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
2
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analysis
fourier-analysis
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0The integral involved has a simple closed form. Shouldn't be too messy. What did you get as integral expression? – 2012-03-03
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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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0Thank you Matt. Exactly that with another normalization (without factor 2). But I know that the result of a Fourier transform should be always continuous and bounded function. Is it in this case? – 2012-03-03
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1@Martin: $\displaystyle\frac{1-e^{-2\pi i\xi}}{2\pi i\xi}=\operatorname{sinc}(\pi\xi)e^{-\pi i\xi}$, so it is continuous and bounded. The Fourier Transform of any $L^1$ function is bounded and continuous. However, the Fourier Transform is often extended to [generalized functions](http://en.wikipedia.org/wiki/Generalized_function), where it is not bounded or continuous. – 2012-03-03
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0Thank you robjohn. This is nice expression and we don't have to find the limit in the zero to show the continuity. – 2012-03-03
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0@user929304 It's not clear to me what you mean. For example, "not differentiable in the usual sense" -- well apart from the points $c,d$ its derivative is 0, or am I missing something? And as for the FT of $\chi_{[c,d]}$: It looks to me like it's decaying. The expression is term one minus term two where each term is of the form one over something exponential times linear. But I might be missing something, it's been a while. – 2015-11-03
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0@RudytheReindeer I think you are right about the fourier transform of f here decaying to 0, because $\chi_{[c,d]} \in L^1$ thus by Riemann-Lebesgue lemma its fourier transform must decay to 0 at infinity. So I guess I need to find my mistake now in the differentiability argument. – 2015-11-03
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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