How fast does $\int_{1 \leq |x| \leq 2}\frac{1}{(1 + N|x_1|)^{10} (1 + |x_2|)^{10}}\, dx$ decay?

Question

Consider the integral (in $\mathbb{R}^2$): $$I_{N} := \int_{1 \leq |x| \leq 2}\frac{1}{(1 + N|x_1|)^{10} (1 + |x_2|)^{10}}\, dx$$ where $x = (x_1, x_2)$. Dominated Convergence implies that $I_{N} \rightarrow 0$ as $N \rightarrow \infty$. Is it possible to extract how fast $I_{N}$ tends to 0? My guess would be $I_{N} \leq CN^{-10}$ for some absolute constant $C$, but I can't seem to bound the integrand since $\{x: 1 \leq |x| \leq 2\}$ contains points with first coordinate equal to 0 or are very small compared to $N$.

Answer 1

Consider the part

$$P_N = \{ x : 1 \leqslant \lvert x\rvert \leqslant 2, \lvert x_1\rvert \leqslant 1/N\}$$

of the annulus. On $P_N$, the integrand is bounded below by $\frac{1}{2^{10}\cdot 3^{10}}$, so

$$I_N \geqslant \frac{1}{6^{10}} \operatorname{vol}(P_N).$$

The volume (or area) of $P_N$ is (approximately) proportional to $N^{-1}$, thus $N\cdot I_N$ is bounded away from $0$.

On the other hand, the change of coordinates $(y_1,y_2) = (Nx_1, x_2)$ gives

$$I_N = \frac{1}{N}\int_{A_N} \frac{1}{(1 + \lvert y_1\rvert)^{10}(1 + \lvert y_2\rvert)^{10}}\,dy \leqslant \frac{1}{N}\int_{\mathbb{R}^2} \frac{1}{(1 + \lvert y_1\rvert)^{10}(1 + \lvert y_2\rvert)^{10}}\,dy,$$

so we find $I_N \in \Theta(N^{-1})$.