# About large z behavior of hypergeometric function $_2F_1(1/2,1/2,1;z)$

The hypergeometric function $_2F_1(\large \frac{1}{2},\frac{1}{2},1;\frac{1-\frac{u}{\Lambda^2}} {2} \large)$ at large $\mid u\mid$ can be approximated by $$-\frac{\Lambda}{\pi} \sqrt{\frac{2}{u}} \ln(\frac{u}{\Lambda^2})$$

How to approximate hypergeometric function at large argument ?

Thanks

• Unless I misunderstand the question, it seems you already gave the answer. Are you looking for a proof then? Nov 29, 2013 at 8:51
• Yes, I do not know how to get it. Nov 29, 2013 at 8:56
• The ${}_2F_1$ series has radius of convergence $1$, so to discuss values $z>1$ you must use the analytic continuation of that series. Aren't there some Gaussian relations that write such a ${}_2F_1$ in terms of another ${}_2F_1$ with other arguments, that will help near $z=\infty$? Nov 29, 2013 at 14:12
• By the way, Maple says ${}_2F_1(1/2,1/2;1;z) = (2/\pi)K(\sqrt{z})$ in terms of complete elliptic integral $K$. Nov 29, 2013 at 14:15
• functions.wolfram.com has formulas for behavior at infinity: functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/… But like the original question, no references or proofs are included. Nov 29, 2013 at 14:23

• It is about $z \sim 0$ in Wiki, and for discussion about $z \sim \infty$, it requires $a-b$ is a integer which is not satisfied here. Nov 30, 2013 at 3:13
• ..."The behavior of the solutions [F(a,b,c,z)] of the hypergeometric equation for large values of $|z|$ can be fully described by means of the formulas for the analytic continuation into the neighborhood of the point at $z = \infty$. Unless $a — b$ is an integer, every solution $u(z)$ can be put into the form $u(z) = \lambda_1z^{-a} +\lambda_2z^{-b} + O(z^{-a-1}) + O(z^{-b-1})$ where $\lambda_1$ and $\lambda_2$ are constants; if $a-b$ is an integer, $z^{-a}$ or $z^{-b}$ has to be multiplied by a factor $\log z$." Dec 1, 2013 at 4:42