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January 19th, 2010, 11:19 AM  #1 
Newbie Joined: Dec 2009 Posts: 6 Thanks: 0  Periodic Integration Proof/Substitution
Let f:R>R be a continuous function. Let T>0 be such that f(x+t) = f(x) for all x. Use an appropriate substitution to prove that for all real numbers a, . ..I was thinking: , but I can't think of a better way to explain it. Also, intuitively, I understand that f(x+T) = f(x), so for a periodic function F(x+T) = F(x), but again can't really translate that into a legible proof. 
January 19th, 2010, 01:33 PM  #2  
Global Moderator Joined: May 2007 Posts: 6,785 Thanks: 707  Re: Periodic Integration Proof/Substitution Quote:
First: there is an integer n and some b between 0 and T so that b=anT. Therefore by periodicity the integral from b to b+T is the same as the integral from a to a+T (the integrands are identical). Second: The integrand from 0 to b is the same as the integrand from T to b+T, so the integral from 0 to T is the same as the integral from b to b+T.  
January 19th, 2010, 03:07 PM  #3 
Newbie Joined: Dec 2009 Posts: 6 Thanks: 0  Re: Periodic Integration Proof/Substitution
Very helpful. Thank you for your insight.

January 19th, 2010, 03:50 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,810 Thanks: 2153 
Each original T should be t. 
January 19th, 2010, 11:08 PM  #5 
Newbie Joined: Nov 2009 Posts: 1 Thanks: 0  Re: Periodic Integration Proof/Substitution
...Hey! You're in Dr. J's class aren't you? Thank goodness someone posted this up, I was having major problems with this too. 

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integration, periodic, proof or substitution 
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