My Math Forum Integral

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 October 28th, 2016, 09:09 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 $\displaystyle 1-e^{2 \pi}$ 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 09:32 AM.
 October 28th, 2016, 09:56 AM #2 Global Moderator   Joined: Dec 2006 Posts: 18,427 Thanks: 1462 Can you post all your working?
October 28th, 2016, 11:14 AM   #3
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Quote:
 Originally Posted by nostradamus1915 Hi guys, I'm losing my mind on this integral. I got the following function: $\displaystyle g(t):t \in [-\pi,\pi[--> \begin{cases} -e^t & \text{if } -\pi \le t < 0 \\ e^t & \text{if } 0 \le t < \pi \end{cases}$
I did an error there, the function is:
$\displaystyle g(t):t \in [-\pi,\pi[--> \begin{cases} -e^{-t} & \text{if } -\pi \le t < 0 \\ e^t & \text{if } 0 \le t < \pi \end{cases}$
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, 02:47 PM #4 Global Moderator   Joined: May 2007 Posts: 6,416 Thanks: 558 The multiplicative factor of $\displaystyle 2\pi$ could be from the precise expression you are using for the Fourier transform. Thanks from nostradamus1915

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