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Let $f(x) = \begin{cases}e^{-x} & ,0<x<1\\0 & ,\text{Otherwise}\end{cases}$

I'm trying to calculate the fourier transform of $xf(x)$, by using the fact that $xf(x) = -\frac{d}{da}f(a x)\bigg|_{a=1}$ and $\mathscr{F}\{f(a x)\} = \frac{1}{a}\hat{f}(\frac{\omega}{a}), \quad a>0$.

The fourier transform should be: $$\mathscr{F}\left\{-\frac{d}{da}f(a x)\bigg|_{a=1}\right\} = - \frac{d}{da}\left(\frac{1}{a}\hat{f}(\frac{\omega}{a})\right)\bigg|_{a=1} $$

This gives the answer: $$\frac{-e+(1+i\omega+\omega^2) \cos(\omega )+i \ (1+i\omega+\omega^2) \sin(\omega )}{e \sqrt{2 \pi } \ (i+\omega )^2}$$

But the correct answer is: $$\frac{-e+(2-i \omega) \cos(\omega)+(2 i+\omega) \sin(\omega)}{e \sqrt{2 \pi \ } (i+\omega)^2}$$

The answer is close to the right one, the difference is $\frac{e^{i\omega-1}}{\sqrt{2 \pi}}$, but why it's not correct?

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The reason is that your derivative $x f(x) = - \frac{d}{da} f(ax)|_{a=1}$ is not quite valid because the function you are differentiating is discontinuous. If you interpret it in the distribution sense, you get an additional term of the form $\frac1e \delta_1(x)$, where $\delta_1$ is the Dirac distribution with mass at $x=1$. The Fourier transform of this extra term is exactly the difference between your solution and the correct one.

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  • $\begingroup$ Thanks, this makes perfect sense. $\endgroup$
    – 0912
    Commented Oct 6, 2012 at 18:22

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