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: 17,440 Thanks: 1312 
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: 6,254 Thanks: 507 
The multiplicative factor of $\displaystyle 2\pi$ could be from the precise expression you are using for the Fourier transform.


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