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If you ask Mathematica to provide the Fourier Transform of a singular functions it is likely to provide an answer that while nearly correct, is technically incorrect and it will do so without a word of warning. Are there any ways to guard for this other than telling the user to beware?

Below I show what happens when you ask for the Fourier Transform of the absolute value function.

Mathematica defines the Fourier transform, $F(\omega)$, of the function $f(t)$, to be: $$F(w) = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^\infty Exp({i w t})\ f(t)\ dt $$ which is a standard normalization. If you ask for the Fourier Transform of $|t|$,

FourierTransform[Abs[t], t, w]

Mathematica returns: $$F(w)=-\frac{\sqrt{2/\pi}}{w^2}$$ which is fine except at $w=0$ where it gives $F(0)=-\infty$. For $w=0$ we have, by definition, $$F(0) = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^\infty |t| \ dt = +\infty \ .$$

Thus it is WAY OFF at $w=0$. The correct transform is of the form:

$$ A \delta(w) - \frac{\sqrt{2/\pi}}{w^2} $$ where $A$ is a divergent integral. I expect that $|t|$ is not the only function for which Mathematica returns an incorrect Fourier Transform with no warning.

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  • $\begingroup$ Usually GenerateConditions should work for this kind of thing. However, at least on V9 FourierTransform[Abs[t], t, w, GenerateConditions -> True] returns unevaluated $\endgroup$
    – Dr. belisarius
    Jan 8, 2015 at 4:37
  • $\begingroup$ With v10: SameQ @@ FourierTransform[{Abs[t], t Sign[t], Sqrt[t^2], Piecewise[{{-t, t < 0}, {t, t >= 0}}]}, t, w] returns True. Although, using GenerateConditions -> True evaluates with t Sign[t]; however, returns the same result, i.e., no conditions. $\endgroup$
    – Bob Hanlon
    Jan 8, 2015 at 5:56
  • $\begingroup$ Using the identity $|t|=\int_0^t Sign[s] ds$ with GenerateConditions also returns unevaluated. I am using Version 10.0.1.0. $\endgroup$
    – JEP
    Jan 8, 2015 at 16:44
  • $\begingroup$ The identity $$|t| = \int_{-\infty}^\infty \frac{1-\cos{\alpha t}}{\alpha^2}\ d\alpha$$ with GenerateConditions also returns unevaluated. It gives the same result as $|t|$ for the transform. $\endgroup$
    – JEP
    Jan 8, 2015 at 17:24
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    $\begingroup$ FourierTransform[Abs[t], t, w] does not exist as a usual Fourier transform because Abs[t] is not integrable over (−∞,∞). Also see Wiki, formula 311. Kammler, David (2000), A First Course in Fourier Analysis, Prentice Hall, presents this formula in Appendix A-6 without any reference. Erdélyi, Arthur, ed. (1954), Tables of Integral Transforms, 1, McGraw-Hill. does not contain it. $\endgroup$
    – user64494
    Oct 5, 2021 at 6:20

1 Answer 1

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Here there is an explanation of why you can dismiss the "missing part", don't really know if is fully mathematically rigurous, but it "sees" right: https://www.thefouriertransform.com/pairs/absT.php Hope this helps.

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    $\begingroup$ This is more appropriate for a comment than an answer. I suggest improving this answer with some code and to also make it “self contained” that is, to not rely on an external link that may, at some point, go dead. $\endgroup$
    – CA Trevillian
    Oct 5, 2021 at 1:13
  • $\begingroup$ Could you explain how to obtain the result of InverseFourierTransform[-Sqrt[2/Pi]/w^2,w,t]? $\endgroup$
    – user64494
    Oct 5, 2021 at 11:08
  • $\begingroup$ Here there is an explanation of why you can dismiss the "missing part", don't really know if is fully mathematically rigurous, but it "sees" right: thefouriertransform.com/pairs/absT.php Hope this helps. (I move it to a comment as indicated by @CATrevillian). I will put what is done in the link, but without proving its validity (its look right to me anyway): 1. first note that $|t| = t\theta(t) - t\theta(t)$ with $\theta(t)$ the unitary step function. $\endgroup$
    – Joako
    Oct 5, 2021 at 15:04
  • $\begingroup$ 2. note that the fourier transform of $\pm t \theta(\pm t)$ equals $\frac{-1}{(2\pi f)^2}+\frac{\delta '(\pm f)}{2}$, so when added to form the fourier transform of $|t|$ the result is $\frac{-1}{2\pi^2 f^2}+\frac{1}{2}(\delta '(f) +\delta '(-f))$, so the rest is to prove that the second part is zero. $\endgroup$
    – Joako
    Oct 5, 2021 at 15:12
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    $\begingroup$ @yarchik I don't know if "knowing" they are odd is enough, since are not functions, and distributions are more complicated mathematical objects. Maybe you can explained better. I don't answer it because I first left the link, were someone else give the explanation. Also, I don't have enough mathematical background to prove is his results is "formaly" right/rigorous, but it seems right to me. I left it in the comments following the request of CA Trevillian and user64494, wanting to help user64494. Best regards. $\endgroup$
    – Joako
    Oct 5, 2021 at 18:57

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