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Difficult integral .

  1. Mar 18, 2003 #1
    difficult integral.....

    I have problems to calculate the integral

    I(dxexp(-ax**2)), where the limits run from -infinite to infinite.
    x**2+1

    and a>0..can someone help?....
     
  2. jcsd
  3. Mar 18, 2003 #2
    Hint: the square of that integral when a=1 is equal to [inte][inte]exp(-x^2-y^2)dxdy. Try and solve that first. :)

    If you can't figure it out, a web search for "Gaussian intregal" will reveal the clever trick.
     
  4. Mar 18, 2003 #3
    -a dx exp x^2
     
  5. Mar 18, 2003 #4

    Tom Mattson

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    I think that elijose's integral is not Gaussian, but that he meant:

    [integral][-inf..+inf](exp(-ax^2)*(x^2+1)^(-1)dx

    which is more difficult.

    Man, I have got to learn how to post math symbols in here.
     
  6. Mar 18, 2003 #5
    -ax^2 exp(arctan(x))
     
  7. Mar 18, 2003 #6

    Tom Mattson

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    No.

    You can take the derivative of your answer to see that it's not correct.
     
  8. Mar 19, 2003 #7
    Oh, hmmm. Have you tried doing the contour integral over the upper half-circle?
     
  9. Mar 19, 2003 #8

    Tom Mattson

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    You could try that. You could also try the following program:

    1. Recognize that the integrand is even, so double the integral and reduce the range of integration from 0 to infinity.

    2. Make the change of variables: u^2=x^2+1. The resulting integrand can be split into two terms, one with integrand exp(-au^2)/u^2 and one with 1/u^2. Don't forget to transform the limits of integration.

    3. The second integral is elementary, while the first can be done by treating it as a function of a. Differentiate with respect to a, and the u^2 from the exponent cancels with the u^2 in the denominator (there's the trick!), and you have a Gaussian.

    If it doesn't work, don't shoot me--I just thought of it. You can always try the contour integral if it doesn't work my way.
     
  10. Mar 19, 2003 #9

    Tom Mattson

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    Oops, my mistake. You get a regular exponential, not a Gaussian. But hey, that's easier anyway.
     
  11. Mar 20, 2003 #10
    it has something to do with probabilities...something linked to Weibull distribution...or something...but I can't find it in a book in which I know it was...wait...a second...
     
    Last edited: Mar 20, 2003
  12. Jan 9, 2009 #11
    Re: difficult integral.....

    Gamma integration.
     
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