# Bound for a certain integral expression

I am working to establish an estimate in $X^{s,b}$ spaces to prove local well-posedness of a certain equation, and I need to consider some sub-cases. In particular, I wish to show that the following integral is bounded in $t$ and $x$, i.e. the following supremum is finite:

$$\sup_{t,x\in \mathbb{R}}\, \frac{\langle x\rangle^{2s}}{\langle t-x^{2\alpha}\rangle^{1-}} \int_{\mathbb{R}^2} \frac{dx_1\;dx_2}{\langle x_1\rangle^{2s} \langle x_2 \rangle^{2s} \langle x-x_1+x_2 \rangle^{2s} \langle t- x_1^{2\alpha}+ x_2^{2\alpha}-(x-x_1+x_2)^{2\alpha} \rangle^{2b}}$$

where the exponent 1- denotes numbers sufficiently close to 1 but less than 1, for $\frac{1-\alpha}{2}< s\leq \frac{1}{2}$, and $\frac{1}{2}<\alpha<1$ and $b>\frac{1}{2}$. Define the notation $\langle z \rangle := (1+z^2)^{1/2}$.

If the above problem is perhaps intractable, I am also happy if the following integral is bounded in $x$ (since then I can do something with the above):

$$\sup_{x\in \mathbb{R}} \,\int_{0<|x_1x_2|\lesssim |x|^{2-2\alpha}} \frac{ \langle x\rangle^{2s}}{\langle x+x_1\rangle^{2s} \langle x+x_2\rangle^{2s} \langle x+x_1+x_2\rangle^{2s}}\; dx_1\,dx_2$$ where similarly $\frac{1-\alpha}{2}< s\leq \frac{1}{2}$, and $\frac{1}{2}<\alpha<1$, and the notation $\langle\cdot\rangle$ as above.

I would be quite happy too if the above holds for other ranges of $s\leq \frac{1}{2}$ for the above $\alpha$.

Thank you.

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Is the region of integration the same as $\{(x_1,x_2): 0<|x_1|\le|x|^{2-2\alpha},\ 0<|x_2|\le|x|^{2-2\alpha}\}$? –  Matt F. Mar 21 at 11:02
@Matt F. : No, the region of integration is more like $\{(x_1,x_2): 0<|x_1|\leq \frac{|x|^{2-2\alpha}}{x_2},\, x_2\in \mathbb{R}\}$. –  digiboy1 Mar 22 at 13:42
What is $x^{2\alpha}$ for $x<0$ ? –  fedja Mar 23 at 23:17
@fedja: for $x<0$, it is $|x|^{2\alpha}$. –  digiboy1 Mar 25 at 3:32