October 28th, 2016, 08:09 AM  #1 
Newbie Joined: Oct 2016 From: italy Posts: 9 Thanks: 0  Fourier Integral With Parseval
Hi guys, I'm losing my mind on this integral. I got the following function: $\displaystyle g(t):t \in [\pi,\pi[> \begin{cases} e^t & \text{if } \pi \le t < 0 \\ e^t & \text{if } 0 \le t < \pi \end{cases}$ And $\displaystyle G(t):t\in R > \begin{cases} g(t) & \text{if } \pi \le t < \pi \\ 0 & \text{if } else \end{cases}$ And I've to solve the following integral: $\displaystyle \int_{\infty}^{+\infty} F[G(t)](w)^2 dw $ This should be kinda easy, I apply Parseval: $\displaystyle \int_{\infty}^{+\infty} F[G(t)](w)^2 dw = \int_{\pi}^{0} G(t)^2 dt + \int_{0}^{\pi} G(t)^2 dt$ I get from all of this $\displaystyle 1e^{2 \pi}$ but the result should be (standing to the book) $\displaystyle 2 \pi (e^{2 \pi}1)$ Ideas? p.s.: F stands for Fourier Transform Last edited by nostradamus1915; October 28th, 2016 at 08:32 AM. 
October 28th, 2016, 08:56 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,844 Thanks: 1566 
Can you post all your working?

October 28th, 2016, 10:14 AM  #3  
Newbie Joined: Oct 2016 From: italy Posts: 9 Thanks: 0  Quote:
$\displaystyle g(t):t \in [\pi,\pi[> \begin{cases} e^{t} & \text{if } \pi \le t < 0 \\ e^t & \text{if } 0 \le t < \pi \end{cases}$ And $\displaystyle G(t)^2$ should be: $\displaystyle g(t):t \in [\pi,\pi[> \begin{cases} e^{2t} & \text{if } \pi \le t < 0 \\ e^{2t} & \text{if } 0 \le t < \pi \end{cases}$ The integrals are: https://www.wolframalpha.com/input/?...)+between(0,pi) The result I find is actually $\displaystyle e^{2 \pi} 1$  
October 28th, 2016, 01:47 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,495 Thanks: 578 
The multiplicative factor of $\displaystyle 2\pi$ could be from the precise expression you are using for the Fourier transform.


Tags 
integral 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Double integral, repeated integral and the FTC  Jhenrique  Calculus  5  June 30th, 2015 03:45 PM 
a new discovery in integral calculus??for integral pros only  gen_shao  Calculus  2  July 31st, 2013 09:54 PM 
integral of double integral in a region E  maximus101  Calculus  0  March 4th, 2011 01:31 AM 
Prove Lower Integral <= 0 <= Upper Integral  xsw001  Real Analysis  1  October 29th, 2010 07:27 PM 
integral of double integral in a region E  maximus101  Algebra  0  December 31st, 1969 04:00 PM 