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I have the following integral $$ I(\varepsilon) = \iint_D \frac{\sqrt{1+|\nabla h(u,v)|^2}}{[(h(u,v)+\varepsilon)^2+u^2+v^2]^2} du dv, $$ where $h$ is a smooth function with $h(0,0)=0 = h_u(0,0) = h_v(0,0)$, $D$ is a disk centered at the origin.

It seems like that the asymptotic of $I(\varepsilon)$ depends only on the first order behavior of $h$ at the origin, so the dominant term of $I(\varepsilon)$ is the same as that with $h\equiv 0$ and $$I(\varepsilon) \sim \frac{\pi}{\varepsilon^2}, \quad \varepsilon \rightarrow 0.$$

Any suggestion on how to prove/disprove it?

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    $\begingroup$ Does "smooth" mean $C^1$? $\endgroup$ – fedja Feb 26 at 23:57
  • $\begingroup$ I was thinking $C^\infty$, or even just $h(u,v) = \kappa_1 u^2 + \kappa_2 v^2$ as a start. $\endgroup$ – TYp Feb 26 at 23:59
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    $\begingroup$ Then split the domain of integration into $u^2+v^2\le\varepsilon^{4/3}$ and the complement. $\endgroup$ – fedja Feb 27 at 0:00
  • $\begingroup$ Do you mean that the dominant asymptotic of $I(\varepsilon)$ comes from restricting the integral to the $\varepsilon^{4/3}$ disc? Why is it so? And would it help proving the $\pi/\varepsilon^2$ asymptotic? Thanks. $\endgroup$ – TYp Feb 27 at 0:23
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    $\begingroup$ $\varepsilon^{2/3}$- disk, that is. Iosif used $3/4$ instead of my $2/3$, but the logic is the same. $\endgroup$ – fedja Feb 27 at 1:55
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$\newcommand{\ep}{\varepsilon}$ This is indeed a matter of splitting the integral. In polar coordinates, for some real $R>0$, \begin{equation} I(\ep) = \int_0^{2\pi}(J_t(\ep)+K_t(\ep))\,dt, \end{equation} where \begin{equation} J_t(\ep):=\int_0^{r_*}\frac{\sqrt{1+|\nabla h|^2}}{\big((h+\ep)^2+r^2\big)^2}\,r\,dr, \end{equation} \begin{equation} K_t(\ep):=\int_{r_*}^R\frac{\sqrt{1+|\nabla h|^2}}{\big((h+\ep)^2+r^2\big)^2}\,r\,dr, \end{equation} $h:=h(r\cos t,r\sin t)$, $\nabla h:=\nabla h(r\cos t,r\sin t)$,
$r_*=r_*(\ep)>0$ varies with $\ep\downarrow0$ so that $$\ep<<r_*<<\ep^{1/2}$$ (e.g., one may take $r_*=\ep^{3/4}$), $a\ll b$ means $|a|=O(b)$, and $a<<b$ means $|a|=o(b)$.

We have $h\ll r^2$ and $|\nabla h|\ll r$, so that (as $\ep\downarrow0$) uniformly in $r\in[0,r_*]$ we have $h\ll r_*^2<<\ep$, $h+\ep\sim\ep$, $(h+\ep)^2+r^2\sim\ep^2+r^2$, $|\nabla h|<<1$, and $\sqrt{1+|\nabla h|^2}\sim1$. So, \begin{equation} J_t(\ep)\sim\int_0^{r_*}\frac{r\,dr}{\big(\ep^2+r^2\big)^2} =\frac1{2\ep^2}\,\int_0^{r_*^2/\ep^2}\frac{ds}{(1+s)^2} \sim\frac1{2\ep^2}\,\int_0^\infty\frac{ds}{(1+s)^2} =\frac1{2\ep^2}. \tag{1} \end{equation} On the other hand, \begin{equation} K_t(\ep)\ll\int_{r_*}^\infty\frac{r\,dr}{(r^2)^2}\le\frac1{r_*^2}<<\frac1{\ep^2}. \end{equation} