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Evaluating the Gaussian integral

  1. Oct 9, 2006 #1
    Hi,

    I'm trying to evaluate the standard Gaussian integral

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

    The standard method seems to be by i)squaring the integral, ii)then by setting the product of the two integrals equal to the iterated integral constructed by composing the two integrals, iii)using Fubini's theorem to turn this into an area integral and iv)then using Fubini's theorem again to turn this back into an iterated integral, this time in polar coordinates.

    Of these, (ii) seems impossible to me. Why would the iterated integral be equal to the product of the two integrals taken separately? Even if this were the case, this does not seem like a result so trivial that you could use it without any justification. Are there other ways to evaluate the integral? Thanks.

    Molu
     
    Last edited: Oct 9, 2006
  2. jcsd
  3. Oct 9, 2006 #2

    dextercioby

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    ii) & not iii) uses Fubini's theorem.

    Daniel.

    BTW, it's "\infty" that gives the symbol [itex] \infty [/itex]
     
  4. Oct 9, 2006 #3
    That's what I wrote, and thanks for the infinity.
     
  5. Oct 9, 2006 #4

    HallsofIvy

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    The fact that
    [tex]\left(\int_{-\infty}^\infty f(x)dx\right)\left(\int_{-\infty}^\infty g(y)dy= \int_{-\infty}^\infty f(x)g(y)dxdy= \int\int_A f(x)g(y)dA[/tex]
    where A is all of R2, is Fubini's theorem.
     
  6. Oct 10, 2006 #5
    Checking again, I see that there are two parts to Fubini's theorem of which (ii) uses one part (product of integrals=double integral) and (iii) and (iv) use the other (double integral=iterated integral). In any case, is this the only way to evaluate the gaussian integral?
     
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