Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to show integral equal to hypergeometric function?

  1. Oct 3, 2013 #1
    Hi,

    I would like to show directly,

    [tex]\int \frac{e^{at}}{e^{it}+e^{-it}}dt=\frac{e^{(i+a) t} \text{Hypergeometric2F1}\left[1,\frac{1}{2}-\frac{i a}{2},\frac{3}{2}-\frac{i a}{2},-e^{2 i t}\right]}{i+a}[/tex]

    I realize I can differentiate the antiderivative to show the relation but was wondering how to integrate the integral directly to obtain the hypergeometric function or are these type integrals evaluated using a different technique? I can express it as:

    [tex]\int \frac{e^{(i-a)t}}{1+e^{2it}}dt[/tex]

    and at least it has the [itex]e^{2it}[/itex] in there.

    How are any integrals expressed in terms of hypergeometric functions?
     
  2. jcsd
  3. Oct 3, 2013 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    How are you defining the "Hypergeometric" functions?

    (And that last integral should be
    [tex]\int\frac{e^{i-a)t}}{1+ e^{-2it}}dt[/tex])
     
  4. Oct 3, 2013 #3

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Even that's not right. It should either be
    $$\int \frac{e^{(a+i)t}}{1+e^{2it}}dt$$
    or
    $$\int \frac{e^{(a-i)t}}{1+e^{-2it}}dt$$
     
  5. Oct 3, 2013 #4
    I'd start by trying a substitution, but I honestly don't know how one goes about integrating that directly.

    Have you considered approaching it by the hypergeometric differential equation?
     
  6. Oct 3, 2013 #5
    Hi,

    I'm defining the hypergeometric function as the analytic continuation of the hypergeometric power series:

    [tex]\mathstrut_2 F_1\left(a,b,c,z\right)=\sum_{n=0}^{\infty} \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!}\quad |z|<1[/tex]

    which I believe is a multi-valued function with a branch-point at z=1 but I'm not very familiar with this function.

    I've considered that but don't know how to approach this integral through that DE but I'll keep working on it.

    Also, thanks for correcting the integral DH. It is:

    [tex]\int \frac{e^{(i+a) t}}{1+e^{2it}}dt[/tex]
     
    Last edited: Oct 3, 2013
  7. Oct 3, 2013 #6
    Why would it be multivalued? It isn't readily apparent to me.

    As to the problem, this is new to me, too. I'll probably piggy-back off of this thread for my own learning. :tongue:
     
  8. Oct 3, 2013 #7
    That's an interesting question in itself. Well, Mathematica says it is although I realize that's not sufficient. However, we can experimentally verify it's multivalued by considering the hypergeometric DE:

    [tex]z(1-z)w''+(c-(a+b+1)z)w'-abw=0[/tex]

    and numerically integrating this equation over a contour which encircles the branch point at z=1, for example over a circle of radius 2, that is, by letting [itex]z=2e^{it}[/itex]. Now when I do that and substitute that into the DE and I did it quickly so not confident I did it error-free, I obtain the DE in terms of t:

    [tex]z(1-z)(-1/4 e^{-2it} w''-iw')+(c-(a+b+1)z)(-i/2 e^{-it}) w'-ab w=0,\quad z=2e^{it}[/tex]

    now, I just arbitrarily let a=1, b=2, c=3, w(0)=2, w'(0)=1 and integrate over 2pi and plot the real (or imaginary) results, the solution does not return to the same point. That is experimentally, a clear sign it's not single-valued although there is always the chance of numerical error.

    Also, getting back to the problem, I'm encouraged by my initial attempts at iterative integration by parts, the solution looks like it's getting there but not quite. I'll work on that approach.
     

    Attached Files:

    Last edited: Oct 3, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How to show integral equal to hypergeometric function?
Loading...