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Normal Distribution Integral

  1. Mar 4, 2008 #1
    1. The problem statement, all variables and given/known data

    I'm having difficulty integrating something,
    click http://en.wikipedia.org/wiki/Normal_distribution
    and under Cumulative distribution function, there is an integral - how do you get to the next line?

    2. Relevant equations

    3. The attempt at a solution

    i have tried the substitution of [tex]t=u-\mu[/tex] which gives

    but can't integrate this.
  2. jcsd
  3. Mar 4, 2008 #2


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    see error functions

    Hi wombat! Welcome to PF :smile:

    All you need is in http://en.wikipedia.org/wiki/Error_function

    Good luck!

    [size=-2](if you're happy, don't forget to mark thread "solved"!)[/size]​
  4. Mar 5, 2008 #3
    Still don't really understand what's what - this to clarify this is what i was originally referring to
    and this is what i got from the error function page
    but i need to integrate from [tex]-\infty[/tex] to x.
    Last edited: Mar 5, 2008
  5. Mar 5, 2008 #4
    Thanks for your help by the way!
  6. Mar 5, 2008 #5


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    An even function!

    Ah! You didn't take account of the fact that the integrand is the same for -t as for t (an "even function"), so you can mutiply the limits by -1:


    is the same as:


    which is (a multiple of) erfc(-x).

    Then the top of the wikipedia page gives you the answer! :smile:

    [size=-2](if you're happy, don't forget to mark thread "solved"!)[/size]​
  7. Mar 6, 2008 #6
    i still don't get it.
  8. Mar 6, 2008 #7
    i know that
    but i don't end up with
  9. Mar 6, 2008 #8
    should i being using this?

    [tex]\int_{-x}^{\infty}e^{-t^{2}/2}dt\qquad[/tex] = [tex]\int_{-x}^{\infty}{e^{-t^{2}}e^{t^{2}/2}}dt\qquad[/tex]
  10. Mar 6, 2008 #9
    what should [tex]\phi[/tex] equal?

    003dabb870f6a1fc0521a85000ea8090.png ?
  11. Aug 29, 2008 #10
    I think in this case, x should be greater than zero.
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