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Gaussian Integral

  1. Jul 23, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider the gaussian distribution shown below

    [tex] \rho (x) = Ae^{-\lambda (x-a)^2 [/tex]

    where A, a, and [itex] \lambda [/itex] are positive real constants. Use [itex] \int^{-\infty}_{+\infty} \rho (x) \,dx = 1 [/itex] to determine A. (Look up any integrals you need)

    2. Relevant equations
    Given in question above

    3. The attempt at a solution
    My plan was to integrate the probability density set it equal to one and then solve for A. The problem is I'm getting stuck on the integration. I started by pulling the constants out of the integral and doing the substitution [itex] u=x-a [/itex] that left me with
    [tex] Ae^{-\lambda} \int^{+\infty}_{-\infty} e^{u^2}\,du [/tex]
    It's been a while since calc II and I can't figure out how to do this one (even though it looks so simple). I also tried looking it up in a integral table but couldn't find it. Any help would be appreciated.
     
  2. jcsd
  3. Jul 23, 2009 #2

    Cyosis

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    You have made an error.

    [tex]
    e^{-\lambda u^2} \neq e^{-\lambda}e^u^2=e^{-\lambda+u^2}
    [/tex]
     
  4. Jul 23, 2009 #3

    tiny-tim

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    Hi DukeLuke! :wink:

    You need the erf(x) function … see http://en.wikipedia.org/wiki/Error_function :smile:

    (btw, there is a way to integrate ∫e-u2du: it's √(∫e-u2du)∫e-v2dv), then change to polar coordinates :wink:)
     
  5. Jul 23, 2009 #4
    Thanks, man am I getting rusty over the summer

    I looked at it but I'm lost on how to use it solve this problem. Could you help me get started?
     
  6. Jul 23, 2009 #5

    D H

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    Have you tried this bit of advise? You even know the relevant keywords (hint: use the title of this thread). Google is your friend.
     
  7. Jul 23, 2009 #6

    Cyosis

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  8. Jul 23, 2009 #7
    [tex] \int_{-\infty}^{\infty} e^{-(x+b)^2/c^2}\,dx=|c| \sqrt{\pi} [/tex]

    Thanks, using the integral above from Wikipedia [itex] c = \frac{1}{\sqrt{\lambda}} [/itex]. From there I get [itex] A = \frac{\sqrt{\lambda}}{\sqrt{\pi}} [/itex].
     
  9. Jul 24, 2009 #8
    Looks correct, studying griffiths' quantum mechanics I see :D
     
  10. Jul 24, 2009 #9
    Yep, thought I would get a head start before the fall semester begins.
     
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