# Fourier transform of $f(x)=x$ if $0<x\leq 1$ and $f(x)=0$ otherwise

What is the Fourier transform of the function defined by $f(x)=x$ on $[0,1]$ and $f(x)=0$ otherwise, i.e., $\hat f(\xi) = \int_\mathbb{R} { e^{-iu\xi} f(u) du }$?

Is there a closed-form? Else, how can I compute it numerically with fft? Fourier/fft are new concepts to me and I am not clear how to use fft.

Thanks

• The FFT algorithm is for calculating the DFT, which corresponds to the continuous FT of a periodic function. Your function is not periodic, so this does not help you. Have you tried integrating by parts? (For caution, you should really only integrate by parts on $(\delta_0,1-\delta_1)$ and then send $\delta_0,\delta_1 \to 0$ at the end.) – Ian Aug 5 '15 at 14:49

Now there's the more subtle question you asked earlier, about whether it is possible to determine this numerically using the DFT. Again, like Ian said, the DFT is for periodic functions. But still if you are careful you can use the DFT for this case. What you have to do is choose some cutoff $R$, and look at the Fourier series of the function which is equal to $f$ on $[-R,R]$, and is periodic otherwise. This can be approximated using the DFT.
• I get the same but the $-1$ where a $1/\xi^2$ is missing in the last row. Could you please help me interpret the fact that we have this explosion around 0 (because of the $1/\xi$ factors)? I would have thought that an L2 function would have an L2 Fourier transform. I seem to be wrong. – user252766 Aug 5 '15 at 17:09
• You dropped a parenthesis or something. What you said can't be right; as pointed out the FT of an $L^2$ function must be in $L^2$, or note that the FT of an $L^1$ function must be continuous. Seems to me that the actual FT is $\frac{e^{i\xi}}{i\xi}+\frac1{\xi^2}(e^{i\xi}-1)$. Look at the first few terms in the Taylor series for the exponential and you see that the terms that blow cancel; this FT is continuous, as it should be. – David C. Ullrich Aug 5 '15 at 17:51