Decay of fractional derivative of Schwartz function Let $\phi$ be a Schwartz function and let $\alpha>0$. I want to analyze the decay as $x\rightarrow\infty$ of: $$\int_\mathbb{R}e^{2\pi i x\xi}|\xi|^\alpha\phi(\xi)\,d\xi$$Heuristically, for $\xi\ll x$, we have one oscillation from $\xi=0$ to $\xi=1/x$ at which point the amplitude is of order $|x|^{-\alpha}$ so the answer should be $|x|^{-\alpha-1}$. However, I have had trouble making this intuition precise. I have tried smoothly partitioning into $|\xi|\ll x$ and $|\xi|\gg |x|$ and separately doing a dyadic decomposition at level $|x|^{-1}$, but the details have not worked out. I am interested a precise solution as well as any heuristic ways of thinking about this sort of integral. If it would help, I can share what I've tried in more detail.

If all the derivatives $\phi^{(k)}(0)=0$ then $|\xi|^a\phi$ is Schwartz thus so is its Fourier transform.
Otherwise let $\phi^{(k)}(0)$ be the first non-vanishing derivative, wlog assume that $a\in (0,1)$ and $l\in 0,1$ then $$h=sign(\xi)^l |\xi|^a\phi-\frac{\phi^{(k)}(0)}{k!} \xi^k sign(\xi)^l |\xi|^a e^{-|\xi|} =O(|\xi|^{a+k+1})$$ is $k+2+\lfloor a\rfloor$-times differentiable in $L^1+\Bbb{C}\delta$ so that for $|x|\ge 1$ $$\mathcal{F}^{-1}(sign(\xi)^l|xi|^a\phi)$$ $$=\mathcal{F}^{-1}(\frac{\phi^{(k)}(0)}{k!} \xi^k sign(\xi)^l |\xi|^a e^{-|\xi|})+\mathcal{F}^{-1}(h)$$ $$=\frac{\phi^{(k)}(0)}{k!}  ((1-2i\pi x)^{-a-k-1}+(-1)^{k+l} (1+2i\pi x)^{-a-k-1}) \Gamma(a+k+1) + (-2i\pi x)^{-k-2-\lfloor a\rfloor}\mathcal{F}^{-1}(h^{(k+2)})$$
$$ = C x^{-a-k-1}+O(x^{-k-2-\lfloor a\rfloor})$$
