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


1 Answer 1


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 $\log\frac 1r$.

  • 1
    $\begingroup$ @Houa Oops, I forgot about $r^\beta$ in my final answer. I edited. So the contribution of $P_2$ is actually $\log(1/r)$. However, it is still $P_1$ that dominates (contributing $\log^2(1/r)$, so $J(r)\asymp r^{-3}\log^2(1/r)$ in full agreement with what you wrote). As to question 1, by my (corrected) formula the integral is $\asymp r^{-1}\int_{\log(1/r)P_{0,1}}e^{-y}dy+\int_{\log(1/r)P_{1,0}}1dy$. From the definition with linear inequalities (just drop $\log(1/r)$ on the RHS), $P_{1,0}=\{0\le y\le 1\}=[0,1]$ and $P_{0,1}=\{y>0,-y\le-1\}=[1,+\infty]$ so you get your $\log(1/r)+1\asymp\log(1/r)$ $\endgroup$
    – fedja
    Jul 29, 2020 at 12:02
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    $\begingroup$ @Houa Is it clearer now? If not, what should I clarify? (I addressed two explicit questions you asked but I suspect more clarification may be still needed, so don't hesitate to ask for it but try to explain what exactly you are confused about) $\endgroup$
    – fedja
    Jul 29, 2020 at 12:25
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    $\begingroup$ (3) Yes, of course. As I said "ignore everything with empty interior". (4) It follows from the physical meaning of your particular problem: since you have a constant term, the integral should converge, and if $\psi$ is unbounded (or the face $F$ on which the maximum is attained is unbounded), you'll certainly just get $+\infty$ immediately. This can happen in the general setting I considered but your particular case is tame. (5) Yes, you can think of it as first rotating the rectangle and then shifting it or you can think of first shifting the origin and then rotating the coordinate frame. $\endgroup$
    – fedja
    Jul 30, 2020 at 14:17
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    $\begingroup$ @Houa (6) Yes, it is related to it (read the previous comment for (3-5); I forgot to address it to you). As to the integral, just integrate term by term. The claim is that if $f(t)$ is any function that equals $p(t)=\sum_k c_kt^k$ ($c_k>0$) near the origin and is always between $0$ and $p(t)$, then $\int_0^\infty f(\delta t)e^{-t}\,dt\approx \sum_k k!c_k\delta^k$ as $\delta\to 0$. $\endgroup$
    – fedja
    Jul 30, 2020 at 14:24
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    $\begingroup$ @Houa That's all right, but note that I also first shifted the origin to the point where the maximum is attained (so I should rather say "making $t=p-\psi(y)$ a new coordinate" if I had used the original $y$, not $y$ with respect to the shifted origin), in which case the shift in the argument of $S_P$ disappears and our formulae agree. I apologize for being somewhat confusing here $\endgroup$
    – fedja
    Jul 31, 2020 at 14:27

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