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FourierTransform can make sense of integrals that diverge according to Integrate.

Integrate[Exp[I w t], {w, -∞, ∞}]
(* Integrate::idiv: Integral of E^(I w) does not converge on {-∞,∞}. *)
FourierTransform[1, w, t, FourierParameters -> {1, 1}]
(* 2 π DiracDelta[t] *)

Integrate and FourierTransform use different theories of integration, but exactly how those theories differ seems to be undocumented. Are there other situations where FourierTransform yields different results versus a formally equivalent Integrate?

J. M.'s missing motivation
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John Doty
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    The conditions for FT to exists are known to be (copied from the web) `1. On any finite interval (a) f(t) is bounded (b) f(t) has a finite number of minima and maxima (c) f(t) has a finite number of discontinuities
    1. f(t) is absolutely integrable,` So the last point is bit more relaxed than integral of the function itself. The function has to be absolutely integrable even if it is possible it is not integrable otherwise, it still will have F.T.
     – Nasser Jan 23 '22 at 18:34
  • @Nasser Isn't "absolutely integrable" more stringent than "integrable"? Also, f[t]=1 is not absolutely integrable, yet FourierTransform yields its Fourier transform, as commonly understood in Fourier analysis. See Bracewell's "The Fourier Transform and Its Applications". – John Doty Jan 23 '22 at 18:48
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    Is this a mathematics question or a symbolic computation question? The difference (for me) is that in the latter, we are not concerned so much with the mathematical validity of the transform or integral as with whether Mathematica will compute an answer. I thought it is the latter you're asking. – Michael E2 Jan 23 '22 at 19:31
  • The Fourier transform FourierTransform[1, w, t, FourierParameters -> {1, 1}] is not defined through the divergent improper integral Integrate[Exp[I w t], {w, -\[Infinity], \[Infinity]}], but in another way see Tempered distribution in Wiki. – user64494 Jan 23 '22 at 19:39
  • @JohnDoty good point above the absolute integrable part. You might want to ask at the math forum if you want more exact mathematical reasoning. These are the conditions I learned at school. But in practice (i.e. in Mathematica) implementation might be more relaxed than these. – Nasser Jan 23 '22 at 19:40
  • @MichaelE2 It is the latter. – John Doty Jan 23 '22 at 19:44
  • @user64494 Yes, that is one way to look at a kind of problem that FourierTransform can solve but Integrate cannot. But that's not my question here. Is there any other kind of problem that FourierTransform can solve but Integrate cannot? – John Doty Jan 23 '22 at 19:49
  • @JohnDoty: See other examples of such type in "Distributions, one-dimensional" section of that Wiki article. – user64494 Jan 23 '22 at 19:55
  • @user64494 That's not the question. We know that FourierTransform uses DiracDelta and relatives in cases where Integrate doesn't. The question is whether FourierTransform can solve other kinds of problems for which Integrate doesn't work. It's a Mathematica question, not a mathematics question. – John Doty Jan 23 '22 at 20:27
  • (1) I think the remarks about the underpinnings of the FT are relevant. What it comes down to is that FourierTransform can use methods such as lookup tables that give results in cases where an integral formulation would diverge. – Daniel Lichtblau Jan 24 '22 at 16:16
  • (2) Not really relevant to this thread but I also remark that one can (often) define a FT as a limiting process involving convergent integrals, wherein a limit of integrals exists even though the integral of the limit does not. A standard case is wherein one introduces a mollifying multiplier Exp[-a*x^2] into the integrand, integrates, and takes a limit as a->0. – Daniel Lichtblau Jan 24 '22 at 16:20
  • @DanielLichtblau Is there any case where FourierTransform can separate a convergent integral into divergent integrals, evaluate them symbolically, and assemble a result? – John Doty Jan 24 '22 at 16:21
  • Offhand I do not know how FourierTransform might deal with a sum, or perhaps create a sum, such that it can handle the summands. My familiarity with the transform code internals is quite dated. (Eventually that will also happen to my familiarity with Integrate. I'm not sure how I feel about that.) – Daniel Lichtblau Jan 24 '22 at 16:24
  • @DanielLichtblau Indeed, simply confine your attention to functions in L^2 space, and the theoretical problems go away. But, you either have to carry around arbitrary time limit and bandwidth parameters, or take a lot of limits, for no practical advantage in many applications. I have some sympathy for the idea, but it's too tedious in practice. – John Doty Jan 24 '22 at 16:27

1 Answers1

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The Fourier transform FourierTransform[1, w, t, FourierParameters -> {1, 1}] is not defined through the divergent improper integral Integrate[Exp[I w t], {w, -∞, ∞}], but in another complex way (see "Tempered distributions and Fourier transform" in Wiki). See other examples of such type in "Distributions, one-dimensional" section of that Wiki article.

PS. It should be noticed that the notation DiracDelta[t] is misleading: the $\delta$-distribution is a functional acting on functions from a certain class, not a usual function of real argument t.

