**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



How to estimate the order of this integral with parameter
It looks like you care only about the order of magnitude (i.e., an answer up to a constant factor), in which case it is fairly easy.


First, ignore all coefficients. Setting them to $1$ just changes the answer at most constant number of times. Now, suppose we have the denominator of the form $\sum\_{(\alpha,\beta)} x^\alpha r^\beta$ where $\alpha$ is a multi-index with real entries and $\beta$ is a real number. The sum is assumed to be finite. Make the change of variable $x\_j=e^{-y\_j}$. Now, at each point, only the maximal term matters (up to a factor that is the total number of terms). In terms of $y$'s, the condition of maximality of $x^\alpha r^\beta$ is $y\_j\ge 0$,
$$
\langle y,\alpha-\alpha'\rangle\le (\beta'-\beta)\log(1/r)
$$
for all $(\alpha',\beta')\ne(\alpha,\beta)$. This domain is just a fixed polyhedron $P\_{\alpha,\beta}$ stretched $\log(1/r)$ times (we keep only those with non-empty interiors in what follows; also I call it a "polyhedron" though, technically, it can be unbounded). Thus,
$$
J(r)\asymp\sum\_{(\alpha,\beta)}r^{-\beta}\int\_{(log\frac 1r)P\_{\alpha,\beta}}e^{\psi\_{\alpha,\beta}(y)}\,dy
$$
where $\psi\_{\alpha,\beta}(y)=\langle \alpha-e,y\rangle$, $e=(1,\dots,1)$.


Now the life becomes straightforward. All you need is to find the order of magnitude of each integral. I'll drop the indices $\alpha,\beta$ for brevity. Let $F$ be the face of $P$ on which $\psi$ attains its maximum $p$ and let $d$ be the dimension of $F$. If $\psi\equiv 0$ (i.e., $\alpha=e$), we just have $F=P$ and $\int\_{(\log\frac 1r)P}e^{\psi}=V(P)\log^d(1/r)$. Consider now the non-trivial situation when $\psi$ is not $0$. Then we can rotate and shrink the coordinate system so that $-\psi(y)$ becomes a new variable $t$. Also we can shift $P$ along this coordinate so that the face $F$ lies on the corresponding coordinate hyperplane $\{t=0\}$. Then the integral in question is just
$$
e^{p\log(1/r)}(\log^{D-1}\frac 1r)\int\_{0}^\infty e^{-t}S\_P(\frac t{\log{1/r}})\,dt
$$
where $S\_P(\tau)$ is the $D-1$-dimensional volume of the cross-section of $P$ by the hyperplane $\{t=\tau\}$. By the general convex geometry nonsense, for small $\tau$, $S\_P(\tau)=v\_d\tau^{D-1-d}+v\_{d-1}\tau^{D-d}+\dots+v\_0\tau^{D-1}$ where $v\_d>0$ and then it becomes smaller (look up "mixed volumes" on Google if you are interested in the details), whence the leading term in the integral becomes $\log^d\frac 1r$ with some coefficient depending on $P$. Thus, the final answer for the integral we are interested in with the factor $r^{-\beta}$ is
$$
\asymp r^{-p\_{\alpha,\beta}-\beta}\log^{d\_{\alpha,\beta}}\frac 1r
$$  

We have several competing terms like that, so the winning one is the one with largest $p+\beta$ and among those the one with the largest $d$.


In your last example $x\_1x\_2+x\_1^2r+x\_2r+x\_1r^2+r^3$ (I ignore $r^3$ that can be carried out and all the coefficients), we have $5$ polyhedra and functionals (I drop the trivial restrictions $y\_1,y\_2\ge 0$):
$$
P\_{1,1,0}=\{-y\_1+y\_2\le 1, y\_1\le 1, y\_2\le 2, y\_1+y\_2\le 3\},
\\
\psi\_{1,1,0}(y)=0 
\\
P\_{2,0,1}=\{y\_1-y\_2\le -1, 2y\_1-y\_2\le 0, y\_1\le 1,2y\_1\le 2\}, 
\\
\psi\_{2,0,1}(y)=y\_1-y\_2
\\
et\ cetera.
$$
Here $P\_{1,1,0}$ dominates and yields $\log^2\frac 1r$ but it may be instructive to find the contribution of $P\_{2,0,1}$. In this case (just draw the picture) $p=-1$, $\beta=1$, $d=1$, so we get $\log\frac 1r$.


