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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.

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    $\begingroup$ Perhaps questions about history would fit better in hsm.stackexchange.com $\endgroup$ – Gerald Edgar Mar 22 '17 at 22:32
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    $\begingroup$ I am strongly reminded of the opening story from Wigner's Unreasonable Effectiveness. $\endgroup$ – Theo Johnson-Freyd Mar 23 '17 at 0:03
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    $\begingroup$ I'm reminded of an anecdote I heard from a professor. When this professor was a graduate student, on the first day of his first Fourier analysis class, the professor walked in, went straight to the chalkboard, and wrote: "$2\pi = \frac{1}{2\pi} = 1$." $\endgroup$ – Neal Mar 23 '17 at 14:58
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    $\begingroup$ This joke is actually written in print, in some notes by L.Nirenberg $\endgroup$ – Piero D'Ancona Mar 23 '17 at 16:46
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    $\begingroup$ There are three major unsolved problems in mathematics: 1. Where to put to $2\pi$ in the Fourier transform, 2. Whether the exponent in the Fourier transform should have the $+i$ or the $-i$, 3. Which factor in the inner product gets the complex conjugate. $\endgroup$ – Michael Renardy Mar 23 '17 at 17:24
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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.

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    $\begingroup$ That's actually really helpful. I often have difficulty remembering the factors of $2 \pi$ in the convolution identity, and now I no longer need to. $\endgroup$ – Adam P. Goucher Mar 23 '17 at 10:39
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    $\begingroup$ In terms of Pontryagin duality, for which there is always a "coordinate-free" Plancherel theorem (and Poisson summation formula) using the dual group, this expresses that ${\rm{d}}x$ is the unique self-dual Haar measure under the associated self-duality of $\mathbf{R}$. (Likewise, that Poisson summation $\sum_{n \in \mathbf{Z}} f(n) = \sum_{n \in \mathbf{Z}} \widehat{f}(n)$ holds without extraneous constants for version 2 expresses the good behavior of making $\mathbf{Z}$ be its own annihilator.) $\endgroup$ – nfdc23 Mar 23 '17 at 14:39
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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.

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  • $\begingroup$ I realize this post is from over 2 years ago, but I've been wondering why the constant gets pinned to $2\pi$. One can show that it must be $2\pi$ by considering test functions, but I wonder if there isn't something deeper going on. Is there a more abstract way to see why it should be $2\pi$? $\endgroup$ – Charles Hudgins Jun 23 at 12:59
  • $\begingroup$ I don't have much detail here, but the deeper reason should be the fact that $x \mapsto e^{2\pi ix}$ has kernel $\mathbb{Z}$ and that $\mathbb{R}/\mathbb{Z}$ has total volume $1$ under Lebesgue measure. $\endgroup$ – WhatsUp Jun 24 at 11:25
  • $\begingroup$ So is the idea roughly that we're implicitly using an isomorphism between $S^1$ and $\mathbb{R}/\mathbb{Z}$ that scales volumes by $\frac{1}{2\pi}$? $\endgroup$ – Charles Hudgins Jun 25 at 1:28
  • $\begingroup$ Yes. It all comes to the identity $e^{2\pi i} = 1$. $\endgroup$ – WhatsUp Jun 26 at 2:01
  • $\begingroup$ Fascinating. I think this is the first time I've really appreciated how remarkable it is that $2\pi$ appears in that equation. $\endgroup$ – Charles Hudgins Jun 26 at 17:16
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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 $\hat{x'}(f) = if\,\hat x(f)$).

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    $\begingroup$ This seems just to beg the question of what "actual frequency" means: is it linear frequency, measured in cycles per second, or angular frequency, measured in radians per second? $\endgroup$ – LSpice Mar 23 '17 at 18:18
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    $\begingroup$ Well, I had acoustic applications in mind, you know, where you have signal in time units and you want to see the pitch (in Hz) in the Fourier transform… Similarly for transmission of electric signals,… $\endgroup$ – Dirk Mar 24 '17 at 17:35
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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.

