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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.

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The integral involved has a simple closed form. Shouldn't be too messy. What did you get as integral expression? – WimC Mar 3 '12 at 11:38
Just 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. – Davide Giraudo Mar 3 '12 at 11:39
up vote 2 down vote accepted

Did you get this?

$$ \mathcal{F} \chi_{[0,1]} (\xi)= \int_{-\infty}^\infty \chi_{[0,1]}(x) e^{-2\pi ix\xi}dx = \int_{[0,1]} e^{-2\pi ix \xi} dx = [\frac{e^{-2\pi ix \xi }}{-2\pi i\xi} ]_0^1 = \frac{e^{-2\pi i \xi }}{-2\pi i\xi} - \frac{1}{-2\pi i\xi} = \frac{1 - e^{-2\pi i \xi}}{2\pi i\xi}$$

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Thank 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? – Martin Mar 3 '12 at 15:10
@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, where it is not bounded or continuous. – robjohn Mar 3 '12 at 15:24
@Martin "Is it in this case?" Have you tried simplifying the answer a little? Hint: multiply and divide by $e^{i\pi\xi}$ and use Euler's identity to write the answer as the product of two continuous functions: $\exp$ and $\text{sinc}$. – Dilip Sarwate Mar 3 '12 at 15:25
@robjohn Thank you! – Rudy the Reindeer Mar 3 '12 at 15:28
Thank you robjohn. This is nice expression and we don't have to find the limit in the zero to show the continuity. – Martin Mar 3 '12 at 16:12

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