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Integral of the S.N.D.F.

  1. May 20, 2007 #1
    I fell very silly posting this, but I am making Normal Distribution tables. I tried to Integrate what seems to be quite a simple function:
    Well... I can't I have tried Integration by Parts, Algebraic Substitution etc. But I cannot do it! Can somebody help me?
  2. jcsd
  3. May 20, 2007 #2


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  4. May 20, 2007 #3
    Ok. But I don't intitutively see why... Can anyone answer this?

    So all tables are created as an approximation - e.g. the Simpson's Rule?
  5. May 23, 2007 #4
    Yes, they are numerical approximations. I can assure you they weren't calculated via Simpson's Rule.

    There is no elementary antiderivative of the pdf. Elementary being sums/products/compositions of 'nice' functions.
  6. May 23, 2007 #5

    Gib Z

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    It can be very hard to prove that a certain function has non-elementary antiderivatives. We know in this case that no known function has its derivative as that integral, so we defined one.
  7. May 23, 2007 #6


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    One could do it by numerical integration, but try a double integral, as in

    [tex]\int_{0}^{x} e^{-x^2} dx\,\int_{0}^{y} e^{-y^2} dy[/tex], where I is each integral.

    combine the two and use the transformation from Cartesian (x, y) to polar coordinates (r, [itex]\theta[/itex]).

    Limits (0,x) and (0, y) become (0, r) and (0, 2[itex]\pi[/itex])
    Last edited: May 23, 2007
  8. May 23, 2007 #7
    I am not too sure about Astronuc's suggestion. But anyway, can someone give me a reasonable function for an approximation. So far I have only come up with tanh(x), with a precding constant...
  9. May 24, 2007 #8


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    The double integral method is good for evaluating the definite integral of exp(-x^2/2) from -infinty to infinity, but it is of no use for integrating over finite limits.

    The double integral method works by transforming the square of the integral into a double integral over a region of the x,y plane. Since the element of area dx dy becomes [tex]r \, dr \, d\theta[/tex] then it follows that the inner of the two dimensional ([tex]dr \, d\theta[/tex]) integral becomes the trivial [tex]\int r e^{-r^2/2} dr[/tex].

    The catch is that the limits of the integration correspond to a square region of the x,y plane, so r is not a constant! Bascially the difficulty just gets tranferred to the outer [tex]d\theta[/tex] integral, so in general this is no solution.

    For the specific case of the integral from -infinity to +infinity however the double integral is over the entire x,y plane and therefore the difficulty with the rectangular limits vanishes. This is the standard method of proving that [tex]\int_{-\infty}^{+\infty} e^{-x^2/2} dx = \sqrt{2 \pi}[/tex]

    Goolging for 'erf approximations' gives several very good approximations in the first few hits. One nice simple one is :

    [tex]\frac{1}{\sqrt{2 \pi}} \int_x^{\infty} e^{-x^2/2} \, \simeq \frac{ e^{-x^2/2} } {1.64 x + \sqrt{0.76 x^2 + 4}} [/tex]
    Last edited: May 24, 2007
  10. May 25, 2007 #9
    I see......
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