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Could anyone explain to me how do we change Fourier transform equation from this [Wiki - look at the top of the page]:

$$ \mathcal{F}(x) = \int\limits_{-\infty}^{\infty} \mathcal{G}(k)\, e^{-2\pi i k x} \, \mathrm{d} k $$

to this [Wiki - check Fourier transform and characteristic function]:

$$ \mathcal{F}(x) = \int\limits_{-\infty}^{\infty} \mathcal{G}(k) \, e^{ikx} \, \mathrm{d} k $$

Where did $-2\pi$ go???

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migrated from Feb 28 '13 at 17:33

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Different authors use different conventions. See e.g. the section about other conventions on the same Wikipedia page. –  Qmechanic Feb 27 '13 at 20:09
Here's another site to learn about Fourier transform… –  raindrop Feb 27 '13 at 20:11
Would it be ok if i would write down an inverse Fourier transform for my second equation like this: $$\mathcal{G}(k) = \int\limits_{-\infty}^{\infty} \mathcal{F}(x) \, e^{-ikx} \, \mathrm{d} x$$ –  71GA Feb 27 '13 at 22:48

4 Answers 4

It's a just a difference in preferred units. See:

In particular, look at the sentence just before equation (7):

Note that some authors (especially physicists) prefer to write the transform in terms of angular frequency $\omega=2\pi\nu$ instead of the oscillation frequency $\nu$...

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There should be a factor $$\frac{1}{2\pi}$$ in the definition of the Fourrier transform and its inverse (pure matter of convention) so you can have $$ \int dk\frac{1}{2\pi}e^{ikx} = \delta(x)$$.

In your case they absorbed it in the exponential by redefining $$k \rightarrow 2\pi k$$.

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Some say that there should be a factor of $1/\sqrt{2 \pi}$ there and not $1/(2 \pi)$ BUT i think that this is just because some use Fourier transform of a Gaussian and if we want to normalize Gauss function $g(x) = a \exp\left[ - \frac{(x-\mu)^2}{2\sigma^2}\right]$, we need to choose $a = 1/\sqrt{2 \pi}$. This is a bit messed up in my head. Could anyone help me clear this up? –  71GA Feb 27 '13 at 22:28

The Fourier inversion theorem says \begin{eqnarray} f(x) = \int_{-\infty}^\infty \frac{1}{2\pi} \int_{-\infty}^\infty f(t) e^{-i\omega t} dt e^{i\omega x} d\omega \end{eqnarray} under sufficient assumptions. The conventional way to extract transforms is to write \begin{eqnarray} \mathscr{F}f(\omega) & = & \frac{1}{2\pi} \int_{-\infty}^\infty f(t) e^{-i\omega t} dt \\ \mathscr{F}^{-1}f(t) & = & \int_{-\infty}^\infty f(\omega) e^{i\omega t} d\omega \ . \end{eqnarray} Here the constant $\frac{1}{2\pi}$ comes from the coefficients of the Fourier series. The Fourier inversion theorem was developed from Fourier series. We can now make a change of variables $\omega = 2\pi w$ to the outer integral on the right hand side of the Fourier inversion theorem and obain \begin{eqnarray} f(x) = \int_{-\infty}^\infty \int_{-\infty}^\infty f(t) e^{-2\pi iwt} dt e^{2\pi iwx} dw \ . \end{eqnarray} We can now extract transforms \begin{eqnarray} \mathscr{F}f(w) & = & \int_{-\infty}^\infty f(t) e^{-2\pi iwt} dt \\ \mathscr{F}^{-1}f(t) & = & \int_{-\infty}^\infty f(w) e^{2\pi iwt} dw \ . \end{eqnarray} We can make an other change of variables $\omega' = -\omega$ to the outer integral in the Fourier inversion theorem to obtain \begin{eqnarray} f(x) = \frac{1}{2\pi} \int_{-\infty}^\infty \int_{-\infty}^\infty f(t) e^{i\omega' t} dt e^{-i\omega' x} d\omega' \ . \end{eqnarray} Again we can extract \begin{eqnarray} \mathscr{F}f(\omega') & = & \int_{-\infty}^\infty f(t) e^{i\omega' t} dt \\ \mathscr{F}^{-1}f(t) & = & \frac{1}{2\pi} \int_{-\infty}^\infty f(\omega') e^{-i\omega' t} d\omega' \ . \end{eqnarray} This explains the different definition. However, it is recommended to stay within one definition in each text.

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You can add a constant and the transform is still valid. Normalization changes. As Penrose states, if you choose the second form, you can normalize multiplying by (2pi)^-1/2 with the benefit of having a symmetry in the anti-transform (that is the transform and the anti-transform have the same form, the only thing that changes is the integration variable)

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