Let $f(x) = \begin{cases}e^{-x} & ,0
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?