1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Homework Helper

    You have made an error.

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


    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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


    User Avatar
    Homework Helper

  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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook