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How do I calculate this integral?

  1. Dec 24, 2016 #1
    1. The problem statement, all variables and given/known data
    We're given the gaussian distribution: $$\rho(x) = Ae^{-\lambda(x-a)^2}$$ where A, a, and ##\lambda## are positive real constants. We use the normalization condition $$\int_{-\infty}^{\infty} Ae^{-\lambda(x-a)^2} \,dx = 1$$ to find: $$A = \sqrt \frac \lambda \pi$$ What I want to find is ##\langle x^2 \rangle##.

    2. Relevant equations
    $$\langle x^2 \rangle = \int_{-\infty}^{\infty} x^2Ae^{-\lambda(x-a)^2} \, dx$$
    Hence, I need to solve the RHS integral.

    3. The attempt at a solution
    I'm really not sure how to solve this integral. Converting to polar seems to produce a very nasty integral. Integration by parts also produces very nasty integrals. I don't think this function is even or odd so the symmetric integration interval won't simplify things. I'm not sure what to do.

    I would appreciate any hints, friends. :) Thank you very much.
     
  2. jcsd
  3. Dec 24, 2016 #2

    BvU

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    Helo BWest, :welcome:

    The hint is: integration by parts (in spite of the nastiness you experienced -- perhaps you can post your steps ?).
     
  4. Dec 24, 2016 #3

    vela

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    Try calculating ##\langle (x-a)^2 \rangle##.
     
  5. Dec 24, 2016 #4

    Ssnow

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    Integration by parts is a good idea, but you must choose well what integrate and what derive in order to simplify the expression ...
     
  6. Dec 24, 2016 #5
    Thank you everyone. I understand how to do this now. I truly appreciate all the responses!
     
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