I was following Yakov Shnir his book about magnetic monopoles, there they derive the monopole spherical harmonics. I will sketch the derivation briefly,
The starting equation is the eigenvalue problem of the total angular momentum,
\begin{align} \hat{\textbf{J}}^2 Y_{\mu lm}(\theta, \phi) &= \hbar^2 \Big\{\frac{-1}{\sin^2 \theta} \Big[ \sin \theta \frac{\partial}{\partial \theta} \big( \sin \theta \frac{\partial}{\partial \theta} \big) + \big( \frac{\partial}{\partial \varphi} - i\mu (1-\cos \theta )\big)^2 \Big]+ \mu^2\Big\} Y_{\mu lm}(\theta, \phi)\\&= \hbar^2\lambda Y_{\mu lm}(\theta, \phi). \end{align}
The $\theta$-parameter can be replaced by the relation $x=\cos\theta$ and the $\phi$ dependence can be eliminated by $Y_{\mu lm}(\theta, \phi)= \exp(i(\mu+m)\phi) P(\theta)$,
\begin{equation} \label{eq: diff eq angular part1} \Big\{ -(1-x^2)\frac{d}{dx^2} + 2x \frac{d}{dx} + \frac{(m+\mu x)^2}{1-x^2} +\mu^2 \Big\} P(x) = \lambda P(x). \end{equation}
To solve the problem we want to rewrite it as the Euler's hypergeometric function. To obtain that form $P(x)$ has to be rewritten as,
\begin{equation} P(x) = (1-x)^{-\frac{\mu +m}{2}} (1+x)^{-\frac{\mu -m}{2}} F(x), \end{equation}
equation (\ref{eq: diff eq angular part1}) will then become,
\begin{equation} (1-x^2) \frac{d^2F}{dx^2} + 2 ( m+(\mu-1)x) \frac{dF}{dx} + (\mu - \mu^2 + \lambda)F = 0. \end{equation} If now the transformation $z= (1+x)/2$ is applied, then we have rewritten the differential equation as the Euler's hypergeometric differential equation,
\begin{align} z(1-z) F'' &+ \big\{ m-\mu + 1 + 2z(\mu-1) \big\} F' + (\mu -\mu^2 + \lambda) F \\ & = z(1-z) F'' + \big\{ c-(a+b+1)z \big\}F' -abF = 0, \end{align}
where $c=m-\mu+1$,$~ab = \mu^2 -\mu -\lambda$ and $a+b+1 = 2(1- \mu)$. The solution of this differential equations is the hypergeometric function $_2F_1(a;b;c;z) $ which is defined as, \begin{equation} _2F_1(a;b;c;z) = \sum_{n=0}^{\infty} \frac{(a)_n(b)_n}{(c)_n} \frac{z^n}{n!} \equiv \sum_{n=0}^{\infty} \alpha_n z^n. \end{equation}
Now my question is the following, they consider that $a$ is negative integer such that the sum terminates, my question now is why? Because if the sum not terminates then will the convergence radius be 1, but since $\theta$ is between $0$ and $\pi$ and by definiton of $z$ you still have a bijection from $z$ to the range of $\theta$.
This indeed doesn't say anything about being square integrable but how could one see this from the hypergeometric function/series? In fact I am searching for an analytical argument why the quantum numbers only differ in integers?
Thanks in advance.