1
$\begingroup$

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 $z\rightarrow 1^-$?

$\endgroup$
2
$\begingroup$

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 $\displaystyle \lim_{z\to 1^-} \frac{F(a,b;a+b;z)}{(1-z)^{c-a-b}}=\dfrac{\Gamma(c)\Gamma(a+b-c)}{\Gamma(a)\Gamma(b)}.$
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.