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Let $\mu$ be a probability measure on $\Bbb R^d$. I met the following two definitions of Fourier transformation of $\mu$ in the textbook:

  1. $\displaystyle \widehat{\mu}(\xi)=\int e^{i\xi\cdot x}d\mu(x)$;
  2. $\displaystyle \widehat{\mu}(\xi)=\int e^{-2\pi i\xi\cdot x}d\mu(x)$.

My question is: what's the difference between these two definitions of Fourier transformation of measure?

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  • $\begingroup$ Do you mean $\widehat{\mu}(\xi)=\int e^{-i\xi\cdot x}d\mu(x)$ in the first case? $\endgroup$ – Sayan Dec 5 '15 at 14:27
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    $\begingroup$ @Sayan: Not necessarilly. That one is yet another convention... $\endgroup$ – Alex M. Dec 5 '15 at 14:39
  • $\begingroup$ No, Sayan. The first case is just as I gave. $\endgroup$ – user47280 Dec 5 '15 at 14:47
  • $\begingroup$ A nice feature of definition 2 is that it removes the need for a prefactor on the inverse transformation. On the other hand every derivative pops out a factor of $-2\pi i\xi$ which can be annoying when working with differential equations. $\endgroup$ – Spencer Aug 15 '16 at 18:59
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Both conventions are used (in fact, Wikipedia list three). Some are more appropriate in an electrical engineering context, some in a quantum-mechanical context and some in pure mathematics. The underlying mathematics are the same, some constant factors will differ, though. They are all correct.

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  • $\begingroup$ Many thanks, Alex M. $\endgroup$ – user47280 Dec 5 '15 at 14:51

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