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I am struggling with this although the question is partially answered a few times before. Here $-\infty < t <\infty$ and I am only interested in $0\leq n \leq 2$. Mathematica gives the FT as $|\omega|^{n-1}$ if $n \neq 1$. Is it strictly correct? My problem is that the FT keeps increasing with $\omega$ for $n>1$. And, for $n=0$ we do not readily get back delta function. If this answer is indeed correct, I have the following two questions which are more important for me.

  1. In realistic complicated physics problems when I clip/regularise the blowing up at $t=0$ by $f(t)=min[|t|^{-n}, 10^{6}]$, I find the FT decays to zero (somewhat like $\exp^{-|\omega|}$). Can we calculate the FT of this function analytically? Would it now stop the monotonic rise in FT?

  2. My main question: Does the FT of a symmetric function (say the one defined in point 1) with narrower peak at $t=0$ have a broader peak at $\omega=0$? In other words, imagine a function $f(t)$ symmetrically falls from a finite peak at $t=0$ to zero at $t \to \infty$. Can we say that FT will also have symmetric fall from $\omega=0$ but the fall will be shallower if the peak in $f(t)$ is narrower? Can we mathematically prove this statement? If it is already established can you please provide a reference?

Does discrete FT (FFT) have any caveats in this regard?

Thanks in advance.

Satadru Bag
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  • Yes, the answer of Mathematica is right, and follows broadly from a scaling argument (see eg https://math.stackexchange.com/questions/3742640/fx-1-lvert-x-rvert2-x-in-mathbbr3-for-the-fourier-transform/3742759#3742759) but $1/|t|^{n}$ for $n>1$ is not locally integrable, so you have to consider its tempered distribution generalization, called principal value of Hadamard finite part. In the critical case $n=1$ you get a logarithm (see here https://math.stackexchange.com/questions/3723136/the-fourier-transform-of-1-p3/3724502#3724502) – LL 3.14 Apr 19 '22 at 18:51
  • The general form when $n\in(0,d)$ in any dimension $d$ is $$ \mathcal{F}(1/(\omega_a|x|^{a})) = \mathcal{F}(1/(\omega_{d-a}|x|^{d-a})) $$ with $\omega_a = 2 \pi^{a/2}/\Gamma(a/2)$. You can find the result in a lot of textbooks, for example "Functional Analysis" from Lieb and Loss. Then you can take derivatives in the sense of distribution to get the other cases $a>d$ or $a<0$. – LL 3.14 Apr 19 '22 at 18:56
  • Thanks. What about my second question when we clip the divergence in $|t|^{-n}$ at $t$ close to zero? – Satadru Bag Apr 20 '22 at 03:23

1 Answers1

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If you cutoff your function like in your point 1., then it becomes a Lipshitz function (its gradient is bounded, but by a large constant). And more regularity of the function implies more decays of the Fourier transform, so yes, this can indeed change completely the behavior at infinity. More precisely, $$ |\nabla f| = n \,|t|^{-n-1} \,\mathbf{1}_{|t|>10^{-6/n}} $$ Since $|\mathcal{F}(g)| \leq \int |g|$ and $|\mathcal{F}(\nabla f)|=|2\pi x \,\mathcal{F}(f)|$, it yields $$ |\mathcal{F}(f)| \leq \frac{n}{\pi\,|x|}\,\int_{10^{-6/n}}^\infty \frac{\mathrm d t}{t^{n+1}} = \frac{10^6}{\pi\,|x|} $$ Notice that this rate of $1/|x|$ is the expected one as your cutoff is not twice differentiable, so I am surprised that you see a $e^{-|x|}$, except if your cutoff at $10^6$ is actually smooth, or if you have some numerical noise. A function like $(10^{-12}+|x|^2)^{-n/2}$ would for instance look perhaps similar to your function but indeed have an exponentially decaying Fourier transform.

Your second point can be easily tested on Gaussian functions for example, for which the Fourier transform is another Gaussian changing exactly as you say. Your question is not precise mathematically speaking but a way to measure this kind of phenomenon is actually the Heisenberg incertitude principle in quantum mechanics. There are a lot of other inequalities of this kind. Usually however, the principle is more the following: local boundedness and regularity of the function implies decay of its Fourier transform, and reciprocally, decay of the function implies boundedness and regularity of its Fourier transform.

LL 3.14
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  • Thank you so much. For completion, can you please tell the FT of $f(t)=\frac{1}{1+|x|^n}$ or $f(t)=\frac{1}{1+(x^2)^{n/2}}$ where $n>0$ ? – Satadru Bag Apr 22 '22 at 03:24
  • Should I revise the question accordingly? – Satadru Bag Apr 22 '22 at 03:25
  • These one exactly I do not know if they have a simple expression. Notice that unless $n$ is even, they are not very regularity. The one that are better known are of the form $1/(1+x^2)^n$ (notice the parentheses). When $n=(d+1)/2$ with $d$ the dimension of the space, this is exactly the exponential (see https://math.stackexchange.com/questions/4220006/computing-the-fourier-transform-of-exponential-decay-in-mathbbr2/4220125#4220125) – LL 3.14 Apr 23 '22 at 19:43
  • For general $n$, they are known as Bessel potentials, have an integral formula similar to one obtained in the linked answer, with the same proof except last step. They have exponential decay. – LL 3.14 Apr 23 '22 at 19:44
  • Great. I wanted to mention in a paper something like "for a generic $f(t)=\frac{1}{1+|x|^n}$ the FT decays exponentially, larger the $n$ is faster the FT decays." Can I cite to some book or article in support? – Satadru Bag Apr 25 '22 at 06:22
  • Hi, I tried my hands on the Bessel potentials but did not succeed. Could you please refer me to some article which can be used to obtain the F.T. of $f(t)=\frac{1}{1+|x|^n}$ or $f(t)=\frac{1}{(1+|x|)^n}$ in terms of Bessel potentials? – Satadru Bag May 03 '22 at 03:09
  • I asked this question separately here: https://math.stackexchange.com/questions/4435550/fourier-transform-of-fx-frac11xn-and-ft-frac1-left1x-rig – Satadru Bag May 03 '22 at 03:10
  • No, larger $n$ does not imply larger decay of the Fourier transform ! And the functions you mention are not regular if $n$ is odd so their FT does not decay exponentially. Only the regularity of a function is important to understand the decay of its FT – LL 3.14 May 07 '22 at 23:06