Let $\sigma(x,y)=\sigma(r),r=\sqrt{x^2+y^2}$ be a rotation invariant probability density funtion(PDF) in the unit disk $\mathbb{D}$, that is $\int_{\mathbb{D}}\sigma(x,y)dxdy=1$, the rotationn invariant property of $\sigma(r)$ means that $\sigma(re^{i\theta_1})=\sigma(re^{i\theta_2})$. Then the 2D Riemann-Lebesgue Lemma tells us that:
$$\int_{\mathbb{D}}\sigma(r)cos(X\cdot t)d X\rightarrow 0 ~~~\text{as}~~t\rightarrow \infty$$
Where $t=(t_1,t_2)$ and $X=(x,y)$ is random variable in the disk of $\mathbb{D}$, and $X\cdot t=xt_1+yt_2$.
My question: Does this have any asymptotic estimate attached to it? i.e. for any PDF $\sigma(r)$. say:
$$\int_{\mathbb{D}}\sigma(r)cos(X\cdot t)d X=O(\frac{1}{|t|}), ~~~\text{as}~~t\rightarrow \infty$$
or if there exists a constant $\beta>0$ depending on $\sigma(r)$ such that
$$\int_{\mathbb{D}}\sigma(r)cos(X\cdot t)d X=O(\frac{1}{|t|^\beta}), ~~~\text{as}~~t\rightarrow \infty$$
we can cite many examples of $\sigma(r)$. such as $\sigma(r)=\frac{3}{2\pi}\sqrt{1-r^2}$, $\sigma(r)=\frac{1}{2\pi}\frac{1}{r}$, $\sigma(r)=\frac{1}{2\pi}\frac{1}{\sqrt{1-r^2}}$, $0<r<1$.
Does the above estimates right or not. Can you prove that?
Any help I would be very grateful. Thank you very much!