MarcoB
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user64494
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    This is not responsive to the question. But note that in real, practical applications of this math, the fact that Dirac δ isn't a "function" according to Dirichlet's definition is trivia. Dirac δ passes the "duck test": it looks and behaves like a function in these applications. The notation is not at all misleading: it leads to correct results, as verified by machines that work and experiments that yield answers consistent with the math. – John Doty Jan 23 '22 at 20:48
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    (I hope I don't come to regret this but) I have to ask why this, and at least two other recent responses from same poster, is/are heavily down-voted. It happens to be correct. Stated differently, FourierTransform can give results that an integral formulation cannot, because the FT exists in cases where the integral formulation of a Fourier transform does not. This is similar to evaluating 1/(1-x) at x=2 vs. evaluation 1+x+x^2+... at x=2. – Daniel Lichtblau Jan 24 '22 at 16:13
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    @DanielLichtblau I explained my reasons for down-voting, and three others appear to agree. The poster is correct in a certain mathematical context, but Dirac δ does not fit well in that context. It fits better with the pragmatic uses of the words "function" and "integral" in applied math (closer to Fourier's understanding than more recent ideas). That usage is not any more problematic than other tricky mathematical concepts like "point" and "line" in applied math. The poster appears not to understand the context in which Dirac δ is useful, and is constantly critical of its uses. – John Doty Jan 26 '22 at 22:59
  • @JohnDoty I don't disagree with the assessment of your last phrase "and is...". But your question is based on a claim that is not correct, and that appears to be the point of this reply. The incorrect assertion involves "FourierTransform yields different results versus a formally equivalent Integrate. In point of fact the integral is not formally equivalent (in this case). In contrast to e.g. Integrate[continuousF[x]*DiracDelta[x],{x,-Infinity,Infinity}], which is well defined as an integral and gives continuousF[0], you used an integrand that diverges. – Daniel Lichtblau Jan 26 '22 at 23:16
  • @JohnDoty What I'm trying to get across (with limited success, granted) is that the math and the applications are not at odds here. The math certainly backs the claim that 1 has an FT, and this of course has applications. But the FT in question cannot be obtained through the integral representation of the FT. So this comprises a case where, as you observed, FourierTransform yields a different result than the (divergent) integral formulation. – Daniel Lichtblau Jan 26 '22 at 23:52
  • @DanielLichtblau Yet Mathematica's own documentation asserts that FourierTransform is defined as an integral. If we accept that definition, then everything I said about formal equivalence is correct. Of course, that depends on what you mean by "integral". – John Doty Jan 27 '22 at 00:42
  • @JohnDoty (1) That line in the documentation is not fully correct. It is correct when the integral exists, using whatever mathematically correct notion of integral you like. This particular integral does not exist. (2) I for one do not propose to change the documentation. – Daniel Lichtblau Jan 27 '22 at 01:00
  • @DanielLichtblau Again, it depends on what you mean by "integral". Some presentations call it a "generalized" integral. I, for one, do not choose to reject a useful concept that avoids an arbitrary distinction between Fourier transforms that are integrals and transforms that aren't. – John Doty Jan 27 '22 at 01:07
  • @JohnDoty We can agree to disagree on this one. "Generalized integral" might make sense here, I'm not certain. Also there are well defined notions of regularized integrals, and maybe it fits into that category as well (I'm not up to speed...). But I would not call the distinction arbitrary. Getting the math right can be important, even (or especially) when the engineering applications preceded the math by a few decades. – Daniel Lichtblau Jan 27 '22 at 14:58
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    @DanielLichtblau I respect your position. I note however that "getting the math right" in applications is more about matching the properties of the abstraction to reality than it is about paying attention to the mathematical details of the abstraction. Nature, not mathematics, decides whether you've gotten it right. Losing track of this by focusing on irrelevant details is not helpful. – John Doty Jan 27 '22 at 15:05
  • @DanielLichtblau It's arbitrary because its roots lie in Dirichlet's definition of a function as a map. We see this is arbitrary because it didn't agree with previous mathematicians' notion of function, it's a poor model of physical "functions", and it conflicts with subsequent ideas on functions like Hilbert spaces and Dirac δ. It should be seen as just as arbitrary as Euclid's fifth postulate. – John Doty Jan 27 '22 at 15:14
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    (1/2) @JohnDoty Are we talking about integration here (in the context of "it's arbitrary")? I think that's a subject I know a little about. A considerable amount of effort was put into getting the Dirac delta "right", as a linear functional. But none of this is relevant to whether a given integral converges. For decades the theory (and practice) of Fourier transforms has gone well beyond what can be integrated, even with extensions that define integrals involving that Dirac delta. – Daniel Lichtblau Jan 27 '22 at 16:26
  • (2/2) @JohnDoty Anyway, not much sense to staking out positions here; we each have some idea of where the other stands. My main point was that I think this response actually got at the heart of why some FourierTransform results work fine even when the corresponding integral representations do not give results. – Daniel Lichtblau Jan 27 '22 at 16:27