**Some introduction:**
Given a homogeneous structure called "dilation" in $R^n$: For $t\geq 0$
$$D_t: R^n\rightarrow R^n$$
$$D_t(x)=(t^{a_1}x_1,...,t^{a_n}x_n)$$
where $1=a_1\leq...\leq a_n$, and $a_i$ are all integers. And we call $Q=a_1+...+a_n$ the homogeneous dimension. In our problem, we only consider when $Q>n\geq 2$.

Now consider the integral: $$J(r)=\int_{[0,1]^n}\frac{dx}{P(x,r)}=\int_{[0,1]^n}\frac{dx}{f_n(x)r^n+f_{n+1}(x)r^{n+1}+...+f_Q(x)r^Q}$$ where $f_k(x)$ satisfies:

(1) $f_k(D_t(x))=t^{Q-k}f_k(x)$ for all $x\in R^n$ and $t\geq0$

(2) $f_k(x)$ is the combination of some positive monomials. (Examples will be shown below)

(3) $f_Q(x)=Constant>0$. (This property follows from other theorems and propositions, but they are too many so I don't describe them here.)

Four examples are the followings:

(ex1) In $R^2$, $D_t(x)=(tx_1,t^2x_2)$, so $Q=3$. And Let $P(x,r)=x_1r^2+r^3$.

(ex2) In $R^3$, $D_t(x)=(tx_1,tx_2,t^2x)$, so $Q=4$. Let $P(x,r)=(x_1+x_2)r^3+r^4$

(ex3) In $R^3$, $D_t(x)=(t^{1}x_1,t^2x_2,t^{3}x_3)$, so $Q=6$. Let $P(x,r)= x_1^3r^3+(x_2+3x_1^2)r^4+5x_1r^5+3r^6$

(ex4) In $R^3$, $D_t(x)=(t^{1}x_1,t^2x_2,t^{3}x_3)$, so $Q=6$. Let $P(x,r)= x_1x_2r^3+(x_2+2x_1^2)r^4+3x_1r^5+r^6$

(You will find that $x_n$ doesn't make effort. In my work $x_n$ do make no sense in the integral but this follows from other theorems, and it doesn't matter here. )

**Problem:** Find the order of $J(r)$ when $r$ goes to $0^+$. Like the following description.

**Attempt and information:** I guess $J(r)=\frac{1}{r^\alpha}I(r)$, where the $\alpha$ is the "critical value", that is:

(i) $\liminf_\limits{r\rightarrow0^+}I(r)>0$.

(ii) for any $\epsilon>0$, $\lim_\limits{x\rightarrow0^+}r^\epsilon I(r)=0$.

I will give the reason why I guess so in the below. I can show that $g_p(r)=r^p J(r)$, then there exists $p_0$ s.t. when $a<p_0$, $\lim_\limits{r\rightarrow0^+}g_a(r)>0$ and when $a>p_0$, $\lim_\limits{r\rightarrow0^+}g_a(r)=0$. But I can't show $\lim_\limits{r\rightarrow0^+}g_{p_0}(r)>0$, that is, I can't show the (i) above. (see https://math.stackexchange.com/questions/3769564/how-to-find-the-critical-index-a-of-xafx) One gave a counterexample for the proposition in that link. But its counterexample *will not* appear in this problem. Because this is a rational fractional integral. The $I(r)$ I guess will be like the combination of $\log$ and $\arctan$.

The four example have the order estimates:

(ex1) We can calculate directly: $$J(r)=\frac{1}{r^2}\ln(1+\frac{1}{r})=\frac{1}{r^2}I(r)$$ where $ I(r)$ satisfies (i)(ii) above.

(ex2) $$J(r)=\frac{1}{r^3}I(r)$$ where $I(r)$ can be calculate or one can use Dominate convergence theorem to estimate that $I(r)$ satisfies (i)(ii)

(ex3) $$J(r)=\frac{1}{r^{3+2/3}}I(r)$$ see https://math.stackexchange.com/questions/3718932/estimate-a-integral-with-parameter

(ex4) $$J(r)=\frac{1}{r^{3}}I(r)$$ First $$J(r)=\frac{1}{r^3}\int_{[0,1]^2}\frac{dxdy}{xy+(y+2x^2)r+3xr^2+r^3}=\frac{1}{r^3}I(r)$$ we can show $I(r)$ satisfies (i)(ii):

(i) change variables: $$I(r)=\int_{0}^{1/r^2}\int_{0}^{1/r}\frac{dxdy}{xy+(y+2x^2)+3x+1}$$ and then obviously.

(ii) for $3>\epsilon>0$ (the part $\epsilon\geq 3$ follows from the part $3>\epsilon>0$),
$$r^\epsilon I(r)=\int_{[0,1]^2}\frac{r^\epsilon}{xy+(y+2x^2)r+3xr^2+r^3}dxdy=\int_{[0,1]^2}h_r(x,y)dxdy=\int_{(0,1)^2}h_r(x,y)dxdy$$
Pointwisely $\lim_\limits{r\rightarrow0^+}h_r(x,y)=0$ in $(0,1)^2$. Now look for a dominating function in $(0,1)^2$:
$$\frac{1}{h_r(x,y)}\geq \frac{xy}{r^\epsilon}+r^{3-\epsilon}\geq C(xy)^{1-\frac{\epsilon}{3}}$$
So $h_r(x,y)\leq \frac{C}{(xy)^{1-\frac{\epsilon}{3}}}$ in $(0,1)^2$, which is integrable. By DCT, we have $I(r)$ satisfying (i)(ii). **But this method doesn't work in other examples like (ex3)**.

Based on the four examples, I guess $$J(r)=\frac{1}{r^\alpha}I(r).$$ But I can't show how to find the critical value $\alpha$ and even it's difficult to show the existence of critical value