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.
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?
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.
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.