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I am trying to understand how to go about the following:

We are given that the Fourier transform of a function $f(r)$ is $F(k)$ where $k$ is a representative wavenumber corresponding to some spatial coordinate $r$. If I am correct, then we can write:

$$F(k) = \int_{-\infty}^{\infty} f(r) e^{-ikr} dr$$

Now, how can one evaluate the Fourier transform of $ g(r) = \frac{d}{dr}(r f(r))$ in terms of $F(k)$? Any suggestions would be appreciated.

Tanmay Agrawal
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    \begin{align} G(k) &= \int_{\mathbb{R}} g e^{-i k r} dr \ &= \int_{\mathbb{R}} r f'(r) e^{-i k r} dr \ &= \int_{\mathbb{R}} i f'(r) \left( -i r e^{-i k r} \right) dr \ &= \int_{\mathbb{R}} i f'(r) \left( \frac{d}{dk} e^{-i k r} \right) dr \ &= i \frac{d}{dk} \int_{\mathbb{R}} f'(r) e^{-i k r} dr \ &= \dots\end{align} – Matthew Cassell May 08 '21 at 07:06
  • Thank you. With your response and the follow-up provided in https://math.stackexchange.com/questions/430858/fourier-transform-of-derivative?rq=1 I was able to understand. Much appreciated. – Tanmay Agrawal May 08 '21 at 07:55
  • @mattos Can you please have a look at the $g(r)$ now? I mistyped the derivative in my original post. – Tanmay Agrawal May 08 '21 at 09:34
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    You need to make sure you type your questions correctly. To solve your problem, just use integration by parts $u = e^{-i k r}, v' = (r f)'$ to get $$G(k) = (r f) e^{-ikr} \bigg\lvert_{\mathbb{R}} + i k \int_{\mathbb{R}} r f e^{-i k r} dr$$ Assuming $f$ decays fast enough so that the evaluated term vanishes i.e $(r f) e^{-ikr} \lvert_{\mathbb{R}} = 0$, if you then follow what I wrote in my other comment above from the second equality onwards you get $$G(k) = -k \frac{d}{dk} F(k)$$ – Matthew Cassell May 08 '21 at 09:52
  • Thank you very much, got it. Appreciate your kind response. – Tanmay Agrawal May 08 '21 at 11:15

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Using the properties of Fourier transforms the identity $g(r) = \frac{\mathrm{d}}{\mathrm{d} r}rf(r) $ becomes $$G(k) = - k\frac{\mathrm{d}}{\mathrm{d} k} F(k), $$ where $G$ is the Fourier transform of $g$.

Khalid Wenchao Yjibo
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  • Can you please explain how you wrote the RHS as $- \frac{d}{dk} k F(k)$? – Tanmay Agrawal May 08 '21 at 07:57
  • Can you please have a look at the g(r) now? I mistyped the derivative in my original post. – Tanmay Agrawal May 08 '21 at 09:36
  • The Fourier tranform transforms derivation into (i times) multiplication by the dual variable, and viceversa. You can have a look for instance here: https://en.wikipedia.org/wiki/Fourier_transform. The new $g$ you have written is not significantly different, the same reasoning apply, see the edited answer. – Khalid Wenchao Yjibo May 09 '21 at 10:46