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Gaussian integral .

  1. Feb 24, 2010 #1
    can some one tell me how to go about solving the gaussian integral
    e^(-x^2) I know it has no elementary integral . but i was told the improper integral from -inf to positive inf can be solved and some said that i haft to do it complex numbers or something and help would be great , this is not a homework question.
  2. jcsd
  3. Feb 24, 2010 #2


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    [tex]\int_{- \infty} ^{\infty} e^{-x^2} dx = 2 \int_0 ^{\infty} e^{-x^2} dx[/tex]

    Write I as ∫e-x2 dx

    If you change the variable x for y and then multiply the two I's you will get

    [tex]I^2 = (\int_0 ^{\infty} e^{-x^2} dx)(\int_0 ^{\infty} e^{-y^2} dy)[/tex]

    this is the same as

    [tex]I^2= \int_0 ^{\infty} e^{-x^2-y^2}dxdy[/tex]

    Convert to polar coordinates now.
  4. Feb 24, 2010 #3

    Why don't you search by yourself before asking?
  5. Feb 24, 2010 #4


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    The Gaussian integral is not one that you can do with contour integrals (complex numbers)*. Typically to evaluate it you consider

    [tex]I = \int_{-\infty}^\infty dx~e^{-x^2};[/tex]

    [tex]I^2 = \int_{-\infty}^\infty dx~e^{-x^2} \int_{-\infty}^\infty dy~e^{-y^2} = \int_{0}^{2\pi} d\theta \int_{0}^\infty dr r e^{-r^2}[/tex]
    and evaluating.

    An intro calculus textbook should treat this example more carefully and rigorously than I've written (identifying [itex](x,y) \in (-\infty,\infty) \mbox{U} (-\infty,\infty)[/itex] with [itex](r,\theta) \in [0,\infty) \mbox{U} [0,2\pi)[/itex] technically takes some care to do).

    *I once read an account of how one could derive the result using a contour integral, but it was not at all a simple (or obvious) integrand that one needed to use in order to get the result.
  6. Feb 25, 2010 #5
    Thanks for your answers
  7. Feb 26, 2010 #6
    Why when i convert the integral to polar form, do the intervals become [itex](r,\theta) \in [0,\infty) \mbox{U} [0,2\pi)[/itex]. Do you prove to me?Thanks.
  8. Feb 26, 2010 #7


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    If you are integrating over "all space", then r and θ must take on all possible values. Hence those integration limits, 0≤r<∞ and 0≤θ≤2π
  9. Feb 26, 2010 #8

    Gib Z

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  10. Feb 27, 2010 #9
    Another way to do this is to consider the integral of sin^n(x) or cos^n(x) and use that to derive Stirling's approximation by first deriving the Wallis product.


    Since the asymptotics of n! can also be obtained directly from its integral representation and then using the saddle point method, this yields a derivation of the Gaussian integral.
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