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DISCLAIMER: This question comes from math.stackexchange (where it has an active bounty). The link is here. UPDATE: the question has been answered on math.stackexchange at the previous link, and the answer is satisfactory.

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).

For reference, I'd like to prove 15.2.3 of https://dlmf.nist.gov/15.2 (where the equality is with the regularized hypergeometric function, that is just the hypergeometric function without a factor $\Gamma(c)$).

To summarize, I have three 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.
    • A last question: how to approach functions in the complex plane that are defined through a parametric integration of functions that may or may not have branch points, depending on the parameter? Even here, I'm more looking for references that a big answer.

Thanks everybody for the time you took reading this.

EDIT: In the comments, it has been asked me to clarify what I'm looking for, and it was pointed that functions with branch cuts are not discontinuous on the cut.

I unfortunately do not know a general framework to discuss functions with branch cuts, so you will have to forgive my lack of precision. I agree that the function is not discontinuous at the cut because it has not to be considered as a function from the complex plane to the complex plane, but as a function from a more general Riemannian surface to the complex numbers. I think my question makes sense even in this framework, but I have to clarify it a bit. As I do not have precise definitions, I will proceed with an example.

The basic example happens by defining the square root of a number $z$ of modulus $r$ and phase $\theta\in[0;2\pi)$ as $$ \sqrt{z}=\sqrt{re^{i\theta}}=\sqrt{r}e^{i\frac{\theta}{2}}. $$ I am basically defining a branch of the global function square root, and I can interpret it as a function on the complex plane, cut with a branch cut going from $0$ to $+\infty$, covering the positive real axis.

What I want to do is to compute, for $x$ real positive \begin{align} \lim_{\epsilon\to0^+}(\sqrt{x+i\epsilon}-\sqrt{x-i\epsilon}). \end{align} To compute that, I can write (the $\arctan$ function is here defined having range in $(-\pi/2,\pi/2)$) $$ \sqrt{x+i\epsilon}=\sqrt{\sqrt{x^2+\epsilon^2}e^{i\arctan{\frac\epsilon x}}}=\sqrt{x}+o(\epsilon),\\ \sqrt{x-i\epsilon}=\sqrt{\sqrt{x^2+\epsilon^2}e^{-i\arctan{\frac{\epsilon}x}+i2\pi}}=-\sqrt{x}+o(\epsilon). $$ From this computation, I conclude that the "discontinuity" of the square root on the cut is given by $2\sqrt{x}$.

I'm looking for the analogue of this computation in the more difficult case in which the function is given by the integral of a function with branch cuts and a parameter (that is the argument of the function), as in my example. The result should include the second equation of this post.

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    $\begingroup$ I must comment that supposed discontinuities along "branch cuts" are phantoms... at most artifacts of conventions... so cannot possibly be "detected". This is already visible in the case of $\sqrt{z}$ and such, where, despite popular conventions, there is in fact no genuine "discontinuity" of the local square root function at the negative-real axis. That's why we can't "detect" it. That is, the local power series there converge with radius of convergence equal to the distance to $0$, ignoring our convention about the "branch cut". So, can you clarify what you're hoping to have explained? $\endgroup$ – paul garrett Feb 23 '18 at 0:17
  • $\begingroup$ I think you don't understand it from the power series. You understand it from the differential equation. $\endgroup$ – Gerald Edgar Feb 23 '18 at 1:09
  • $\begingroup$ paul garrett, thanks for your comment. I do not know how to treat functions with branches in a formal setting, for now I just have to develop some intuition and computation skill. I have expanded my question to try and address your question. Is it clear? $\endgroup$ – Salvatore Baldino Feb 23 '18 at 14:27
  • $\begingroup$ Gerald Edgar, that would be the first time for me in which I study the analytic properties of the solution of a differential equation from the equation itself. But maybe it will be the best approach, as I almost never will have a function to work on, just the equation. I will look for this approach, but I'd also be interested in an answer based on manipulations of the integral that I wrote. Thanks for your hint. $\endgroup$ – Salvatore Baldino Feb 23 '18 at 14:29
  • $\begingroup$ I would like to point out that the question has received a satisfactory answer on Math Stackexchange. I will state it in the question for anyone who is interested in the answer. $\endgroup$ – Salvatore Baldino Mar 13 '18 at 11:01

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