Let $0<\alpha<1$. First, we have an estimate $$\int_0^\infty \frac{e^{ikx}}{x^\alpha} dx = O\left(\frac{1}{k^{1-\alpha}}\right), \quad k\to\infty,$$ obtained by substituting $kx=x'$. It is notable that the power-law singularity at $x=0$ gives the fractional leading order term. This is in contrast with the standard stationary phase lemma, which gives $O(1/k^n)$ for $n\in\mathbb N$. (I would also appreciate if someone gives any references that deals with fractional leading term due to singularity.)
Motivated from this, consider the following situation. Let $f:(0,\infty) \to \mathbb R$ be a $C^1$ function that satisfies $$\lim_{x\to 0^+} f(x) = 0.$$ Hence, we can continuously extend $f$ to the domain $[0,\infty)$. Furthermore, assume that $f$ has compact support. One example of such $f$ is $f(x) = x^\beta \chi(x)$, where $\beta>0$ and $\chi(x)$ is a bump function supported near 0.
My question: do we have $$\int_0^\infty \frac{e^{ikx}}{x^\alpha} f(x)dx = o\left(\frac{1}{k^{1-\alpha}}\right) ?$$
If $f$ can be written as $f(x)=xg(x)$ using a non-singular function $g$ with well-defined $g(0)$, then the problem is trivial by integrating by parts. However, if we cannot assume this, I cannot figure out the answer.
I found a counterexample: $$\int_0^\infty \frac{e^{ikx}}{x^\alpha} e^{\frac ix - x} dx =O\left(\frac{1}{k^{3/4-\alpha/2}}\right),$$ since this is not necessarily of $O(1/k^{1-\alpha})$. The proof of this counterexample is in Proving the asymptotic expansion $\int_0^\infty \frac{e^{ikx}}{x^\alpha} e^{\frac ix - x} dx =O\left(\frac{1}{k^{3/4-\alpha/2}}\right) $.