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Gaussian Integral

  1. May 3, 2009 #1
    EDIT: meant to post this is the math forums, if you can remove this I'm going to switch it over

    1. The problem statement, all variables and given/known data
    Solve:

    In = [tex] \int_{0}^{\infty} x^n e^{-\lambda x^2} dx [/tex]


    2. Relevant equations



    3. The attempt at a solution
    So my teacher gave a few hints regarding this. She first said to evaluate when n = 0, then consider the cases when n = even and n = odd, comparing the even cases to the p-th derivative of Io.

    For the Io case, I evaluated it and obtained [tex] I_o = \frac{1}{2} \sqrt{\frac{\pi}{\lambda}} [/tex]

    Now, for the "p-th" derivative of Io, i got

    [tex] \frac{d^p}{d \lambda^2} I_o = \frac{\prod_{p=1}^p (1 - 2p)}{2^{p+1}} \sqrt{\pi} \lambda^{-\frac{(2p + 1)}{2}} [/tex]

    I don't see how this related to n = 2p (even case) where

    I2p = [tex] \int_0^\infty x^{2p} e^{- \lambda x^2} dx [/tex]

    And even when I do figure this out, does this all combine into one answer, or is it kind of like a piecewise answer?

    Any help with what to do with the even/odd cases would be greatly appreciated

    Thanks
     
  2. jcsd
  3. May 5, 2009 #2
    Hello, I will give some hints here, and probably I will rearrange this quick reply in 24 hours.

    \int_0^\infty e^{-ax^2}dx = \frac{1}{2} \sqrt{\frac{\pi}{a}}

    \int_0^\infty x e^{-ax^2}dx = \frac{1}{2a}

    \int_0^\infty x^n e^{-ax^2}dx = \frac{(n-1)!}{2(2a)^{n/2}} \sqrt{\frac{\pi}{a}}

    \int_0^\infty x^{2n+1} e^{-ax^2}dx = \frac{n!}{2 a^{n+1}}

    \int_0^\infty x^{2n} e^{-ax^2}dx = \frac{(2n-1)!}{2^{2n+1} a^n} \sqrt{\frac{\pi}{a}}
     
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