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From this page:

enter image description here enter image description here

So I'm trying to get that result in Mathematica. From the fourier formula above, we can see that {a, b} = {1,-1}.

enter image description here

However, the code below returns a different result.

FourierTransform[Cos[2 \[Pi] A t], t, w, FourierParameters -> {1, -1}]
(*\[Pi] DiracDelta[2 A \[Pi] - w] + \[Pi] DiracDelta[2 A \[Pi] + w]*)

Is there anything wrong here?

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  • $\begingroup$ The result is true, its linked pseudo-proof is not: that improper integral diverges. The Fourier transforms of distributions are not defined as improper integrals (see "Tempered distributions" section in Wiki). I am sure those Fourier transforms are implemented as table values in Mathematica. $\endgroup$
    – user64494
    May 22 at 5:49

1 Answer 1

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I think the confusion is between frequency in cycles per second or Hz and frequency in radians per second. In the page you link to the frequency is in Hz. Thus to match this convention the FourierParameters have to be {0, -2 [Pi]} Try this

FourierTransform[Cos[2 \[Pi] A t], t, f, 
 FourierParameters -> {0, -2 \[Pi]}]

(* 1/2 DiracDelta[A - f] + 1/2 DiracDelta[A + f] *)

Hope that helps

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