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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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    $\begingroup$ 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 Bollinger Apr 22 '12 at 13:10
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    $\begingroup$ 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. $\endgroup$ – user29679 Apr 22 '12 at 13:18
  • $\begingroup$ Yes exactly Anuragsn , if i know what FT of $f(x)$ is , how do i find FT of $xf(x)$. :) $\endgroup$ – Theorem Apr 22 '12 at 13:25
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    $\begingroup$ Oh... I get it now. But it is already answered... And you should post such stuff to mathematical forums... $\endgroup$ – Pygmalion Apr 22 '12 at 13:42
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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 $$FT[xf(x)] = \int_{-\infty}^{\infty} x f(x) e^{i k x} dx $$ $$ = \int_{-\infty}^{\infty} \frac{1}{i} \frac{\partial}{\partial k} \Big[ f(x) e^{i k x} \Big] dx = -i \frac{\partial}{\partial k} \int_{-\infty}^{\infty} f(x) e^{i k x} dx$$ which means $$FT[xf(x)] = -i \frac{\partial f(k)}{\partial k} $$

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    $\begingroup$ I would add that this is in keeping with the general principle that polynomials transform to differential operators and vice-versa. $\endgroup$ – Neal Apr 22 '12 at 18:58

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