Consider the function $$ f(x_1,x_2)=|x_1x_2|^{-\alpha/2}\int_{\mathbb{R}} \frac{e^{it(x_1+u)}-1}{i(x_1+u)} \frac{e^{it(x_2-u)}-1}{i(x_2-u)} |u|^{-\beta}du. $$ It is known that $f(x_1,x_2)\in L^2(\mathbb{R^2})$ if $$ -1/2<\alpha<1, \quad-1/2<\beta<1,\quad \alpha+\beta>1/2. $$ The question is to find an expression for the $L^2$-defined Fourier transform $$ \hat{f}(w_1,w_2)=\int_{\mathbb{R^2}} f(x_1,x_2)e^{-ix_1w_1-ix_2w_2} dx_1dx_2. $$
When $\alpha$ and $\beta$ are nonnegative, using the facts (1) $\int_0^t e^{isx}ds=\frac{e^{itx}-1}{ix}$, (2) $\int_{\mathbb{R}}e^{ixw} |x|^{a} dx=c|w|^{-a-1}$ for $a\in(-1,0)$ and (3) $\int_{\mathbb{R}}e^{ixw} dx=2\pi\delta(w)$, I have guessed (probably can be justified by some regularization argument) that $\hat{f}(x_1,x_2)$ is up to some constant the following
$\alpha>0$, $\beta>0$: $$ \int_0^t\int_0^t |u-v|^{\beta-1}|u-x_1|^{\alpha/2-1} |v-x_2|^{\alpha/2-1} du dv. $$
$\alpha>1/2$, $\beta=0$: $$ \int_0^t |u-x_1|^{\alpha/2-1}|u-x_2|^{\alpha/2-1} du. $$
- $\alpha=0$, $\beta>1/2$: $$ |x_1-x_2|^{\beta-1} I\{0<x_1,x_2<t\}. $$
But what about the cases: $\alpha>1/2$, $\beta<0$ and $\alpha<0$, $\beta>1/2$? I don't even know how to heuristically guess the answer.