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Consider a Gaussian Hypergeometric function $_2F_1(a,b;c;z)$. What condition should $a,b,c$ satisfy in order for $$ \lim_{z\to\infty}{}_2F_1(a,b;c;-|z|^2)=0\ ?\tag{A} $$

Using eq. 32 from the MathWorld article Hypergeometric Function, $$ _2F_1(a,b;c;z)=\frac{1}{(1-z)^a}{}_2F_1\left(a,c-b;c;\frac{z}{z-1}\right) $$ it seems to me that $(\mathrm A)$ will be satisfied if and only if $a>0$ and $_2F_1(a,c-b;c;1)<\infty$. Moreover, using $$ _2F_1(a,b;c;1)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)} $$ the condition $_2F_1(a,c-b;c;1)<\infty$ is equivalent to $$ \frac{\Gamma(c)\Gamma(b-a)}{\Gamma(c-a)\Gamma(b)}<\infty $$ that is, $c\not\in-\mathbb N$ and $(b-a)\not\in-\mathbb N$. Is this analysis correct?

Moreover, if any of $a>0$, $c\not\in-\mathbb N$ or $(b-a)\not\in-\mathbb N$ is not satisfied, then will $(\mathrm A)$ diverge? or may it approach a non-zero finite value?

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  • $\begingroup$ Some basic numerical analysis seems to confirm this, but I remain unconvinced, and I'd like some confirmation. Thanks! $\endgroup$
    – AccidentalFourierTransform
    Commented Mar 30, 2017 at 14:42
  • $\begingroup$ The conditions you found are definitely sufficient. $\endgroup$
    – Antonio Vargas
    Commented Mar 31, 2017 at 0:30
  • $\begingroup$ I agree with your numerical analysis, but I will also point out that the identity for $_2F_1(a,b,c,1)$ is valid only for $\mathfrak{Re}(c-a-b)>0$. More disturbing, is that the Atlas of Functions (K. Oldham, J. Myland, & J. Spanier, An Atlas of Functions, Ch. 60, Springer) says that $F(a,b,c,-\infty)=0$ for $a<0$ and $b<0$. That's disturbing in light of my own numerical calculations. $\endgroup$
    – Cye Waldman
    Commented Mar 31, 2017 at 17:43

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