December 3rd, 2012, 06:21 AM  #1 
Senior Member Joined: Oct 2011 From: India ???? Posts: 224 Thanks: 0  A Challenging Integral
Prove that: is Euler's Constant. 
December 3rd, 2012, 06:25 AM  #2  
Math Team Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions  Re: A Challenging Integral Quote:
viewtopic.php?f=15&t=36221  
December 3rd, 2012, 11:51 PM  #3  
Senior Member Joined: Oct 2011 From: India ???? Posts: 224 Thanks: 0  Re: A Challenging Integral Quote:
 
December 4th, 2012, 05:55 AM  #4  
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: A Challenging Integral Quote:
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  
December 4th, 2012, 07:14 AM  #5 
Math Team Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions  Re: A Challenging Integral
I think you made a mistake in the substitution .

December 4th, 2012, 10:48 AM  #6 
Senior Member Joined: May 2011 Posts: 501 Thanks: 6  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. 
December 4th, 2012, 11:59 PM  #7  
Senior Member Joined: Oct 2011 From: India ???? Posts: 224 Thanks: 0  Re: A Challenging Integral Quote:
 
December 5th, 2012, 01:29 AM  #8 
Math Team Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions  Re: A Challenging Integral
I would know it involves the digamma or psi function , but I don't quite get the differentiating part. 
December 5th, 2012, 05:56 AM  #9 
Senior Member Joined: May 2011 Posts: 501 Thanks: 6  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. 
December 5th, 2012, 06:18 AM  #10 
Senior Member Joined: Oct 2011 From: India ???? Posts: 224 Thanks: 0  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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