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


Does anyone know whether there is a known function/distribution that corresponds to the limit:

$\lim_{\epsilon\rightarrow0^+} \mathfrak{Re}\left[f(x+i\epsilon) - f(x-i\epsilon)\right]$

when $f(z)={}_2F_1(\frac12+\mu,\frac12-\mu,1,z)$? I think this is essentially the definition of a hyperfunction.

In particular I am interested in the case where $\mu$ and $x$ are both purely imaginary. You can plot it in Mathematica and see the function for small values of epsilon. When $x$ is imaginary a nice function emerges, when $x$ is real there seems to be a Dirac delta function behaviour at the origin.

I wonder whether there is a standard reference where I could look for distributional expressions of hyperfunctions other than the common ones like the Dirac delta or the Heaviside step function. (The context here is that of quantum field theory on curved manifolds - the commutator typically turns out to be a hyperfunction.)

share|cite|improve this question
According to the usual definition, the branch cut of the hypergeometric function is on the real axis extending from 1 to infinity. So there should not be any jump on the imaginary axis. – Michael Renardy Apr 1 '11 at 19:44
what is $\mu$? for $\mu\in1/2\mathbb{N}$ $f$ should be just a polynomial and the expression would be 0. – Marcel Bischoff Apr 2 '11 at 10:13
There may simply not be an expression involving functions less general than the ${}_2F_1$ itself. Perhaps what you are really after is not a name for this function (hyperfunction), but rather some specific properties of it. Support? Singular support? Asymptotics? You may want to refine your question. – Igor Khavkine Apr 2 '11 at 17:55

Your Answer


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

Browse other questions tagged or ask your own question.