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We know that Fourier Transform of $e^{ixt}$, where $x$ is a real parameter, $t\in \mathbb R$ is

$$\int_{-\infty}^{+\infty} e^{ixt} e^{-i \omega t} dt=\int_{-\infty}^{+\infty} e^{ixt-i \omega t} dt=\int_{-\infty}^{+\infty} e^{i(x- \omega )t} dt=2 \pi \delta(x- \omega)$$ The result is the same if you consider the following integral? $$\int_{-\pi}^{+\pi} e^{ixt} e^{-i \omega t} dt$$

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$$ \color{#00f}{\large\mbox{No}}.\quad\mbox{The last one is equal to}\quad \left\lbrace\begin{array}{lcl} 2\,{\sin\left(\left[x - \omega\right]\pi\right) \over x - \omega} & \mbox{if} & x \not= \omega \\[3mm] 2\pi & \mbox{if} & x = \omega \end{array}\right. $$

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  • $\begingroup$ costant "i" missing? $\endgroup$ – Mark Jul 22 '14 at 6:54
  • $\begingroup$ @Mark There is nothing missing. It's fine. The result is a $\large\tt\mbox{real number}$ since $\displaystyle\large\int_{-\pi}^{\pi}\exp\left({\rm i}\left[x - \omega\right]t\right)\,{\rm d}t = \int_{-\pi}^{\pi}\cos\left(\left[x - \omega\right]t\right)\,{\rm d}t$. Thanks. $\endgroup$ – Felix Marin Jul 22 '14 at 19:54
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No, for $x\neq \omega$ you have: $$\int_{-\pi}^{+\pi} e^{ixt} e^{-i \omega t} dt=2i\frac{e^{i\pi(x-\omega)}-e^{-i \pi(x-\omega)}}{2i(x-\omega)} =2i\frac{\sin(\pi (x-\omega))}{x-\omega} \neq 2 \pi \delta(x- \omega)$$

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  • $\begingroup$ You have an extra $\Large{\rm i}$ before the symbol $\large\not=$. $\endgroup$ – Felix Marin Jul 22 '14 at 19:58

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