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

Question

There are several conventions for the definition of the Fourier transform on the real line.

1 . No $2\pi$. Fourier (with cosine/sine), Hörmander, Katznelson, Folland.

$ \int_{\bf R} f(x) e^{-ix\xi} \, dx$

2 . $2\pi$ in the exponent. L. Schwartz, Trèves

$\int_{\bf R} f(x) e^{-2i\pi x\xi} \, dx$

3 . $2\pi$ square-rooted in front. Rudin.

${1\over \sqrt{2\pi}} \int_{\bf R} f(x) e^{-ix\xi} \, dx$

I would like to know what are the mathematical reasons to use one convention over the others?

Any historical comment on the genesis of these conventions is welcome. Who introduced conventions 2 and 3? Are they specific to a given context?

From the book of L. Schwartz, I can see that the second convention allows for a perfect parallel in formulas concerning Fourier transforms and Fourier series. The first convention does not make the Fourier transform an isometry, but in Fourier's memoir the key formula is the inversion formula, I don't think that he discussed what is now known as the Plancherel formula. Regarding the second convention, Katznelson warns about the possibility of increased confusion between the domains of definition of a fonction and its transform.

Answer 1

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.

Answer 2

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.

Answer 3

Just an elaboration on the comments by nfdc23 and KConrad. I think differences between conventions 1 and 3 or the one mentioned by Alexander of putting $\frac{1}{2\pi}$ in front of the direct transform are just a matter of taste, but convention 2 stands out.

If instead of $\mathbb{R}$ you work on the field of $p$-adic numbers, $\mathbb{Q}_p$ then the standard Fourier transform is $$ \widehat{f}(\xi)=\int_{\mathbb{Q}_p} e^{-2i\pi\{\xi x\}_p}\ f(x)\ dx $$ where $\xi x$ is just the product of the two $p$-adic numbers and $\{\cdots\}_p$ is the polar part defined as follows. An element $x\in\mathbb{Q}_p$ has unique convergent series representation as $$ x=\sum_{n\in \mathbb{Z}} a_n p^n $$ where the "digits" $a_n$ belong to $\{0,1,\ldots,p-1\}$ and only finitely many of them are nonzero for negative $n$. Then one sets $$ \{x\}_p=\sum_{n<0} a_n p^n\ . $$ The definition of the Fourier transforms simply reflects the fact $\mathbb{Q}_p$ is its own Pontryagin dual. However, what this abstract statement means concretely is that all additive characters are of the form $x\mapsto e^{-2i\pi\{\xi x\}_p}$ for $\xi\in\mathbb{Q}_p$. I don't know how one would avoid putting the $2\pi$ in the exponential in this context. Since number theorists need to work adelically and combine $\mathbb{R}$ with all the $\mathbb{Q}_p$, it makes sense, if only for esthetic reasons, to treat everybody the same and put the $2\pi$ in the exponential for $\mathbb{R}$ too. Of course, this opens other cans of worms like: what is the missing polar part for $\mathbb{R}$? why isn't $e^{-\pi x^2}$ the indicator function of a subring of $\mathbb{R}$? etc.

Answer 4

The book Numerical Recipes in C explains (p. 496) that the authors (after clearly having had to work with every possible convention) have solidly fixed on $$ H(f) = \int_{-\infty}^{\infty} h(t) e^{2 \pi i ft} dt $$ $$ h(t) = \int_{-\infty}^{\infty} H(f) e^{-2 \pi i ft} df $$ (almost convention 2) which have some obvious symmetry advantages, and agree with the convention in harmonic analysis, and lead to fewer multiples of $2 \pi$ overall in various formulas, and make the Fourier transform unitary in $L^2$ (see Folland, Real Analysis, p. 244).

Answer 5

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 $\hat{x'}(f) = if\,\hat x(f)$).

Answer 6

There are no mathematical reasons. It is a question of convenience. If you use the first definition, the inverse transform will have $1/2\pi$. If you use the third definition, the inverse transform will have $1/\sqrt{2\pi}$ (and will be very similar to the direct transform). If you use the second definition, none will have any $2\pi$ in front. So the convenience of the second and third definition is a symmetry between the direct and inverse transform.

Finally your list misses one more possible definition used in some books: with $1/2\pi$ in front of direct transform. Then the inverse one has no multiple in front. (For example, the textbook by Folland, Fourier Analysis and its applications).

To conclude: these $2\pi$'s are unavoidable, where to place them is a matter or taste and convenience. Same applies to Fourier series, of course. I've seen options 1, 2, 3 in textbooks.

Answer 7

Not an answer but an elegant approach (due to Treves) to avoid this problem, also in higher dimensions. I attended a series of his lectures at a conference in Edinburgh about 50 years ago. Since this involved many formulae containing Fourier direct and inverse transforms, he announced at the beginning that he would use a standard convention often used by physicists and set all constants equal to "1". The factors (various positive or negative powers of multiples of $\pi$) in the equations disappeared, the lectures went more smoothly for the audience (and, presumably, for the lecturer) and it was a matter of a few minutes' thought to choose one's own favourite convention and recreate the precise form of a concrete formula in the (unlikely) case of this being required.

Answer 8

My take is : "If you use the third definition, the inverse transform will have $\sqrt\frac{1}{2\pi}$ factor and will be very similar to the direct transform)." (lifted from answer there). I'd like to preserve time frequency duality, in the sense both should get essentially same treatment, at-least for real valued functions. But the FT is complex! So I'd like to look only at functions that are real, even symmetric, so that both function and its FT are real even symmetric functions, and both inverse and forward formulae are very similar with a $\sqrt\frac{1}{2\pi}$ factor. Of course which one gets +i and -i is just a matter of pure convention.