integral binomial differential

**hurz** · December 13th 2011, 04:33 AM

Hiow to evaluate
$\int_0^1{dx \frac{d^{2l-1}}{dx^{2l-1}} \[ (x^2-1)^{2l} \]$
Using the binomial theorem
$\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?
Thanks.

**jens** · December 14th 2011, 02:50 AM

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

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

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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