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Tricky Integral

  1. Mar 8, 2004 #1
    Given

    [tex]P(x)= \frac{1}{\sigma \sqrt{2\pi}} e ^ \frac { -(x - \mu )^2}{2 \sigma ^2 }[/tex]

    This is of course the normal distribution curve. When [tex] \mu = 0[/tex] and [tex] \sigma = 1[/tex] I can integrate this from minus infinity to positive infinity no problem using polar coordinates and a bit of multivariable calculus. The question I have, is, is it at all possible to do this if one leaves [tex] \mu[/tex] and [tex]\sigma[/tex] in as generic parameters? I would think so, but I'm not sure. No need to give a worked through example, just, is it possible at all to fit it to some form? Or is it just that with those parameters set to 0 and 1 that this is an integrable function?
     
  2. jcsd
  3. Mar 8, 2004 #2
  4. Mar 9, 2004 #3

    HallsofIvy

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    Certainly. Just do a little creative substitution:
    Let [tex]u= \frac{x-\mu}{\sqrt(2)\sigma^2}[/tex].

    Then [tex] du= \frac{1}{\sqrt(2)\sigma^2}dx[/tex]

    [tex]P(x)= \frac{1}{\sigma \sqrt{2\pi}} e ^ \frac { -(x - \mu )^2}{2 \sigma ^2 }[/tex]

    becomes [tex]P(u)= \frac{1}{\sqrt{\pi}} e ^{-u^2}[/tex]

    which you can integrate from negative to positive infinity as usual.
    (That's why you probably have only seen that case.)
     
  5. Mar 9, 2004 #4
  6. Mar 16, 2004 #5
    can it be integrated from a number to anohter... say -2 and 2?
     
  7. Mar 17, 2004 #6
    Numerically, yes.
    Analytically, no (unless you count functions like Erf(x), which is cheating in my view).
     
  8. Mar 17, 2004 #7
    Well wait a minute. Wouldn't you just change the bounds on the integral with respect to radius from 0 to infinity to 0 to 2? The theta integral would remain the same, and thus you'd get the analytical answer?
     
  9. Mar 17, 2004 #8
    Maybe I misunderstood allergic's question. I thought that he wanted to know

    [tex]\int_0^2 \frac{1}{\sqrt{\pi}} e ^{-u^2} \,{\rm d}u[/tex]

    I don't think there is an analytical expression for that. Numerically it's easy (0.497661).
     
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