MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The Gauss theorem $${_2F_1}(a,b;c;1)=\frac{\Gamma(c-a)\Gamma(c-b)}{\Gamma(c)\Gamma(c-a-b)}$$ allows to compute the analytic continuation of ${_2F_1}(a,b;c;1)$ for $a+b>c$ when the series definition diverges. The same can be done for $_3F_2(1)$ via Thomae relations. My question is how to find analytic continuation for $_4F_3(1)$ to the subdomains of $\mathbb{C}^7$ where the sum of the upper parameters is greater than the sum of the lower parameters, so that the series diverges. I know that no direct analogues of Thomae relations exit in this case so the formulas may be more complicated. I have seen some work by Allen Miller and other authors giving transformations for $_4F_3(1)$, but these transformations leave the excess (total upper parameters minus total lower parameters) invariant, so that divergent series is transformed into divergent series. The same question, of course, pertains to $_pF_{p-1}(1)$ with $p>4$...

Any help is highly appreciated.

share|cite|improve this question
Anton, thank you for correction! – Dmitri Jan 7 '11 at 1:42

The documentation to Christian Krattenthaler's HYP package contains many contiguous relations for arbitrary $_pF_q$s. In particular, C20 on page 17 of looks as though it might help.

The URL for the whole HYP package is

share|cite|improve this answer
Ira, thanks a lot for the links! Contiguous relations do provide a transition from divergent series to convergent, however in my case the excess is some indeterminate positive $n$ so that I get as finite sum on the right hand side which I was trying to avoid by finiding a compact transformation a la Thomae relations... – Dmitri Jan 7 '11 at 1:48

Dmitri, I am confused by your way to continue analytically to a point rather than to a domain. First of all, the Gauss summation formula is valid only if $\operatorname{Re}(c-a-b)>0$, so that your value at 1 is only a formal quantity assigned to the right-hand side (the ratio of gamma functions) when $a+b>c$. "The same can be done for ${}_3F_2$ via Thomae's transformations" is not correct by the same reason. So, your question is not about analytic continuation: you cannot continue the Gauss series at 1, viewed as an analytic function of three variables $a$, $b$ and $c$, through the barrier $\operatorname{Re}(c-a-b)=0$.

If you look for a former way of assigning some reasonable values to the hypergeometric functions $ {}_p F _{\substack{p-1}} $ at 1, it is naturally to play with the classical integral representations: the $(p-1)$-fold integral due to Euler or the complex Barnes integral (expressing the hypergeometric functions as Meijer's $G$-functions). All these can be found in the monographs of W.N. Bailey or L. Slater on hypergeometric functions; alternative sources are Andrews--Askey--Roy and Whittaker--Watson. However, this can never be used in deriving identities/transformations of the hypergeometric functions because of validity issues.

share|cite|improve this answer
Wadim, thank you for clarification! I admit I have misunderstood analytic continuation due to very poor knowledge of several complex variables. Nevertheless, looking at each parameter separately is the following argument wrong? Fix some complex $c\ne{0,-1,-2,\ldots}$ and $b$. Take $$f_1(a)={_2F_1}(a,b;c;1),~~\Re(a)<\Re(c-b)$$ and $$f_2(a)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}$$. The function $f_2(a)$ is meromorphic in the whole complex $a$-plane and we have $f_1(a)=f_2(a)$ in the half-plane $\Re(a)<\Re(c-b)$. Is this not called analytic continuation by definition? – Dmitri Jan 7 '11 at 2:00
Dmirti, thanks for clarification. It's indeed an analytic continuation to a meromorhpic function. And if you divide your $f_1(a)$ by $\Gamma(c-a-b)$, then it can be continued analytically to an entire function... So, your question is about how to "see" the poles of hypergeometric functions at 1 using the summation and transformation theorems. Because the latter are quite "rare", using the integral representations I mention is an efficient way to do the job. Another way is to introduce a dummy integer parameter $-k$ and compensate it in the denominator and then take the limit $k\to+\infty$. – Wadim Zudilin Jan 7 '11 at 3:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.