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Feb 23, 2022 at 16:33 comment added Daniel Lichtblau Not clear what is expected to happen, but note that FT of product is not going to be product of FTs.
Feb 23, 2022 at 15:20 comment added user64494 @yarchik: If I state a non-trivial claim, I try to give an accessible reference to it.
Feb 23, 2022 at 12:51 comment added user64494 @yarchik: Sorry, again ungrounded $$\int dx f(x) g(x) = \int dt \mathcal{F}[f](t) \mathcal{F}[g](-t).$$ It should be noticed that InverseFourierTransform[Convolve[Exp[-Abs[t]], I/t, t, y], y, t] returns the input.
Feb 23, 2022 at 12:47 comment added user64494 @yarchik: Sorry, don't understand you. Ungrounded statements do not make a good impression. Can you elaborate your previous comment, giving us a possible answer to the question? TIA.
Feb 23, 2022 at 12:13 comment added user64494 @yarchik: Wiki says "The Fourier transform translates between convolution and multiplication of functions. If f(x) and g(x) are integrable functions with Fourier transforms f̂(ξ) and ĝ(ξ) respectively, then the Fourier transform of the convolution is given by the product of the Fourier transforms f̂(ξ) and ĝ(ξ) (under other conventions for the definition of the Fourier transform a constant factor may appear).".
Feb 23, 2022 at 12:05 comment added user64494 @yarchik: Can you ground your claim giving us an accessible reference? TIA.
Feb 23, 2022 at 12:00 answer added user64494 timeline score: 3
Feb 23, 2022 at 11:38 history asked MBolin CC BY-SA 4.0