Hypergeometric function at $z=1$

Question

There is a nice formula for the value of the hypergeometric function ${}_2 F_1(a,b,c,z)$ at $z=1$ when $\Re{(c)}>\Re(b+a)$ given for example at https://en.wikipedia.org/wiki/Hypergeometric_function

Is there some formula for what happens when $\Re{(c)}\leq\Re(b+a)$. Presumably, the function diverges but is there a known asymptotic behavior as $z\rightarrow 1^-$?

Answer 1

The (Gauss) hypergeometric function $F(a,b;c;z):={_2}F_1(a,b;c;z)$ is the subject of Chapter 15 of the DLMF. In particular section 15.4(ii) describes the asymptotic behavior as $z\to 1^{-}$ in all cases:

• If $\Re(c)>\Re(a+b)$, then $F(a,b;c;1)=\dfrac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}.$
• If $c=a+b$, then $\displaystyle \lim_{z\to 1^-} \frac{F(a,b;a+b;z)}{-\ln(1-z)}=\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}.$
• If $\Re(c)=\Re(a+b)$ and $c\neq a+b$, then $$\displaystyle \lim_{z\to 1^-} (1-z)^{a+b-c}\left(F(a,b;c;1)-\dfrac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}\right)=\dfrac{\Gamma(c)\Gamma(a+b-c)}{\Gamma(a)\Gamma(b)}.$$
• If $\Re(c)<\Re(a+b)$, then $\displaystyle \lim_{z\to 1^-} \frac{F(a,b;c;z)}{(1-z)^{c-a-b}}=\dfrac{\Gamma(c)\Gamma(a+b-c)}{\Gamma(a)\Gamma(b)}.$