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.

  • $\begingroup$ Where did this problem come from? $\endgroup$ Commented Mar 28, 2014 at 1:40
  • 2
    $\begingroup$ From computing an expected number of isolated nodes of a causet in (1+1) de Sitter space-time. Not sure this will help :) $\endgroup$
    – Alex
    Commented Mar 28, 2014 at 2:15

1 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.

  • $\begingroup$ This is great! Thanks a lot for your help. $\endgroup$
    – Alex
    Commented Mar 31, 2014 at 14:23
  • $\begingroup$ I went to the transformation page. There are lots of similar transformations; but which one exactly you mean? $\endgroup$
    – hossayni
    Commented Sep 11, 2014 at 12:53
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    $\begingroup$ @hossayni I found the reason of controversy mentioned in my previous comments. So the formula to use is indeed 15.8.2. It is equivalent to my formula (1) because of the nonstandard normalization dlmf.nist.gov/15.1#E2 used in 15.8.2. $\endgroup$ Commented Sep 11, 2014 at 20:12
  • $\begingroup$ @O.L. Thanks for the time you spent $\endgroup$
    – hossayni
    Commented Sep 12, 2014 at 10:21
  • $\begingroup$ @Startwearingpurple: Nice method and results! We are looking for expressions similar to your Eq.(1) for $_1F_2(x)$,$_2F_3(x)$, and $_3F_4(x)$. I was not been able to final these expressions from dlmf.nist.gov. Best $\endgroup$
    – mike
    Commented May 11, 2016 at 4:14

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