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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

    rock.freak667

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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
    http://en.wikipedia.org/wiki/Gaussian_integral

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

    Mute

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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]
    then,

    [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

    Redbelly98

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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.

    http://en.wikipedia.org/wiki/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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