Suppose $f(x)$ is a probability density on $\mathbb{R}$. Let $\varphi(t)=\int e^{itx}f(x)dx$ denote the Fourier transform (characteristic function). It is well-known that if $\int |x|^p f(x)dx<\infty$, then $\varphi^{(p)}(t)= i^p \int x^p e^{itx} f(x) dx $. If in addition $\varphi^{(p)}\in L^1$, then through inverse Fourier transform $$ (ix)^pf(x)=\frac{1}{2\pi} \int e^{-ixt} \varphi^{(p)}(t)dt $$ one has the bound $$|f(x)|\le \frac{1}{2\pi|x|^p} \int |\varphi^{(p)}(t)|dt,$$ and hence one obtains the tail decay rate $|f(x)|=O(|x|^{-p})$ as $|x|\rightarrow\infty$.
I wonder if there is a generic way to obtain an estimate better than $O(|x|^{-p})$ by working with suitable conditions on $\varphi(t)$. I have this question because for "nice" $f$ (e.g., $f$ decays like a power function) to satisfy $\int |x|^p f(x)dx<\infty$, one typically needs $|f(x)|=o(|x|^{-p-1})$ in order to have integrability for large $|x|$. Hence there is a gap between the expected order of decay $o(|x|^{-p-1})$ and the order $O(|x|^{-p})$ obtained from $\varphi$.
Another perspective is that for the inverse Fourier transform, one has by Riemann Lebesgue Lemma that $\int e^{-ixt} \varphi^{(p)}(t)dt$ tends to zero when $|x|$ tends to infinity. So it seems possible to extract extra decay rate under suitable conditions.
