Asymptotics of the Fourier transform of a non-analytic function involving an exponential

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While reading this article, the I was puzzled by the following statement, used several times in the article. Consider a function $f(x)$ whose Fourier transform is $$ \hat{f}(k) = \exp\left(-c {|k|^\alpha}\right), $$

for some $c>0$. Then the author claims that, "The asymptotics at large x are determined by the first non-analytical term in the series expansion of the exponent", which is $c|k|^\alpha$.

Can anyone tell me where this comes from? No sources are cited for this step in the paper.

The author then goes on to write $$ f(|x|\to \infty) \approx \int_{-\infty}^\infty -c |k|^\alpha \exp(ikx) \frac{dk}{2\pi}, $$ which of course is not defined as such, but he uses Abel summation in the form $$ \int_0^\infty \exp(-it)t^{\alpha+2m} dt = (-1)^{m+1} i e^{-i(\alpha \pi/2)} \Gamma(\alpha+2m+1), $$ to conclude that $$ f(|x|\to \infty) \approx \frac{const.}{|x|^{\alpha+1}}, $$ with $const. = \frac{c}{\alpha} \sin(\pi\alpha/2)\Gamma(\alpha)/\pi$. Can anyone explain to me what is going on in the first step where the approximation happens? I'm willing to put up with the ill-defined integral in the second step if I can understand that.

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For $c >0,\alpha > 0, \alpha\not \in 2 \Bbb{Z}$, $\phi \in C^\infty_c(\Bbb{R}), \phi \ge 0,\hat{\phi}(0)=1$, $m\alpha> n > \alpha + 1$, $\hat{\phi}^{(l)}(0)=0$ for $1\le l\le n$, as $x\to \infty$, due to the decay of the Fourier transform of $C^n(\Bbb{R})$ functions whose derivatives are integrable $$\int_{-\infty}^\infty e^{-c |k|^\alpha} e^{ikx}dk = \int_{-\infty}^\infty e^{-c |k|^\alpha} \hat{\phi}(k) e^{ikx}dk+O(x^{-n})$$ $$= \int_{-\infty}^\infty \sum_{l=1}^m \frac{(-c |k|^\alpha)^l}{l!} \hat{\phi}(k) e^{ikx}dk+O(x^{-n})$$

With the convolution theorem $$ = \sum_{l=1}^m \frac{(-c)^l}{l!} \phi\ast \mathcal{F}^{-1}[|k|^{\alpha l}](x)+O(x^{-n})$$

With a bit of complex analysis and theory of distributions we get that for $x\ne 0$, $\mathcal{F}^{-1}[|k|^{\alpha l}](x)$ has a closed-form $=h(\alpha l) |x|^{-\alpha l-1}$ so that
$$\int_{-\infty}^\infty e^{-c |k|^\alpha} e^{ikx}dk =\sum_{l=1}^m \frac{(-c)^l}{l!} \phi\ast h(\alpha l) |x|^{-\alpha l-1} + O(x^{-n})$$ $$= \sum_{l=1}^m \frac{(-c)^l}{l!} h(\alpha l) (|x|^{-\alpha l-1}+O(x^{-\alpha l-2})) + O(x^{-n})$$