I am going through this paper, and I am having trouble understanding page 20. I am still learning my way around managing multi valued complex functions, so I'd like your help in understanding what's happening there.

I have the definition of the hypergeometric series $_2F_1$ as $$ _2F_1(a,b;c;z)=\sum_{n=0}^\infty \frac{(a)_n(b)_n}{(c)_n}\frac{z^n}{n!}. $$ Here $(a)_n$ is the rising Pochhammer symbol, $(a)_n=\Gamma(a+n)/\Gamma(a)$ (well defined whenever $a$ is not a negative integer or zero). By elementary computations, the radius of convergence of this series is 1. It is then said that $_2F_1$ can be continued analitically in the whole complex plane along any curve not passing through $[1;+\infty)$, and $1$ itself is a branching point and the function has a cut on the previous segment.

Furthermore, equation $2.115$ of the paper computes the discontinuity when crossing the branch cut as (with $x\geq4/3$ to have the real part of the argument bigger than one) $$ _2F_1\left(\frac16,\frac56;1;\frac{3x}4+i\epsilon\right)-_2F_1\left(\frac16,\frac56;1;\frac{3x}4-i\epsilon\right)=i\, _2F_1\left(\frac16,\frac56,1;1-\frac{3x}4\right). $$ I am trying to understand those results. It is not clear to me how to understand from the power series form the behaviour on the boundary of the disk of convergence. It is clear to me that the series in $z=1$ diverges, as the coefficients do not go to zero, so there should be no analytical continuation.

A way to study that function would be its integral representation: $$ _2F_1(a,b;c;z)=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_0^1t^{b-1}(1-t)^{c-b-1}(1-zt)^{-a}dt. $$ Here it is understood that $\arg t=0=\arg(1-t)$. From this representation, it is clear to me that the function can have branches whenever $a$ is not a positive integer, and if $z\in[1;\infty)$ we have $0$ in the integration path, so we can have multiple branches. And that's exactly my case. The problem is that I can't use this form to prove the previous equation about the discontinuity: all I manage to write is (here I write z=x with x real number bigger than one, and I am neglecting some numerical factors that can easily be reinserted) $$ \lim_{\epsilon\to0}((1-xt-i\epsilon t)^{-a}-(1-xt+i\epsilon t)^{-a})=-(1-e^{-2i\pi a})(1-xt)^{-a}. $$ This is different from the line that I would like to prove (as an example, I do not see why I should change the argument, but I also see that with that argument changing the function is at least evaluated in a point where well defineteness is guaranteed).

To summarize, I have two questions:

  • How to prove the discontinuity from the integral form?
  • More in general: how to treat power series in order to understand if their singularities are poles or branch cuts, and the discontinuity of their analytic continuations? I know that this is a very broad topic, and I'd also like some references to continue my study.

Thanks everybody for the time you took reading this.

  • $\begingroup$ Why do you say "whenever $a$ is not a \textit{negative} integer"? Is it not the case that there is no branching if $a$ is any integer, positive or negative? $\endgroup$
    – Tuneer
    Mar 29, 2022 at 16:01

1 Answer 1


When computing the limits of $(1-z\,t)^{-a}$, you need to take into account that the limits differ only for $t>1/z$. So the integral has to be split into two, and then indeed the limit from above can be computed as $${_2F_1}(a,b;c;z+0i) = \frac {\Gamma(c)} {\Gamma(b) \Gamma(c-b)} \bigg(\\ \int_0^{1/z} t^{b-1} (1-t)^{c-b-1} (1-z\,t)^{-a} dt +\\ e^{2 \pi i a} \int_{1/z}^1 t^{b-1} (1-t)^{c-b-1} (1-z\,t)^{-a} dt \bigg)$$ Then, making the change of variables $t=(1-1/z)\tau+1/z$, we have $${_2F_1}(a,b;c;z+0i) - {_2F_1}(a,b;c;z-0i) =\\ \frac {\Gamma(c)} {\Gamma(b) \Gamma(c-b)} \left( e^{2\pi i a} - 1 \right) z^{1-c} (1-z)^{-a} (z-1)^{c-b} \times\\ \int_0^1 \tau^{-a} (1 - \tau)^{c-b-1} (1-(1-z)\tau)^{b-1} d\tau=\\ \frac {2 \pi i \,\Gamma(c)} {\Gamma(a) \Gamma(b) \Gamma(c-a-b+1)} z^{1-c} (z-1)^{c-a-b} {_2F_1}(1-a,1-b;c-a-b+1;1-z).$$

  • $\begingroup$ You have my upvote and my gratitude. $\endgroup$ Mar 7, 2018 at 0:49

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