Limit of the hypergeometric function

Question

I dont' have experience with hypergeoemtric functions, but need to compute the following limit:$$\lim_{z\rightarrow0+}F\left(1,\alpha;\frac{\beta}{z};\frac{\gamma}{z}\right),$$ where $\alpha$ is non-integer real and $\beta$ and $\gamma$ are purely imaginary parameters. It seems the limit exists and finite. I tried to use an integral representation and a few standard transformations (such as Pfaff), but could not get the result. Any help would be appreciated.

Answer 1

1. Before taking the limit, it is useful to apply the transformation \begin{align}F(a,b;c;x)&=\frac{\Gamma(c)\Gamma(b-a)}{\Gamma(b)\Gamma(c-a)}(-x)^{-a} F\left(a,1-c+a;1-b+a;x^{-1}\right)+\\ &\,+\frac{\Gamma(c)\Gamma(a-b)}{\Gamma(a)\Gamma(c-b)}(-x)^{-b} F\left(b,1-c+b;1-a+b;x^{-1}\right). \tag{1} \end{align} In that way in your case one obtains hypergeometric functions with the argument tending to zero and one of the parameters tending to infinity so that their product remains finite.

2. Next use the limit $$\lim_{\Lambda\rightarrow\infty}F\left(a,b\Lambda;c;\frac{x}{\Lambda}\right)= {}_1F_1\left(a;c;bx\right).$$ (very easy to get from the series representation of $_2F_1$).

3. One also needs the asymptotics $\frac{\Gamma(\Lambda)}{\Gamma(\Lambda-a)}\sim \Lambda^a$ as $\Lambda\rightarrow\infty$.

Combining these formulas, we can compute the limiting value you are interested in. The only nontrivial thing is to correctly keep track of complex phases in the expressions like $(-z)^b$.

For example, if $\beta\in i\mathbb{R}_{>0}$ and $\gamma\in i\mathbb{R}_{<0}$, after some simplifications the limiting value is given by $$\lim_{z\rightarrow +0}F\left(1,\alpha;\frac{\beta}{z};\frac{\gamma}{z}\right)=e^{-\beta/\gamma}\left(-\frac{\beta}{\gamma}\right)^{\alpha}\Gamma\left(1-\alpha,-\frac{\beta}{\gamma}\right),$$ where $\Gamma(\nu,x)$ denotes the incomplete gamma function.