# Asymptotic form of the Gauß Hypergeometric function 2F1 for three parameters approaching infinity

I am trying to find the leading order expression in an expansion for large $\Delta$ of ${}_2F_1\left(\frac{\Delta}{2},\frac{\Delta+1}{2},\Delta,z^{-2}\right)$, where $z\in\mathbb{C}$.

The only helpful relation I could find in the literature Higher transcendental functions by H. Bateman and A. Erdelyi, p.77. There, I found:

If we define $\xi$ by $y\pm\sqrt{y^2-1}=e^{\pm\xi}$, where the upper sign is for $\text{Im}\,{y}>0$ and the lower sign for $\text{Im}\,y<0$, then for large $\lambda$ we have
\begin{align*} &\left(\frac{y}{2} - \frac{1}{2}\right)^{-a-\lambda}\cdot{}_2 F_1(a+\lambda,a-c+1+\lambda,a-b+1+2\lambda,2(1-y)^{-1})\\ &\qquad = \frac{2^{a+b}\Gamma(a-b+1+2\lambda)\Gamma\left(\frac{1}{2}\right)\lambda^{-\frac{1}{2}}}{\Gamma(a-c+1+\lambda)\Gamma(c-b+\lambda)}\cdot e^{-(a+\lambda)\xi}(1-e^{-\xi})^{-c+\frac{1}{2}}(1+e^{-\xi})^{c-a-b-\frac{1}{2}}[1+\mathcal{O}(\lambda^{-1})]. \end{align*}

For $\lambda=\frac{\Delta}{2}$, $a=0$, $b=1$, $c=\frac{1}{2}$ and $y=1-2z^2$ we get \begin{equation*} \left(-z^2\right)^{-\frac{\Delta}{2}}{}_2F_1\left(\frac{\Delta}{2},\frac{\Delta+1}{2},\Delta,z^{-2}\right) = \frac{2\Gamma(\Delta)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{\Delta+1}{2}\right)\Gamma\left(\frac{\Delta-1}{2}\right)\sqrt{\frac{\Delta}{2}}}\cdot e^{-\frac{\Delta}{2}\xi}\left(1+e^{-\xi}\right)^{-1}[1+\mathcal{O}(\Delta^{-1})]. \end{equation*}

This can be further simplified by using relations for the Gamma function: \begin{equation*} \frac{\Gamma(\Delta)}{\Gamma\left(\frac{\Delta+1}{2}\right)}=\frac{2^{\Delta-1}\Gamma\left(\frac{\Delta}{2}\right)}{\sqrt{\pi}} \end{equation*} and \begin{equation*} \frac{\Gamma\left(\frac{\Delta}{2}\right)}{\Gamma\left(\frac{\Delta-1}{2}\right)}=\frac{\sqrt{\Delta}}{\sqrt{2}}+\mathcal{O}\left(\Delta^{-\frac{1}{2}}\right) \end{equation*} to get \begin{equation*} {}_2F_1\left(\frac{\Delta}{2},\frac{\Delta+1}{2},\Delta,z^{-2}\right) = \left(-z^2\right)^{\frac{\Delta}{2}}2^\Delta e^{-\frac{\Delta}{2}\xi}\left(1+e^{-\xi}\right)^{-1}[1+\mathcal{O}(\Delta^{-1})]. \end{equation*} My question its still if there is no nicer form, that in particular avoids the two cases in the definition of $\xi$.

Thanks in advance for any help!

• But how is one supposed to determine the sign in the definition of $\xi$ if it is not according to the same condition. My understanding was that both the definition and the identity you stated both depend on the imaginary part condition. Commented Mar 6, 2014 at 18:39
• You are right. I deleted my comment. Commented Mar 6, 2014 at 22:08

In fact your function is elementary and very simple, for its explicit form look at Brychkov,Marichev,Prudnikov, Integral and Series, vol.3 : $$F(a,a+1/2;2a;z)=\frac{1}{\sqrt{1-z}}\left(\frac{2}{1+\sqrt{1-z}}\right)^{2a-1}.$$