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Problem with a gaussian integral.

  1. May 30, 2005 #1
    Hey, i've been learning about gaussian integrals lately. And i'm now stuck in one part. I am now trying to derive some kind of general formula for gaussian integrals

    [tex] \int x^n e^{-\alpha x^2} [/tex]

    for the case where n is even. So they ask me to evaluate the special case n=0 and alpha=1. So its [tex] I= \int^{\infty}_{-\infty} e^{-x^2} dx [/tex]. When i square this integral, they said that its [tex] I^2= ({\int^{\infty}_{-\infty} e^{-x^2} dx})({\int^{\infty}_{-\infty} e^{-y^2} dy}) [/tex] with both x and y as according to them, i have to use a different variable for the first and second integral factors.

    Why is this so? I have limited calc background. So i was wondering you guys could help me out. Thanks alot.....
     
    Last edited: May 30, 2005
  2. jcsd
  3. May 30, 2005 #2
    I think they're trying to get you to use the gamma function.

    [tex]\Gamma(z)=2{\int^{\infty}_{0}dte^{-t^2}t^{z-1} [/tex]

    Those two integrands you have are even and the gamma func has useful properities for situations like this.
     
  4. May 30, 2005 #3

    Dr Transport

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    By using different variables for the squared version, you can convert to different coordinate systems, i.e. to cylindrical coordinates. What you typed above is the way you prove that [tex] \int^{\infty}_{-\infty} e^{-x^2} dx = \sqrt{\pi} [/tex]
     
  5. May 30, 2005 #4

    HallsofIvy

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    "When i square this integral, they said that its [tex] I^2= ({\int^{\infty}_{-\infty} e^{-x^2} dx})({\int^{\infty}_{-\infty} e^{-y^2} dy}) [/tex]"

    No, they didn't say that! You put in the "when I square this integral" yourself- they didn't square the integral.

    What they did say was: If [tex]I= \int^{\infty}_{-\infty} e^{-x^2} dx[/tex], then it is also true that [tex]I= \int^{\infty}_{-\infty} e^{-y^2} dy[/tex] because that is just a change in the dummy variable.

    It is then true that [tex]I^2= I*I= \(\int^{\infty}_{-\infty} e^{-x^2} dx\)\(\int^{\infty}_{-\infty} e^{-y^2} dy\)[/tex]- just multiplying two different ways of writing the same thing.

    Of course, the really important thing is that fact that that product of integrals can be written as a double integral:
    [tex](\int^{\infty}_{-\infty} e^{-x^2} dx)(\int^{\infty}_{-\infty} e^{-y^2} dy= \int_{y= -\infty}^{\infty}\int_{x=-\infty}^{\infnty}e^{-(x^2+ y^2)}dxdy[/tex].
     
    Last edited: May 30, 2005
  6. May 30, 2005 #5
    ahhhh so i see, a double integral. Thanks alot. Maybe i misinterpreted what they said, cos' they made it sound that changing the dummy variable was a must.
     
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