I am not able to get the Fourier Transform of $xf(x)$ if $<f(x)>$ is the Fourier transform of $f(x)$ .
BTW i tried using convolution theorem but didn't work out .
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1$\begingroup$ Vedananda, could you perhaps expand on that a bit? It's not completely clear what you question is, at least not to me. $\endgroup$– Terry BollingerCommented Apr 22, 2012 at 13:10
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2$\begingroup$ Are you asking how to find FT of xf(x) given FT of f(x) is known? Its been while I read Fourier transform but I think the derivation can be found in any book. $\endgroup$– user29679Commented Apr 22, 2012 at 13:18
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$\begingroup$ Yes exactly Anuragsn , if i know what FT of $f(x)$ is , how do i find FT of $xf(x)$. :) $\endgroup$– TheoremCommented Apr 22, 2012 at 13:25
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1$\begingroup$ Oh... I get it now. But it is already answered... And you should post such stuff to mathematical forums... $\endgroup$– PygmalionCommented Apr 22, 2012 at 13:42
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1 Answer
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If the Fourier-transform of $f(x)$ is $$FT[f(x)] \equiv f(k) = \int_{-\infty}^{\infty} f(x) e^{- i k x} dx$$ then $$FT[xf(x)] = \int_{-\infty}^{\infty} x f(x) e^{- i k x} dx $$ $$ = \int_{-\infty}^{\infty} i \frac{\partial}{\partial k} \Big[ f(x) e^{-i k x} \Big] dx = i \frac{\partial}{\partial k} \int_{-\infty}^{\infty} f(x) e^{-i k x} dx$$ which means $$FT[xf(x)] = i \frac{\partial f(k)}{\partial k} $$
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1$\begingroup$ I would add that this is in keeping with the general principle that polynomials transform to differential operators and vice-versa. $\endgroup$– NealCommented Apr 22, 2012 at 18:58
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1$\begingroup$ is there a typo on the second line? on the first line it states that the transform of f(k) is the integration of f(x) and "k" appears in the exponent, but on the second line "k" still appears in the denominator but the transform was applied to "xf(x)". $\endgroup$– quantifCommented Apr 26, 2020 at 14:42
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$\begingroup$ I know it is possible to proceed like the previous answer if $f \in \mathbb{L}^1$ or $f \in \mathcal{S}$ (Schwartz space). Is it possible to apply the same result to other kind of functions, for example $f(x)=\frac{x}{1+x^2}$? $\endgroup$ Commented May 19, 2020 at 19:22