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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

This question came from our site for active researchers, academics and students of physics.

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

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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