Evaluating $\lim_{\epsilon\to 0^{+}}\ \frac{ _2F_1\left( \tfrac{1}{2} - \nu, \tfrac{1}{2} + \nu; \epsilon; y \right) }{\Gamma(\epsilon)}$

Question

For $\nu \in \mathbb{C}$ and negative $y<0$ is there a way to compute the limit $$ f(\nu,y) \equiv \lim_{\epsilon \to 0^{+}} \ \frac{ _2F_1\left( \tfrac{1}{2} - \nu, \tfrac{1}{2} + \nu; \epsilon; y \right) }{\Gamma(\epsilon)} $$ in terms of simpler special functions (ideally not hypergeometrics...)? So far, I thought about writing this in terms of the series $$ f(\nu,y) \ = \ \sum_{n=0}^\infty \frac{\Gamma(\frac{1}{2} - \nu + n )\Gamma(\frac{1}{2} + \nu + n )}{(n-1)! n!} \ y^n $$ I think that this series converges at least for $-1 < y < 0$, but I am also interested in values of $y \leq -1$.

(EDIT: I know that $_2F_1(a,b;c;z)$ has poles at $c=0,-1,-2,\ldots$, this makes it hard to learn anything about this limit using Mathematica)

Answer 1

We may express \begin{equation} _2F_1\left( \tfrac{1}{2} - \nu, \tfrac{1}{2} + \nu; \epsilon; y \right) =\frac{\Gamma\left(\epsilon\right)}{\Gamma\left(\tfrac{1}{2} - \nu\right)\Gamma% \left(\tfrac{1}{2} + \nu\right)}\sum_{s=0}^{\infty}\frac{\Gamma\left(\tfrac{1}{2} - \nu+s\right)\Gamma\left(\tfrac{1}{2} + \nu+s% \right)}{\Gamma\left(\epsilon+s\right)s!}y^{s} \end{equation} and thus, \begin{align} \lim_{\epsilon \to 0^{+}} \ \frac{ _2F_1\left( \tfrac{1}{2} - \nu, \tfrac{1}{2} + \nu; \epsilon; y \right) }{\Gamma(\epsilon)}&=\frac{1}{\Gamma\left(\tfrac{1}{2} - \nu\right)\Gamma% \left(\tfrac{1}{2} + \nu\right)}\sum_{s=1}^{\infty}\frac{\Gamma\left(\tfrac{1}{2} - \nu+s\right)\Gamma\left(\tfrac{1}{2} + \nu+s% \right)}{\Gamma\left(s\right)s!}y^{s}\\ &=\frac{y}{\Gamma\left(\tfrac{1}{2} - \nu\right)\Gamma% \left(\tfrac{1}{2} + \nu\right)}\sum_{t=0}^{\infty}\frac{\Gamma\left(\tfrac{3}{2} - \nu+t\right)\Gamma\left(\tfrac{3}{2} + \nu+s% \right)}{\Gamma\left(t+2\right)t!}y^{t}\\ &=\frac{y\Gamma\left(\tfrac{3}{2} - \nu\right)\Gamma% \left(\tfrac{3}{2} + \nu\right)}{\Gamma(2)\Gamma\left(\tfrac{1}{2} - \nu\right)\Gamma% \left(\tfrac{1}{2} + \nu\right)} \,_2F_1\left(\tfrac{3}{2} - \nu,\tfrac{3}{2} +\nu;2;y \right)\\ &=y\left( \frac{1}{4}-\nu^2 \right)\,_2F_1\left(\tfrac{3}{2} - \nu,\tfrac{3}{2} +\nu;2;y \right) \end{align} From this representation of the associated Legendre function \begin{equation} _2F_1\left(a,b;\tfrac{1}{2}(a+b+1);z\right)=\left(-z(1-z)\right)^{% \frac{(1-a-b)}{4}}\,P^{(1-a-b)/2}_{(a-b-1)/2}\left(1-2z\right) \end{equation} with $a=3/2+\nu,b=3/2-\nu$ we can express \begin{equation} f(\nu,y)=\left( \frac{1}{4}-\nu^2 \right)\sqrt{\frac{-y}{1-y}}P^{-1}_{\nu-1/2}\left( 1-2y \right) \end{equation} where $1-2y>0$.