In this paper the authors discuss the following function defined on the interval $\left(0,\dfrac{\pi}{2}\right)$
\begin{align}\Phi(x)&=B_1(\sin x)^{-\nu+1/2}(\cos x)^{\sigma+1/2}\left\{\dfrac{\Gamma(1+\sigma)\Gamma(\nu)}{\Gamma(\zeta^\omega_{\nu,\sigma})\Gamma(\zeta^{-\omega}_{\nu,\sigma})}{_2F_1}(\zeta^{\omega}_{-\nu,\sigma},\zeta^{-\omega}_{-\nu,\sigma},1-\nu;\sin^2x) \right. \\ &+ \left. \dfrac{\Gamma(1+\sigma)\Gamma(-\nu)}{\Gamma(\zeta^{\omega}_{-\nu,\sigma})\Gamma(\zeta^{-\omega}_{-\nu,\sigma})}(\sin x)^{2\nu} {_2 F_1}(\zeta^{\omega}_{\nu,\sigma},\zeta^{-\omega}_{\nu,\sigma},1+\nu,\sin^2x)\right\}\tag{154}\end{align}
Here $\sigma$ and $\nu$ are parameters with $\nu\neq 0$ and $\nu\notin \mathbb{N}$. Moreover the quantity $\zeta^{\omega}_{\nu,\sigma}$ is defined by the equation $$\zeta^{\omega}_{\nu,\sigma}=\dfrac{\nu+\sigma+1+\omega}{2}.$$
Now the authors claim that if $\nu \geq 1$ then $\Phi(x)$ is not square integrable unless $\omega$ is a real number so that $\Gamma(\zeta^\omega_{\nu,\sigma})$ or $\Gamma(\zeta^{-\omega}_{\nu,\sigma})$ diverges and the first term in (154) is set to zero.
On the other hand they claim that if $0< \nu < 1$ then $\Phi(x)$ is square integrable for all $\omega\in \mathbb{C}$.
Where these results come from? They mention this in the end of page (31) in a way that it seems these are easy things to see from inspection, but I confess I'm missing it. How to understand when this $\Phi(x)$ is square integrable and get the conclusions the authors found?