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Difficult integral involving Hypergeometric functions

  1. May 12, 2010 #1
    Hey,

    I want to compute integrals of the following form
    [tex] I= \int_{-i \infty}^{i\infty} (1-x^2)^\frac{d-1}{2} \prod_{i=1}^4 _2F_1(a_i,b_i,c_i;\frac{1-x}{2}) dx [/tex]

    where [tex]a_i,b_i,c_i[/tex] are constants and [tex]c_i\in \mathbb{N}[/tex].
    d is a positive integer.

    For odd d I know that the integral will be zero by calculus of residues, with the right complex half plane as integration contour, as the constants are such that the Hypergeometric functions decay fast enough.

    For even d I cannot do this as I get a branch cut from 1 to infinity on the positiv real axis.

    Does somebody see a way, how i can explicitly integrate this?
    Thanks
    betel
     
    Last edited: May 12, 2010
  2. jcsd
  3. May 12, 2010 #2
    Can you post the values of a,b,and c please. Also, I would just start out with a square root and one hypergeometric function and just integrate it numerically, then fit the numeric answer to some analytical set up such as a square contour with the branch cut from 1 to infinity. Perhaps it would reduce to only the branch-cut integrals along the real axis. Just a quess though.
     
  4. May 12, 2010 #3
    The values are
    [tex]a_i = \frac{1}{2} + c_i + i\mu, b_i= a_i-2 i\mu[/tex]
    and [tex]c_i\in \mathbb{N},\mu\in \mathbb{C}[/tex] arbitrary (aside from [tex]\sum_i k_i\in 2\mathbb{N}[/tex]).

    I dont see, how I can numerically integrate this without putting explicit values for my parameters.

    You are right, that after some contour shuffling I can re-express the integral in terms of the integrals along the cut. But I dont see how to calculte them either.
     
  5. May 12, 2010 #4
    Hi Betel. What's wrong with putting explicit values and integrating it numerically at least to get an approximation to the answer for one particular case? Then you can compare it to the numerical results along the real axis. If that agrees, then you could focus on trying to obtain some explicit expression for just one integral along the real axis since the other is just some constant factor times the other.

    Another possibility is to integrate it directly over the imaginary axis by expressing the root as a binomial expansion, form the Cauchy product of both series, then integrate term by term and use the antiderivative inside the unit circle. Then form the binomial expansion of (-1)^{1/2}(z^2-1)^{1/2}, and do the same outside the unit circle. Not sure though if that is valid however.

    Got another one. For the case of square root and one hypergeometric function, we could write the integral as:

    [itex]\int \sqrt{1-z^2}\sum_{n=0}^{\infty}a_n(1-z)^n dz[/itex]

    and it seems at least a finite number of these can be integrated explicitly. I wonder how accurate the results would be if for example, we used only ten terms and then compared it to the numerical result?

    I realize these I'm suggesting may appear to you what, a little pathetic maybe but often when confronted with a tough problem, one needs to first look at a simpler version then build it back up. I've found that to be a very successful approach in practical mathematics. :)

    edit: alright I made a mistake: the series representation for the hypergeometric function 2F1 is valid only for |z|<1. Sorry.
     
    Last edited: May 12, 2010
  6. May 12, 2010 #5
    Well, we could use analytic continuation to find a series expansion for larger z. Then each F will be expressed as a sum of two F.

    I just found a paper calculating the case for 3 F (or rather equivalent function). I'll try to work it through and see if I can adapt their result.
     
  7. May 12, 2010 #6
    Can you post a link to that paper? I'd be interested to see how it's done. If not, it's ok. I'm just curious. :)
     
  8. May 12, 2010 #7
    The paper is
    http://arxiv.org/abs/0901.4223
    The hypergeometric functions have been converted to Legendre Polynomials. Haven't quite gotten through it yet.
     
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