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Analytic solution to an exponential integral

  1. Aug 10, 2009 #1
    Hello,

    I am trying to find an analytic solution to the following:

    [tex]\int_{-1}^{1}\exp(-p\sqrt{1-x^{2}}-qx)dx[/tex]

    where [tex]p,q > 0.[/tex]

    Does anyone have any ideas? Thanks.
     
  2. jcsd
  3. Aug 10, 2009 #2
    Hi! Welcome to PF!

    Could you possibly tell me what exp supposed to mean ?

    Regards.
     
  4. Aug 10, 2009 #3
    Taylor series expansion of the exponential.
    http://www.efunda.com/math/taylor_series/exponential.cfm

    Wouldn't take care of the square root, But with your range (-1,1) You could probably get away with only the first few terms. This would be an approx. though.
     
  5. Aug 10, 2009 #4
    Try, the substitution method.

    [tex]t=e^{-p\sqrt{1-x^{2}}-qx}[/tex]

    Regards.
     
  6. Aug 10, 2009 #5
    I see no reason to think there is a closed-form expression for this.
     
  7. Aug 10, 2009 #6
    Substitute x =sin(t) and take a look at the integral representations of Bessel functions, e.g., http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html" [Broken]
     
    Last edited by a moderator: May 4, 2017
  8. Aug 11, 2009 #7
    Thanks Count Iblis,

    I think the result does involve a Bessel function. Related to this, does anyone know how the result

    [tex]\int_{0}^{2\pi}\exp(x\cos\theta + y\sin\theta)d\theta = 2\pi I_{0}(\sqrt{x^2+y^2})[/tex]

    is obtained? [tex]I_{0}[/tex] is the modified Bessel function of the first kind.
     
    Last edited by a moderator: May 4, 2017
  9. Aug 11, 2009 #8

    Add up the sin and cos, like:

    x sin(theta) + y cos(theta) = sqrt(x^2 + y^2) cos(theta + phi)

    The value of phi doesn't matter, because the integral is over an entire period. You can substitute theta = u - phi and then the integral over u will be from minus phi to 2 pi -phi, but that is the same as integrating over y from zero to 2 pi. So, you get rid of the phi this way.

    And then it is just a matter of using the integral definition of I_{0}
     
  10. Aug 11, 2009 #9

    Excellent, thanks Count Iblis!
     
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