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November 16th, 2012, 05:58 PM   #1
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A challenging integral

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November 17th, 2012, 04:43 AM   #2
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Re: A challenging integral



This can be evaluated with the Beta:

Let

This gives:



This means that

So,

But,

So, we have

Now, using the identity: , we happily arrive at:

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November 17th, 2012, 04:56 AM   #3
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Re: A challenging integral

Really , I thought that can't be solved without using residues.

Nice Approach [color=#0000BF]galactus[/color].
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