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I have the following Gaussian Hypergeometric function: $$f(y) =~ _2F_1\left(1,2;2+\frac{2}{\alpha};1-\frac{C}{y^\alpha}\right), $$ where $\alpha>0$ and $C>0$. I want to find the function $g(y)$ such that $$ \lim_{y\to\infty} \frac{f(y)}{g(y)} = 1. $$ I guess $g(y)$ has the form of $g(y) = Dy^E$ ($f(y)$ increases super-linearly but not exponentially). Is there any references about this asymptotic equivalence?

Note that $f(y)$ does not converge to $_2F_1\left(1,2;2+\frac{2}{\alpha};1\right) = -\frac{\alpha+2}{\alpha-2}$ because the function $_2F_1\left(1,2;2+\frac{2}{\alpha};x\right)$ is not contiuous at $x=1$.

cocobear
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1 Answers1

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Fact: If $\Re(c-a-b)<0$ , then $$ \lim_{z\to 1-} \frac{{}_2F_1(a,b;c;z)}{(1-z)^{c-a-b}} = \frac{\Gamma(c)\Gamma(a+b-c)}{\Gamma(a)\Gamma(b)}. $$

So, if $\alpha>2$, we obtain $$ {}_2{F_1}\left( {a,b,c;z} \right) = {\left( {\frac{C}{{{y^\alpha }}}} \right)^{\frac{2}{\alpha } - 1}}\frac{{\Gamma \left( {2 + \frac{2}{\alpha }} \right)\Gamma \left( {1 - \frac{2}{\alpha }} \right)}}{{\Gamma (1)\Gamma (2)}} = \frac{{2\pi (\alpha + 2)}}{{{\alpha ^2}\sin\left( {\frac{{2\pi }}{\alpha }} \right)}}{C^{\frac{2}{\alpha } - 1}}{y^{\alpha - 2}}. $$

KJ Choi
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