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Where F is the fourier transform, how can you show that $$\mathcal F(x\cdot f(x)) = −i \frac{d\mathcal F}{dw}.$$

I understand that you are meant to apply the inverse transform to the left hand side, but I just can not seem to get it to work.

Many thanks, Rachel.

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Actually, for a simple case (when f(x) is a continues function and xf(x) is absolutly integrable on the real line), you can just insert the differentiation inside the integral, that is $$ \frac{d}{dy} \int_{-\infty}^{\infty} f(x)e^{-ixy}dx = \int_{-\infty}^{\infty} \frac{d}{dy} (f(x)e^{-ixy})dx = \int_{-\infty}^{\infty} -ixf(x)e^{-ixy} dx = -i \int_{-\infty}^{\infty} xf(x)e^{-ixy} dx $$ since it can be shown that both $$ f(x)e^{-ixy} $$ and $$ -ixf(x)e^{-ixy} $$ are continues, and the integral $$ \int_{-\infty}^{\infty} -ixf(x)e^{-ixy} dx $$ converges uniformly (for every y) .

for functions that are not continues, other techniques are required, like Lebesgue's dominated convergence theorem.

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i think this is what you're asking...
integrating by parts we have (assuming $f\to 0$ at $\pm\infty$) $$\mathcal{F}(-if'(x))(y)=\int -if'(x)e^{-ixy}dx=\int f(x)ye^{-ixy}dx=y\mathcal{F}(f(x))(y)$$ so that the fourier transform of $f'$ is $iy$ times the fourier transform of $f$

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