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Homework Help: (x^4)e^(-x^2/a^2) integral

  1. May 20, 2010 #1
    1. The problem statement, all variables and given/known data

    There should be a gamma coefficient. I've put it in using latex but it doesn't seem to be showing in the preview...

    2. Relevant equations


    3. The attempt at a solution

    I know that the answer is (3/4)*gamma*a^4 (I can't seem to get Latex to work any more... sorry :/), my lecturer did all the steps leading up to the integral on the board then went straight to the answer so I'm guessing it's a definite integral.
    I know that without the constant coefficient and with x^2 instead of x^4 the answer is (1/2)(pi*a^6)^0.5, it's given on a formula sheet. I have no idea what to do for x^4 though...
    Both integrals are between + and - infinity.
    Any help would be greatly appreciated.

    Edit: I see the gamma is showing now. If you would like me to put the other couple of solutions in then let me know and I'll have another go
    Last edited: May 20, 2010
  2. jcsd
  3. May 20, 2010 #2


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    Welcome to PF!

    Hello duckie! Welcome to PF! :smile:

    (have a pi: π and a gamma: γ and an integral: ∫ and try using the X2 tag just above the Reply box :wink:)
    (there's been problems with previewing latex since the last server update … it doesn't affect the final result, and you can eliminate it by clicking "Refresh" on your browser with every preview :wink:)

    Essentially, you're trying to integrate x4e-x2

    if you keep integrating by parts, you should get down to ∫-∞ e-x2 dx, which you presumably know how to do. :smile:
  4. May 20, 2010 #3
    Re: Welcome to PF!

    Thanks :smile:

    I had a go at it this way earlier by setting u = x4 and dv = e-x2 but that ended pretty quickly as v = ∫-∞ e-x2 dx = π1/2... right?
  5. May 20, 2010 #4


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    No that is wrong. You can't enter the boundary values when you go from dv to v. The product of uv in the partial integration needs to be evaluated as a whole.

    Hint: [tex]x^4 e^{-x^2}=x^3 \left(xe^{-x^2}\right)[/tex]
  6. May 20, 2010 #5


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    Hello duckie! :smile:

    Think laterally …

    try xe-x2 :wink:
  7. May 20, 2010 #6
    Oh, do I use the substitution u = x2?

    Edit: Just did it this way, I have an answer now! Which is more that I've had for a long while... though apparently it's wrong. On integrating x4e-x2 I got 3π1/2/4.
    Though when I put the 1/a2 term in the exponential I ended up with a41/2/4. Which is logical as far as I can see. Though in the original question there is an additional coefficient of 1/a which will mean my result cancelling to an a3 term...
    Is the answer I was given wrong or have I lost a factor of a somewhere?
    Edit 2: Don't worry, I see where I lost it :smile:

    Thanks for all the help!
    Last edited: May 20, 2010
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