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December 3rd, 2012, 06:21 AM   #1
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A Challenging Integral

Prove that:


is Euler's Constant.
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December 3rd, 2012, 06:25 AM   #2
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Re: A Challenging Integral

Quote:
Originally Posted by Etyucan
Prove that:


is Euler's Constant.
You can post it here if you wish .

viewtopic.php?f=15&t=36221
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December 3rd, 2012, 11:51 PM   #3
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Re: A Challenging Integral

Quote:
Originally Posted by zaidalyafey
Quote:
Originally Posted by Etyucan
Prove that:


is Euler's Constant.
You can post it here if you wish .

viewtopic.php?f=15&t=36221
No, thanks.
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December 4th, 2012, 05:55 AM   #4
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Re: A Challenging Integral

Quote:
Originally Posted by Etyucan
Prove that:


is Euler's Constant.
Without parentheses this is ambiguous. Do you mean or ? And, of course, there is a problem at x= 0. ln(0) is not defined so you really mean the limit as x goes to 0. Ignoring that for the moment, and assuming you meant the first of the two integrals, an obvious first step is to let u= ln(x) so that du= dx/x and . Then the integral becomes .

Now, what do you know about Euler's constant? In particular is there an integral representation for it?
http://en.wikipedia.org/wiki/Euler%E2%8 ... i_constant
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December 4th, 2012, 07:14 AM   #5
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Re: A Challenging Integral

I think you made a mistake in the substitution .
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December 4th, 2012, 10:48 AM   #6
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Re: A Challenging Integral

We can start with the identity:



Now, differentiate w.r.t a. Note the digamma functions involved due to the derivative of Gamma.



Now, let and this simplifies to:



But,


So, we get:

The identity at the top I used at the start would be a good one to prove as well. I have to go now, though.
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December 4th, 2012, 11:59 PM   #7
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Re: A Challenging Integral

Quote:
Originally Posted by galactus
We can start with the identity:



Now, differentiate w.r.t a. Note the digamma functions involved due to the derivative of Gamma.



Now, let and this simplifies to:



But,


So, we get:

The identity at the top I used at the start would be a good one to prove as well. I have to go now, though.
Thank You!!!
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December 5th, 2012, 01:29 AM   #8
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Re: A Challenging Integral

I would know it involves the digamma or psi function , but I don't quite get the differentiating part.
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December 5th, 2012, 05:56 AM   #9
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Re: A Challenging Integral

The derivative of Gamma is

So, it involves that. Those solutions may be able to be reduced saomewhat if we look at them.
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December 5th, 2012, 06:18 AM   #10
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Re: A Challenging Integral

I found out a way to calculate the integral





Now, we can use Euler's Reflection Formula to express sin in terms of gamma function





and finally from the duplication rule,

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