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October 28th, 2016, 09:09 AM   #1
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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.
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October 28th, 2016, 09:56 AM   #2
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Can you post all your working?
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October 28th, 2016, 11:14 AM   #3
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Quote:
Originally Posted by nostradamus1915 View Post
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 $\displaystyle e^{2 \pi} -1$
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October 28th, 2016, 02:47 PM   #4
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The multiplicative factor of $\displaystyle 2\pi$ could be from the precise expression you are using for the Fourier transform.
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