ \frac{a_{m+2}}{a_m} = \frac{(m+\sigma-n)(m+\sigma+n+1)}{(m+\sigma+2)(m+\sigma+1)} \to 1. $$
Therefore by the ratio test, the radius of convergence of the power series is $1$. But the only finite singular points of the differential equation are $\pm 1$, so there must be a singularity at one (or both) of these points. Since the solutions we have derived have definite parity, it suffices to examine $x=1$. 
Because one solution is even and the other odd, any linear combination of them certainly has a solution at one of $x=\pm 1$, since cancelling one singularity will still keep the other one around. Hence it suffices to exhibit the singular behaviour of any solution in the neighbourhood of $x=1$.
Applying the first part of the Method of Frobenius again at the regular singular point $x=1$, we find the indicial equation is $\sigma^2=0$. Working through the Method of Frobenius then implies that there is an analytic solution $b_0+b_1 (x-1)+\dotsb$, but there must also be a logarithmic solution $C(b_0+b_1(x-1)+\dotsb)\log{(1-x)}+(c_0+c_1(1-x)+\dotsb)$. Because we know the solutions we derived above have a singularity, they must include this logarithmic term.
So any nonpolynomial solution of Legendre's equation with definite parity has a logarithmic singularity at $x=1$, and any nonpolynomial solution without definite parity has a logarithmic singularity at at least one of $x=\pm 1$.
