# Asymptotic expansion of the hypergeometric function

Given a function of the form:

$$f(z)= \text{ }_2F_1\left(a,b ;c;\frac{1 - z}{2}\right)$$

is there a way to find to find the asymptotic form of this function in the limit $z \rightarrow 1$ ?

Let $x=\frac{1-z}2$ $$\, _2F_1\left(a,b;c;\frac{1-z}{2}\right)=\, _2F_1\left(a,b;c;x\right)=1+\frac{a b }{c}x+\frac{a (a+1) b (b+1) }{2 c (c+1)}x^2+O\left(x^3\right)$$ Now, replace $x$.
• How do I put in the limit $z \rightarrow 1$? x is simply zero in this limit. – cord Jan 28 '18 at 8:53
• @cord. Replace $x$ by $\frac{1-z}2$ to get the asymptotics. – Claude Leibovici Jan 28 '18 at 9:05
We may use its integral representation ((16) here) which shows that: $$f(z) = \frac{\Gamma(c)}{\Gamma(b) \Gamma(c-b)} \int_0^1 \frac{t^{b-1} (1-t)^{c-b-1}}{(1-t\frac{1-z}{2})^a} dt$$ As this is an integral with a positive integrand we may now interchange limit and integral sign, which yields: $$\lim_{z\rightarrow 1} f(z) = \frac{\Gamma(c)}{\Gamma(b) \Gamma(c-b)} \int_0^1 t^{b-1} (1-t)^{c-b-1} dt$$ Continuing as suggested in the comments we find that the integral on the rhs is the Beta function which equals $\frac{\Gamma(b) \Gamma(c-b)}{\Gamma(c)}$ thus the result is one.
• The remaining integral is a Beta function, which cancels with all of the gamma functions. The final result is just $1$. – Antonio Vargas Jan 28 '18 at 8:36