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Thread: integral binomial differential

  1. #1
    Junior Member
    Mar 2011

    integral binomial differential

    Hiow to evaluate
    $\displaystyle \int_0^1{dx \frac{d^{2l-1}}{dx^{2l-1}} \[ (x^2-1)^{2l} \]$
    Using the binomial theorem
    $\displaystyle \int_0^1{dx \frac{d^{2l-1}}{dx^{2l-1}} \[ \sum_{k=0}^{2l} \frac{(2l)!}{k!(2l-k)!} x^{4l-2k} (-1)^k \]$

    l is a non negative integer.

    I don't know how to proceed. Is there another way?
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  2. #2
    Junior Member
    Oct 2011

    Re: integral binomial differential

    Move the differential operator into the sum. You'll need to evaluate

    $\displaystyle \frac{d^{2l-1}}{dx^{2l-1}} x^{4l-2k}$

    which you can easily do by induction. Some terms of the sum may vanish while doing this, I didn't check. Finally, swap the sum and integral. Evaluate the integral. And you should be done.
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