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    $\begingroup$ The notion of "polar part" expresses in terms of coset representatives the canonical identification of $\mathbf{Q}_p/\mathbf{Z}_p$ with the $p$-primary part of the torsion subgroup $\mathbf{Q}/\mathbf{Z}$ of the group $\mathbf{R}/\mathbf{Z}$ that is identified with $S^1$ via $t \mapsto e^{2\pi i t}$. The intrinsic significance of the implicit associated self-duality of $\mathbf{Q}_p$ is to make the compact open subring $\mathbf{Z}_p$ be its own exact annihilator (and correspondingly makes the Haar measure ${\rm{d}}x$ giving it volume 1 be the associated self-dual measure on $\mathbf{Q}_p$). $\endgroup$ – nfdc23 Mar 23 '17 at 14:48
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    $\begingroup$ It is not only for esthetic reasons that one puts $2\pi$ in the exponential for $\mathbb{R}$ too. The point is that, globally one needs to deal with characters on $\mathbb{A}$, the ring of adeles, and for applications in number theory and automorphic representations, one would like to choose an additive character that factors through $\mathbb{A}/\mathbb{Q}$. If one fixes normalizations on $\mathbb{Q}_p$ as you did, then the choice on $\mathbb{R}$ is forced. $\endgroup$ – WhatsUp Mar 23 '17 at 16:09
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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).

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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.

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    $\begingroup$ The 1st and 3rd cases correspond to a common identification of $\mathbf{R}$ with its Pontryagin dual ${\rm{Hom}}_{\rm{cont}}(\mathbf{R},S^1)$ under which $\mathbf{Z}$ is not its own exact annihilator and the standard Lebesgue measure of $\mathbf{R}$ (determined by counting measure on $\mathbf{Z}$ and volume-1 measure on the compact $\mathbf{R}/\mathbf{Z}$) is not its own dual measure in the "coordinate-free" language of Pontryagin duality (with which there are no $2\pi$'s). But the 2nd case defines a self-duality for which $\mathbf{Z}$ is its own exact annihilator and ${\rm{d}}x$ is self-dual. $\endgroup$ – nfdc23 Mar 23 '17 at 5:36
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    $\begingroup$ @coudy: In Fourier series, one can also put the $2\pi$ in any place. I've seen choices 1,2,3 in the books. In Fourier transform $1/2\pi$ in front is used in a popular text Folland, Fourier Analysis and its applications. $\endgroup$ – Alexandre Eremenko Mar 23 '17 at 13:29
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    $\begingroup$ The comment by @nfdc23 explains why number theorists prefer the 2nd convention. In the 2nd convention $e^{-{\pi}x^2}$ is self-dual (equal to its own Fourier transform) and in the 3rd convention $e^{-x^2/2}$ is self-dual. That shows a link between the 3rd convention and probability, so probabilists probably prefer the 3rd convention. There is no self-dual $e^{-cx^2}$ for an $c > 0$ in the 1st convention. $\endgroup$ – KConrad Mar 23 '17 at 13:37
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    $\begingroup$ Here's the "most general" situation. If the Fourier transform is defined by $\widehat{f}(y) = a\int_{\mathbf R} f(x)e^{-ibxy}\,dx$ for nonzero real $a$ and $b$ then the inversion formula reads $f(x) = (b/(2\pi{a}))\int_{\mathbf R} \widehat{f}(y)e^{ibxy}\,dy$ and for $c > 0$ the Fourier transform of $e^{-cx^2}$ is $a\sqrt{\pi/c}e^{-(b^2/(4c))y^2}$, so $e^{-cx^2}$ is self-dual if and only if $c= |b|/2$ and $a = \sqrt{|b|/(2\pi)}$. $\endgroup$ – KConrad Mar 23 '17 at 13:59
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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.

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    $\begingroup$ Yet Trèves uses convention 2 in his books. I am wondering what explains the success of the $\pi = 1$ trick? Maybe this just amounts to redefining the exponential as $e^{2\pi x}$ in convention 2. At least with Fourier series it is customary to define $e_n(x) = e^{2\pi i n x}$ to alleviate notations. $\endgroup$ – coudy Mar 23 '17 at 8:29
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    $\begingroup$ The convention to use $e^{2\pi i n x}$ for Fourier series expresses the fact that the self-duality of $\mathbf{R}$ determined by convention 2 identifies the discrete subgroup $\mathbf{Z}$ with its counting measure as Pontryagin dual to the compact quotient $\mathbf{R}/\mathbf{Z}$ with its volume-1 measure (quotient measure from the unique self-dual measure ${\rm{d}}x$ on $\mathbf{R}$). $\endgroup$ – nfdc23 Mar 23 '17 at 15:00
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    $\begingroup$ Does this really deserve to be called 'elegant'? It can also simplify formulæ dramatically if you don't have to worry about signs, but declaring all signs to be $+1$ and leaving it up to the reader to reconstruct them is surely not a good idea. $\endgroup$ – LSpice Mar 23 '17 at 18:20
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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.

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