I'm covering the Hydrogen atom wavefunctions $\psi(r,\theta,\phi)$ which are separated into three functions $R(r), \Theta(\theta), \Phi(\phi)$. ((The function $Y(\theta,\phi)$ is what we call a spherical harmonic)). $\theta$ runs from $0$ to $\pi$ and $\phi$ runs from $0$ to $2\pi.$
We require $\psi$ to be normalizable, meaning
$$ \int |\psi(r,\theta,\phi)|^2 \mathrm{d}^3\mathbf{r} = \int_0^{2\pi}\int_0^{\pi}\int_0^{\infty}|\psi|^2r^2\sin\theta dr d\phi d\theta =1. $$
If the quantum numbers $l,m$ in the differential equation for $\Theta(\theta)$ are both zero, $\Theta(\theta)$ obeys:
$$ \frac{1}{\Theta}\left[\sin\theta \frac{d}{d\theta} \left(\sin\theta \frac{d\Theta}{d\theta}\right)\right]=0 $$ This is a 2nd order DE, so there's two solutions: $\Theta=1$ (the physical one) and $\Theta = A \ln(\tan(\theta/2))$ - how do we prove that the latter is unphysical?
I don't think that we can prove that on the grounds that it blows up at $\theta=0$ and $\theta=\pi$, since the integral $\int_0^\pi \ln^2(\tan(\theta/2))\sin\theta d\theta$ is bounded and guarantees $\psi$ is normalized. ((Graph of integrand: https://www.desmos.com/calculator/mgsbwnnqqd))
There must be some other condition on $\Theta(\theta)$ that lets us discard this solution. What am I missing?
Could it be something similar to how we require $\Phi(\phi)=\Phi(\phi+2\pi)$?
