Fourier transform 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???
 A: It's a just a difference in preferred units. See:
http://mathworld.wolfram.com/FourierTransform.html
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 $\nu$... 

A: 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 $$k \rightarrow 2\pi k$$.
A: 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.
A: 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)  
