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I need some help... I am asked to prove the following property of the Fourier transform, when $F[f(x)]=\widetilde{f}(x)$, where $F[f(x)]$ is the Fourier transform of $f(x)$:

$$F[ \widetilde{f}(x) ]= \frac{f(-k)}{2 \pi}$$

We know that: $F[ \widetilde{f}(x) ]=\int_{- \infty}^{+ \infty}{ \widetilde{f}(x) e^{-i k x}}dx$.

But how can I prove this? I got stuck.. :/ Could you give me a hint?

Mary Star
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  • Compare the formulas of the forward and reverse transform. They are very similar! – fgp Apr 12 '14 at 17:59
  • @fgp When I start with $F[ \widetilde{f}(x) ]=\int_{- \infty}^{+ \infty}{ \widetilde{f}(x) e^{-i k x}}dx$, what can I replace at $ \widetilde{f}(x)$?? – Mary Star Apr 12 '14 at 18:15
  • What you are doing is finding the fourier transform $\tilde f$ of the function $f$, and then applying the fourier transform to $\tilde f$ again. If you applied the reverse transform the second time, you'd obviously get back to $f$. So write down both $F[\tilde f]$ and $F^{-1}[\tilde f] = f$ and compare... – fgp Apr 12 '14 at 18:20

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Then $\int_{-\infty}^\infty \bar {f}(x)e^{-ikx}dx=\int_{-\infty}^{\infty} \bar {f}(x)e^{i(-k)x}dx=f(-k)$

I realise this is out by a $2\pi$ but you get the idea. (Your definition of a fourier transform needs a $2\pi $ somewhere).

Ellya
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  • I did the following: $$F[\widetilde{f}(x)]=\int_{-\infty}^{+\infty}{\widetilde{f}(x) e^{-ikx}}dx=2 \pi \frac{1}{2 \pi}\int_{-\infty}^{+\infty}{\widetilde{f}(x) e^{ix(-k)}}dx=2 \pi f(-k)$$ But the property is: $F[\widetilde{f}(x)]=\frac{f(-k)}{2 \pi}$.. What have I done wrong?? – Mary Star Apr 12 '14 at 21:26
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    The proper fourier transform is $F(f)=\frac{1}{\sqrt 2\pi}\int_{-\infty}^\infty f(x)e^{-i kx}dx $ – Ellya Apr 12 '14 at 21:38
  • But in my notes there are the following formulas: $$\text{Inverse Fourier Transform: }f(x)=\frac{1}{2 \pi} \int_{-\infty}^{+\infty}{\widetilde{f}(k) e^{ikx}}dx$$ $$\text{Fourier Transform: }\widetilde{f}(k)=\int_{-\infty}^{+\infty}{f(x) e^{-ikx}}dx$$ Using these formulas, how can I prove the property? – Mary Star Apr 13 '14 at 15:58
  • Interesting, the apperances of $\pi $ seem to change in fourier transforms, ive also seen them in the exponential, but I would stick with your notes. – Ellya Apr 13 '14 at 16:01
  • But the two formulas you have been given seem quite inconsistent. – Ellya Apr 13 '14 at 16:07
  • What do you mean?? – Mary Star Apr 13 '14 at 16:11
  • Well normally if there is a "$2\pi$" in one then there should be a "$2\pi $" in the other. – Ellya Apr 13 '14 at 16:14