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What's up with this integral

  1. Feb 18, 2007 #1
    how did gauss integrate this function
    to obtain the values of the normal distribution?:confused:
     

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  3. Feb 18, 2007 #2

    arildno

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    By a sequence of clever tricks:

    1. Let us consider the integral
    [tex]I=\int_{-\infty}^{\infty}e^{-x^{2}}dx[/tex]
    Now, since a dummy variable's name is irrelevant, we may write:
    [tex]I^{2}=(\int_{-\infty}^{\infty}e^{-x^{2}}dx)^{2}=(\int_{-\infty}^{\infty}e^{-x^{2}}dx)(\int_{-\infty}^{\infty}e^{-x^{2}}dx)=(\int_{-\infty}^{\infty}e^{-x^{2}}dx)(\int_{-\infty}^{\infty}e^{-y^{2}}dy)[/tex]
    2. By Fubini's theorem, we have:
    [tex]\int_{-\infty}^{\infty}e^{-x^{2}}dx\int_{-\infty}^{\infty}e^{-y^{2}}dy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-x^{2}}e^{-y^{2}}dxdy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^{2}+y^{2})}dxdy[/tex]
    3. Switching to polar coordinates, [itex]x=r\cos\theta,y=r\sin\theta,0\leq{r}\leq\infty,0\leq\theta\leq{2\pi}[/itex], we get:
    [tex]\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^{2}+y^{2})}dxdy=\int_{0}^{\infty}\int_{0}^{2\pi}re^{-r^{2}}d\theta{d}r[/tex]
    which is easily integrated to the value [itex]\pi[/tex]

    4. Hence, we have [itex]I^{2}=\pi\to{I}=\sqrt{\pi}[/itex]
     
    Last edited: Feb 18, 2007
  4. Feb 18, 2007 #3

    HallsofIvy

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    But since the question was
    The answer is "numerically". In general it is impossible to find a closed form (elementary) anti-derivative but one certainly can do a numerical integration- that's how the values in a "Gaussian distribution" table are calculated.
     
  5. Feb 18, 2007 #4

    arildno

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    Blarrgh, you read it weller than me!
    I LIKE those tricks..:frown: :cry:
     
  6. Feb 18, 2007 #5
    :cool:
    arildno,you are a mean green integration machine :approve:
     
    Last edited: Feb 19, 2007
  7. Feb 18, 2007 #6
    But whats happend if I want to know the integral from 0 to x. specifically from 0 to 2
     
  8. Feb 18, 2007 #7

    arildno

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    then you should read HallsofIvy's post once more.
     
  9. Feb 19, 2007 #8

    HallsofIvy

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    You could look it up in a table of the "Error function", Erf(x). That is the "anti-derivative" of [itex]e^{-x^2/2}[/itex] and is derived, just as I said, by numerical integration.
     
  10. Feb 19, 2007 #9
  11. Feb 20, 2007 #10

    Gib Z

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    If you encounter a table of values for the Error function, don't quickly take that as your value for the integral.

    [tex]Erf(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2} dt[/tex]
     
  12. Feb 20, 2007 #11

    HallsofIvy

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    Right. You will need to use the given values of [itex]\mu[/itex] and [itex]\sigma[/itex] to convert to the "standard z" variable.
     
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