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!

  • $\begingroup$ 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. $\endgroup$
    – physicus
    Commented Mar 6, 2014 at 18:39
  • $\begingroup$ You are right. I deleted my comment. $\endgroup$ Commented Mar 6, 2014 at 22:08

1 Answer 1


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

Unfortunately I can not give a reference to the exact formula, this vol. 3 is the only one I have not an English version.

Also sorry for may be somehow akward post -- it was my first answer here, be patient.

  • 1
    $\begingroup$ Please consider writing out this elementary and very simple form in your answer, since the text books.google.com/… might not be easily accessible for all MO users. It would make for a better and more useful MO answer. $\endgroup$
    – Todd Trimble
    Commented Apr 5, 2014 at 19:18
  • $\begingroup$ @NeilStrickland: you are right, I will delete my comment. I wanted to express that I am not satisfied with this criptic answer, but choose the wrong option. $\endgroup$ Commented Apr 6, 2014 at 5:28

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