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Can anyone help with this integral?

  1. Jul 19, 2008 #1
    Hello
    I need to evaluate this integral:

    [tex]\int_0^1e^{ax^2}erf(bx)dx[/tex]

    where "erf" is the error function:[itex]erf(x)=\frac{2}{\sqrt{\pi}}\int_0^x\exp(-t^2)dt[/itex].
    I can't find this integral in Gradshteyn, Abramowitz, Prudnikov, etc.
    Neither Mathematica nor Maple do this integral (definite or indefinite).
    The best I can do is by series...
    Can anyone help with this integral?

    Thanks
     
    Last edited by a moderator: Jul 19, 2008
  2. jcsd
  3. Jul 20, 2008 #2

    CompuChip

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    So it is very likely that there is no closed form solution. So what kind of an answer would you like then?
     
  4. Jul 20, 2008 #3
    If a happens to be equal b^2, then Mathematica gives the answer

    [tex]
    \frac{b}{\sqrt{\pi}}\operatorname{HypergeometricPFQ}[\{1,1\},\{\frac{3}{2},2\},b^2]
    [/tex]
     
  5. Jul 25, 2008 #4
    I'm looking for a closed form... with a not necessarily equal b.
     
    Last edited: Jul 25, 2008
  6. Jul 25, 2008 #5
    Yes, I knew it.
    This is in Prudnikov tables.
    But a not necessarily equal b.
    Thanks
     
    Last edited: Jul 25, 2008
  7. Jul 25, 2008 #6
    I'm looking for a closed form.
    Thanks
     
  8. Jul 29, 2008 #7

    CompuChip

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    So if you checked several references an none of them gives one, maybe it doesn't exist :)
    You can define it, though.
    RicardoP's function
    [tex]\mathcal P(a, b) :=
    \int_0^1e^{ax^2} \operatorname{erf}(bx)dx
    [/tex]
    Then the closed form solution is
    [tex]\mathcal P(a, b)[/tex]
     
    Last edited: Jul 29, 2008
  9. Jul 29, 2008 #8
    You may be able to do a change of variables to make the limits from 0 to infinity, then try some sort of complex contour integral. I haven't tried this yet but its worth a shot.
     
  10. Jul 30, 2008 #9
    I can change x=1/y, and the interval goes from 1 to \infty, but not from 0 to \infty, or -\infty to \infty.

    I tryed several changes of variable...
    I changed the order of integration (this is a double integral).
    I tryed complex contour integral too, but without sucess.

    Thanks!
     
  11. Jul 30, 2008 #10
    This integral is very alike the integrals in the article:
    "Integration of a Class of Transcendental Liouvillian Functions with Error-Functions, Part II."
    Paul H. Knowles, J. Symb. Comp. 16 (1993) 227.
    http://dx.doi.org/10.1006/jsco.1993.1042
    I'm working on it...
     
    Last edited: Jul 30, 2008
  12. Aug 23, 2008 #11
    The "Cumulative Bivariate Normal Probability Function" is defined by:
    [tex]L(h,k,\rho)=\int_h^\infty dx\int_k^\infty dy\;g(x,y,\rho)[/tex]
    where
    [tex]g(x,y,\rho)=\frac{1}{\sqrt{1-\rho^2}}Z(x)Z\left(\frac{y-\rho x}{\sqrt{1-\rho^2}}\right)[/tex]
    is the "Bivariate Normal Probability Function" and
    [tex]Z(x)=\frac{\exp(\frac{-x^2}{2})}{\sqrt{2\pi}}[/tex]
    is the "Normal Probability Function", in Abramowitz & Stegun's notation.

    With these definitions we have the answer
    [tex]\int_0^1 dx\exp(-ax^2)\mathrm{erf}(bx)=\frac{1}{2}\sqrt{\frac{\pi}{a}}\mathrm{erf}(\sqrt{a})-\frac{1}{\sqrt{a\pi}}\arctan\left(\frac{\sqrt{a}}{b}\right)+2\sqrt{\frac{\pi}{a}}L\left(\sqrt{2a},0,\frac{-b}{\sqrt{a+b^2}}\right)[/tex]

    I wanna thank Axel for the hint
    http://groups.google.com.au/group/sci.math.symbolic/browse_thread/thread/ddbbb1f3ab27b3c4
     
  13. Aug 23, 2008 #12
    Is this any more useful than using RicardoP's function as suggested by CompuChip?
    Granted, you can now make use of some known properties of this "Cumulative Bivariate Normal Probability Function"...
     
  14. Aug 23, 2008 #13
    obvious...
     
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