Asymptotic expansion of the hypergeometric function

Question

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 $z \rightarrow 1$ ?

Answer 1

Almost by the definition.

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 $x$.

Answer 2

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.