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We are supposed to be able to show that the Fourier transform of $-ixf(x)$ is equal to $c'(w)$ without using knowledge of real analysis.

Trying to transform the above yields the following integral: $\frac{-i}{2\pi}\int_{-\infty}^{\infty} x f(x) e^{-i w x} dx$

I've never seen such an integral before, and assuming f(x) has 1 or more x's, and not knowing whether its component(s) can be merged with something else in the integral we can't use integration by parts as I know it with 2 components. how would one go on solving this type of problems?

Update: $c$ is the fourier transform of $f$

jibo
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1 Answers1

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If you (all computations here are strictly formal - this is not intended to be rigorous) differentiate $$c(w) = \frac{1}{2\pi} \int_{-\infty}^\infty f(x) e^{-iwx}\, dx$$ you get $$c'(w) = \frac{-i}{2\pi} \int_{-\infty}^\infty x f(x) e^{-i w x} \, dx.$$ Compare this to the formula you provided.

Umberto P.
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  • I think your convention is different: $c$ itself will have a minus sign in the exponent, which is where there is a discrepancy of $-1$ between the two formulae. – Ian Oct 22 '15 at 16:43
  • Yeah, I was guessing about $c$ based on the formula provided. I will change it. – Umberto P. Oct 22 '15 at 16:55
  • I believe the lack of dx at the end of the first integral is unintentional? Ok so we are basically partially differentiating c w.r.t. omega $(w)$, and the function $f(x)$ appears to not be a function of $w$. @Ian: Thx Ian that is correct, the minus sign should be there. – jibo Oct 23 '15 at 08:08