A question about Fourier transform of function of the type $Q(x)(1+P(x))^{z}$

For simplicity, consider in $\mathbb{R}^3$, and the Fourier transform of the following function $$f=(x_1+x_2+x_3)(1+|x|^2+x_1^2(x_2^2+x_3^2)+x_2^2x_3^2)^{-t+is},~~ \frac12<t<1,~~s\in \mathbb{R}.$$

Note that the function is not in $L^1$, in fact $f$ behaves like $|x|^{-2t+1}$ at $\infty$. So the Fourier transform is taken in the sense of tempered distribution.

Note also that the function is not radial symmetric, however $f$ is symmetric in the sense that if any of the variables are interchanged, one obtains the same function. I'm interested in the quantitative behavior of $\hat{f}$ near the origin. Especially, for what values of $\sigma_i\ge 0$,$i=1,2,3,$ such that $$\hat{f}(\xi)\leq C|\xi_1|^{-\sigma_1}|\xi_2|^{-\sigma_2}|\xi_3|^{-\sigma_3},~~|\xi|<1.$$ The following is an "unsuccessful" approach. First let's estimate the Fourier transform of $f(x)=1/{(1+|x|^2+x_1^2(x_2^2+x_3^2)+x_2^2x_3^2)^{t+is}}$. Then one can write $$f(x)=\frac{1}{(B^2+A^2x_1^2)^{t+is}},$$ where$A^2(x_2, x_3)=1+x_2^2+x_3^2$, $B^2(x_2, x_3)=A^2+x_2^2 x_3^2$. Recall that \begin{align}\label{equ4.3} \mathcal{F}^{-1}(\frac{1}{(1+|x|^2)^z})(\xi)=\pi^{-\frac{n}{2}}2^{-(\frac{n}{2}+z-1)}\frac{1}{\Gamma(z)}|\xi|^{z-\frac{n}{2}}K_{\frac{n}{2}-z}(|\xi|), \end{align} where $K_v$ denotes the modified Bessel function of the second kind. $K_v$ satisfies \begin{align}\label{equ4.4} \frac{d}{dz}(z^{-v}K_v(z))=z^{-v}K_{v+1}(z), ~~~ \forall v\in \mathbb{C}. \end{align}

Combine the above two formula and by scaling, one has \begin{align}\label{equ4.6} I(\xi)&\triangleq \int_{\mathbb{R}^3}{e^{ix\cdot\xi}f(x)dx}\nonumber\\ &=\pi^{-\frac{1}{2}}2^{\frac12-z}\frac{1}{\Gamma(z)} \int_{\mathbb{R}^{2}}{e^{ix'\cdot\xi'}B^{1-2z}A^{-1}(\frac{B}{A}|\xi_1|)^{z-\frac12}K_{\frac12-z}(\frac{B}{A}|\xi_1|)dx'}\nonumber\\ &=\pi^{-\frac{1}{2}}2^{\frac12-z}\frac{|\xi_1|^{z-\frac12}}{\Gamma(z)} \int_{\mathbb{R}^{2}}{e^{ix'\cdot\xi'}A^{-2z}(x')(\frac{B}{A})^{\frac12-z}K_{\frac12-z}(\frac{B}{A}|\xi_1|)}dx', \end{align} In order to take out the factor $\frac{B}{A}$ from $K_{\frac12-z}(\frac{B}{A}|\xi_1|)$, we can use the following multiplication theorem of Bessel function \begin{align}\label{equ4.7} \lambda^{\nu}K_{\nu}(\lambda z)=\sum_{l=0}^{\infty}\frac{(-1)^l}{l!}(\frac{(\lambda^2-1)z}{2})^lK_{\nu-l}(z),~~~\lambda>0. \end{align} However, the estimate is fine with respect to $\xi_1$, but it get trouble with $\xi'$ since now it becomes \begin{align} I(\xi)&=c_z\sum_{l=0}^{\infty}\frac{(-1)^l}{2^ll!}(\int_{\mathbb{R}^{n-1}}{e^{ix'\cdot\xi'}\frac{x_2^{2l}x_3^{2l}}{A^{2(z+l)}}} ~~~dx')|\xi_1|^{z-\frac12+l}K_{\frac12-z-l}(|\xi_1|) \end{align} It seems that the integral involved in the series above is too singular that doesn't imply the convergence. I'm not sure how to overcome this issue.

• Do you require $P(x) \geq 0$ for all $x$? In general the function $f$ is not defined on some $x$. Think about $f(x) = (1 - x^2)^{-1}$. – WhatsUp May 30 '16 at 13:18
• P=0, f=1, $\hat f=\delta$ – Piero D'Ancona May 30 '16 at 15:41
• @WhatsUp, I did require $P\ge 0$. Fixed it. Thanks. – Tomas May 31 '16 at 6:28
• @PieroD'Ancona, Thanks, I'm more interested in the case that $P\ne 0$. – Tomas May 31 '16 at 6:30