My qustion is about the Fourier transform of the characteristic function $\chi_{[0,1]}$. How can I find what it is? The problem is I got something really messy, so I think I didn't get it right.
Fourier transform of the characteristic function
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analysis
fourier-analysis
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0Just apply the definition: you have to find $\int_{\mathbb R}e^{itx}\chi 1_{[0,1]}(t)dt$ so it reduces to $\int_0^1e^{itx}dt$. Now yo just have to compute this integral. – 2012-03-03
1 Answers
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Did you get this?
$ \mathcal{F} \chi_{[0,1]} (\xi)= \int_{-\infty}^\infty \chi_{[0,1]}(x) e^{-2\pi ix\xi}dx = \int_{[0,1]} e^{-2\pi ix \xi} dx = \left[ \frac{e^{-2\pi ix \xi }}{-2\pi i\xi} \right]_0^1 = \frac{e^{-2\pi i \xi }}{-2\pi i\xi} - \frac{1}{-2\pi i\xi} = \frac{1 - e^{-2\pi i \xi}}{2\pi i\xi}$
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0@user929304 I see. I don't know if it makes sense, this is beyond my modest knowledge of Fourier transforms. It's not very useful but: Surely the decaying properties of $\mathcal F (\chi_{[c,d]}$ and $\mathcal F (\chi_{[0,1]}$ are almost identical? – 2015-11-03