limit of $\int_{[0,n]\times \mathbb[0,n]}e^{-(x^3+y^2)}dxdy$, as $n\to\infty$

Question

I have to compute the limit as $n\to\infty$ of the following: $$\int_{[0,n]\times \mathbb[0,n]}e^{-(x^3+y^2)}dxdy$$ To study the integrability I have though to say: $$\int_{[0,n]\times \mathbb[0,n]}e^{-(x^3+y^2)}dxdy=\int_{\mathbb{R}^+\times \mathbb{R}^+}e^{-(x^3+y^2)}\chi_{[0,n]\times[0,n]}dxdy=$$ From Fubini then I have $$\int_{0}^{\infty}\int_{0}^{\infty}e^{-(x^3+y^2)}dxdy=\int_{1}^{\infty}\int_{1}^{\infty}e^{-(x^3+y^2)}dxdy+\int_{0}^{1}\int_{0}^{1}e^{-(x^3+y^2)}dxdy$$ In $[0,1]$ there are no problems of convergence. In $[1,\infty]$ I can say $e^{-x^3-y^2}\leq e^{-x^2-y^2}$, so: $$\int_{1}^{\infty}\int_{1}^{\infty}e^{-(x^3+y^2)}dxdy\leq \int_{1}^{\infty}\int_{1}^{\infty}e^{-(x^2+y^2)}dxdy$$ for this last integral I have thought to use polar coordinates to prove the convergence...but I can't fix the interval of $r$ and $\theta$, considering $x>1$ and $y>1$.

Questions: 1) Extremes of integration in Polar coordinates
2)My idea is right or there is a better way of working?

Answer 1

If the only thing you are concerned about is integrability, then without knowing how to evaluate the Gaussian integral $\iint e^{-(x^2+y^2)}\,dx\,dy$, you can use the simple bound $e^x \ge 1+x^m/m!$ for any $m\ge 1$ and $x\ge 0$ to get $$ e^{-x^3-y^2} \le e^{-(x^2+y^2)} \le \frac{1}{1+(m!)^{-1}(x^2+y^2)^m}, \qquad x,y\ge 1. $$ You should be able to verify in an elementary fashion that the latter function is integrable on $\mathbb R^2$ provided we choose $m>2$ (say) by dividing $\mathbb R^2$ into level sets of $x^2+y^2$.

Answer 2

You have that for fixed $y$, $x\mapsto e^{-x^3-y^2}$ is integrable on $\mathbb R_{+}$ and $$\int_{0}^{\infty} e^{-x^3-y^2}\mathrm d x = \left(\int_0^{\infty} e^{-x^3}\mathrm d x\right) e^{-y^2}.$$ Now the function $$y\mapsto \left(\int_0^{\infty} e^{-x^3}\mathrm d x\right) e^{-y^2}$$ is clearly integrable on $\mathbb R_{+}$. This proves that $(x, y) \mapsto e^{-x^3-y^2}$ is integrable on $\mathbb R_{+}^2$.