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Is there a reason why most people use the normalization $$ \hat{f}(x) = \int_{-\infty}^{\infty} e^{- 2 \pi i t x} \cdot f(t) d t $$ instead of $$ \hat{f}(x) = \int_{-\infty}^{\infty} e^{2 \pi i t x} \cdot f(t) d t \text{ } ? $$ I've just finished writing a paper where I've used the second definition, and I'm wondering now, if I should re-write it. The paper is in analytic number theory.

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  • $\begingroup$ Sure. Using this normalization, $\hat{f}(x)$ is the "coefficient" of $e^{2 \pi i tx}$ (generalizing the obvious notation from Fourier series). $\endgroup$
    – Qiaochu Yuan
    Commented May 28, 2012 at 4:02
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    $\begingroup$ There is a natural reason to use the first convention if you think about its origin from Fourier series: we tend to like writing a Fourier series as $f(x) = \sum_{n \in {\mathbf Z}} c_ne^{2\pi inx}$, not as $\sum_{n \in {\mathbf Z}} c_n e^{-2\pi inx}$; the coefficient of $e^{2\pi inx}$ is reasonable to index by $n$, not $-n$. Then one has the formula $c_n = \int_0^1 f(x)e^{-2\pi i{n}x}\,dx$. Notice the appearance of the sign in this coefficient formula, which is the Fourier transform $\widehat{f} \colon {\mathbf Z} \rightarrow {\mathbf C}$: $\widehat{f}(n) = c_n$. $\endgroup$
    – KConrad
    Commented May 28, 2012 at 4:05
  • $\begingroup$ Thus the first normalization in your question appears by analogy. On the other hand, there are sometimes good reasons to choose the Fourier transform using your second convention. For instance, in Tate's thesis the first convention can lead to some weird signs when you try to align his general calculations with classical calculations. $\endgroup$
    – KConrad
    Commented May 28, 2012 at 4:08

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I think I've come to the personal concensus, that with the choice of notation $$ \hat{f}(x) = \int_{-\infty}^{\infty} e^{- 2 \pi i t x} f(t) d t $$ we are extracting information from $f$, while with the notation $$ \hat{f}(x) = \int_{-\infty}^{\infty} e^{2 \pi i t x} f(t) d t $$ we are generating information from the function $f$.

For example, if $f$ were the probability density of a distribution I would go with the second notation to denote the characteristic function of the distribution. This aligns with the standard notation for the fourier transform of a probability distribution (check "characteristic function" on wiki). Anybody with me on this?

P.S: I've seen both conventions used in my field. I believe the choice has to do with the perspective you take on the role of $f$. Do you want to understand $f$ better or do you want to generate things from $f$ and are actually interested in $\hat{f}$ rather than $f$?

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There are many variations of the definition of the Fourier transform and its inverse that are all essentially equivalent. For example, the factor of $2\pi$ often appears as $1/(2\pi)$ in front of the integral in one of the transform pair, or as $1/\sqrt{2\pi}$ in front of both integrals in the transform pair.

Researchers in different fields have often adopted a particular convention. If you want to publish in a journal where such a convention is well established, then you will probably find that it is easier to get your paper accepted for publication if you follow the convention used by people working in that field.

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  • $\begingroup$ I doubt the acceptance or not of the paper is going to turn on the choice of sign in the Fourier transform! $\endgroup$
    – KConrad
    Commented May 28, 2012 at 4:01
  • $\begingroup$ Well, if you get the "wrong" answer for some constant... $\endgroup$
    – Igor Rivin
    Commented May 28, 2012 at 4:12
  • $\begingroup$ The point is that some reviewer might well ask to have the paper rewritten using the common convention and it's quite likely that an editor would agree that the author should do this. If this happens it will slow down the process of getting your paper accepted. $\endgroup$
    – Brian Borchers
    Commented May 28, 2012 at 4:22

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