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How to solve an integral with the Inverse error function

  1. May 16, 2014 #1
    Hi,
    this is not a homework and my problem is much bigger for me to give full details here. I came across this integral

    [itex]\mathcal{I}(\xi)=\int^{\xi_c}_{\xi}{\rm d}\xi^\prime\exp\left[\sqrt{2}\sigma\,{\rm Erf}^{-1}\left(1-\frac{8\pi}{3}{\xi^\prime}^3\right)\right][/itex]


    where Erf[itex]^{-1}[/itex] is the inverse error function and

    [itex]\xi_c=\left[\frac{3}{8\pi}\left(1-{\rm Erf}\left(\frac{\sigma^2-\sqrt{2}\sigma\,{\rm Erf^{-1}(2\beta-1)}}{\sqrt{2}\sigma}\right)\right)\right]^{1/3}[/itex]

    with [itex]0\le\beta\le1[/itex].

    I would like get an analytical approximation but I can't figure out a way to do that, even with software like Mathematica. I tried solving the integral numerically and I find a reliable solution, however, I'm mostly interested in points where [itex]\xi\to\xi_c^-[/itex], and here the inverse error function diverges.

    Do you have any ideas on how to approximate this integral?
     
  2. jcsd
  3. May 17, 2014 #2

    Simon Bridge

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    Use a taylor series?
     
  4. May 17, 2014 #3
    Thank you for your answer. But I don't understand how I can Taylor expand where the integrand diverges.
    Also, do you see a way to normalize the integral and/or make it more simple to solve it numerically.
    What I find challenging is that for every value of beta I have to numerically solve the integral. It would be better to normalize somehow the result with beta and then do the integral only once.
     
  5. May 17, 2014 #4

    Simon Bridge

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    The inverse errorfunction can be defined as a Maclaurin Series.
    That seems like a way to get rid of some awkwardness there, but I don't know for sure that it is the best way to go.

    But I don't immediately see how to make it much simpler besides that.
     
  6. May 17, 2014 #5
    Thanks, the problem seems very complicated and I think I have to resort to a numerical integration.
    However, that would be easier if one could write

    [itex]\mathcal{I}(\xi,\beta)=f(\beta)\mathcal{I}^{\prime}(\xi)[/itex]

    so this way I would have to integrate only once. Do you see a way to possibly achieve that?
     
  7. Feb 2, 2017 #6
    I understand this is quite an old post but: any luck with computing this integral? I have come across something similar recently.
     
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