January 8th, 2012, 11:29 AM  #1 
Member Joined: Jan 2012 Posts: 86 Thanks: 0  integrals I evaluated (which can be done in numerous ways) and then tried to argue that the answer is valid for imaginary by using the uniqueness theorem and Morera's theorem from complex analysis. Unfortunately I got nowhere with that. What's another way to evaluate these integrals? 
January 8th, 2012, 02:56 PM  #2 
Senior Member Joined: Dec 2011 From: Argentina Posts: 216 Thanks: 0  Re: integrals
Ok I'd go with: (same for ) And consider like you an arbitrary complex number Solving this with Wolfram I get that the expression is: Now plug in z = to get And taking and we get that the integrals are I do not know how far this is correct but do it yourself and check if there is something wrong. The case and gives the correct Fresnel integral values: 
January 8th, 2012, 03:37 PM  #3 
Senior Member Joined: Dec 2011 From: Argentina Posts: 216 Thanks: 0  Re: integrals
I'll illustrate how to solve the Laplace Transform (so I don't leave holes in the solution): Change the variable Express the polinomials as: Change the variable in the first integral Change the variable One last variable change 
January 8th, 2012, 04:14 PM  #4 
Senior Member Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0  Re: integrals Let us first rewrite with S and C the Fresnel integrals. Yes, same result as Weiler!!! 
January 8th, 2012, 05:21 PM  #5 
Member Joined: Jan 2012 Posts: 86 Thanks: 0  Re: integrals
Thanks. I've posted these on other forums, but no one could offer much help. Weiler You evaluated that Laplace transform as z were a real parameter. Is that justified? 
January 8th, 2012, 05:37 PM  #6 
Senior Member Joined: Dec 2011 From: Argentina Posts: 216 Thanks: 0  Re: integrals
What steps are you talking about? I solved the integral pretty much with integral theorems and considering to be complex. Btw, nice one wnvl, trigonometry always helps. You used , right? What I don't understand is the notation . It means you changed variables? 
January 8th, 2012, 05:53 PM  #7 
Senior Member Joined: Dec 2011 From: Argentina Posts: 216 Thanks: 0  Re: integrals
wnvl, just realized you solution and mine are the same only you worked in the real line and I worked in the complex plane. (See how if you take Re and Im in my steps it is pretty much the same.)

January 8th, 2012, 07:02 PM  #8 
Member Joined: Jan 2012 Posts: 86 Thanks: 0  Re: integrals
I'm not sure what the notation means and I'm assume he didn't mean for those two cosine functions to become sine functions when he rewrote the integral, but if you integrate over instead of simple substitutions at the end lead to the answer.

January 9th, 2012, 03:08 AM  #9  
Senior Member Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0  Re: integrals Quote:
Quote:
 
January 9th, 2012, 03:31 AM  #10 
Senior Member Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0  Re: integrals
I first tried to solve it using complex contour integration, but that didn't work without doing first the goniometric simplifications.


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