Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

1 Answer 1

up vote 2 down vote accepted

Did you get this?

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

share|improve this answer
    
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
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, 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! –  Matt N. 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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.