Consider the following integral: $$J(u)=\int_{0}^{u^2}\int_{0}^{u}\frac{1}{x^3+y+3x^2+5x+3}dxdy$$ My question: I want to separate it into two parts: $$J(u)=u^aI(u)$$ where $a\geq 0$ and $I(u)$ is "smaller" than polynomials, that is, $I(u)$ satisfies the followings:
(1) $\liminf\limits_{u\rightarrow+\infty}I(u)>0$. (2) for any $p>0$, $\lim\limits_{u\rightarrow+\infty}\frac{I(u)}{u^p}=0$
For example, $I(u)=\log(u)$, or $\arctan u$ etc.
My attempt: I first guess $a=0$, that is, $J(u)$ itself satisfies $I(u)$. But I can't show that property (2) above. Then I guess $a=1$, that is, $J(u)=uI(u)$, I can show that property (2) by change the variables $(\frac{x}{u},\frac{y}{u^2})=(w,s)$ and using Dominated convergence theorem. But this case I can't show the property (1)... And now I use computer to calculate, I guess $a$ may be between $(0,1)$.
Some tips:
(i) if you change the variables, $(\frac{x}{u},\frac{y}{u^2})=(w,s)$ may be better. this is called dilation and it is a kind of "homogeneous"
(ii) If I change $x^3$ into $xy$ in $J(u)$, I can show that actually $a=0$, by Dominated convergence theorem.
Background: Given a dilation in $R^n$: $$D_t(x)=(t^{c_1}x_1,...,t^{c_n}x_n)$$ then consider $P(x,r)=r^nf_n(x)+...+r^Qf_Q(x)$, where $Q=m_1+...+m_n$, $f_k(x)$ satisfies $f_k(D_t(x))=t^{Q-k}f_k(x)$, and $f(x)>0$ are the combinations of positive monomials ($x_i>0$). The origin type of the denominator of the $J(u)$ is $x^3r^3+(y+3x^2)r^4+5xr^5+3r^6$. And the dilation is $D_t(x)=(t^{1}x,t^2y,t^{3}z)$. I just take $1/r=u$ and do some simplification.