The $2\pi$ in the definition of the Fourier transform

The version number 2 is the only one that makes the Fourier transform both a unitary operator on $L^2$ and an algebra homomorphism from the convolution algebra in $L^1$ to the product algebra in $L^\infty $.

It is not, however, of widespread use in analysis as far as I know. From the point of view of semiclassical analysis, it amounts roughly speaking to consider Planck's constant $h $ rather than $\hslash=h/2\pi $ as the semiclassical parameter (or as the constant set to one in quantum systems). This is somewhat differing from the common practice in physics.

I would like to add the point of view from a more general aspect.

In the general situation, the Fourier transform can be defined for any locally compact abelian group $G$.

Let $G$ be a locally compact abelian group. Let $\hat{G} = \mathrm{Hom}_{\mathrm{cont}}(G, \mathbb{S}^1)$ be the Pontryagin dual of $G$, where $\mathbb{S}^1 = \{z\in\mathbb{C}: |z|=1\}$ is the "circle group". Pontryagin duality asserts that $\hat{\hat{G}} = G$. We write $\langle \cdot, \cdot \rangle$ for the canonical pairing $G \times \hat{G} \rightarrow \mathbb{C}$.

If one fixes Haar measures $dg$ and $dh$ on $G$ and $\hat{G}$, respectively, then for any Schwartz function $f:G\rightarrow\mathbb{C}$, its Fourier transform is defined by: $$\hat{f}:\hat{G}\rightarrow\mathbb{C}, \hat{f}(h)=\int_G f(g)\langle g, h\rangle dg.$$

The Fourier transform of a Schwartz function $\phi$ on $\hat{G}$ is defined by the same formula, i.e. $$\hat{\phi}(g)=\int_\hat{G} \phi(h)\langle g, h \rangle dh.$$

The Fourier inversion formula states that, if one chooses the Haar measures $dg$ and $dh$ properly, then we have $\hat{\hat{f}}(g) = f(-g)$. When this happens, the two Haar measures are called dual to each other.

Now in the case of $G = \mathbb{R}$, the interesting thing is that we actually have $G \simeq \hat{G}$. The isomorphism can be given by a pairing $\langle x, y \rangle = e^{iaxy}$, where $a$ can be chosen to be any nonzero real number.

However, in this particular case, one naturally wants both Haar measures to be the Lebesgue measure, i.e. the measure of the interval $[0, 1)$ is equal to $1$. If one demands that the Lebesgue measure is dual to itself, then the constant $a$ is pinned down to $2\pi$ (or $-2\pi$, which are equivalent).

This somehow explains the expression 2.

The version 2 is also popular among electrical engineers as the variable $\xi$ is then the actual frequency. For an electrical engineering view on the Fourier transform, I can recommend the lecture notes The Fourier Transform and its Applications by Brad Osgood.

Also, this notion of frequency explains that electrical engineers sometimes use variable names that seem a bit odd for mathematicians: A (complex valued) signal with angular frequency $\omega$ is $\exp(i\omega t)$ and written in linear frequency $f = \omega/(2\pi)$ it is $\exp(i2\pi ft)$. Hence, the Fourier transform of a signal $x(t)$ ($t$ in seconds) may look like $\hat x(f) = \int_{-\infty}^\infty \exp(i2\pi ft)x(t) dt$ ($f$ in $\mathrm{Hz}$).

One downside of this convention is that the handy rule of differentiation gets a bit more complicated (namely $\hat{x'}(f) = 2\pi i f\, \hat x(f)$ instead of $\hat{x'}(f) = if\,\hat x(f)$).