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I am not able to get the Fourier Transform of $xf(x)$ if $&ltf(x)>$ is the Fourier transform of $f(x)$ .
BTW i tried using convolution theorem but didn't work out .

draks ...
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Theorem
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    Vedananda, could you perhaps expand on that a bit? It's not completely clear what you question is, at least not to me. – Terry Bollinger Apr 22 '12 at 13:10
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    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. –  Apr 22 '12 at 13:18
  • Yes exactly Anuragsn , if i know what FT of $f(x)$ is , how do i find FT of $xf(x)$. :) – Theorem Apr 22 '12 at 13:25
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    Oh... I get it now. But it is already answered... And you should post such stuff to mathematical forums... –  Apr 22 '12 at 13:42

1 Answers1

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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 $$FT[xf(x)] = i \frac{\partial f(k)}{\partial k} $$

zabop
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Vijay Murthy
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    I would add that this is in keeping with the general principle that polynomials transform to differential operators and vice-versa. – Neal Apr 22 '12 at 18:58
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    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)". – quantif Apr 26 '20 at 14:42
  • 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 $f(x)=\frac{x}{1+x^2}$? – Cleto Pereira May 19 '20 at 19:22