What is the fourier transform of $xf'(x)$?

Question

What is the Fourier transform of $xf'(x)$? I have tried using integration by parts, and I only end up with a long messy equation that gives no results.

Answer 1

Use properties of Fourier transform: assuming $$\mathcal{F}\{f(x)\}=F(w),$$

$$\mathcal{F}\{\frac{d}{dx}f(x)\}=jwF(w)$$ $$\mathcal{F}\{-jxg(x)\}=\frac{d}{dw}G(w)$$

where $g(x)=f'(x)$ in here. Therefore, $$\mathcal{F}\{xf'(x)\}=j\frac{d}{dw}\left(jwF(w)\right)=-(F(w)+w\frac{d}{dw}F(w))$$