Limit of the hypergeometric function 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. 
 A: *

*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.

*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 $_2F_1$).

*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. 
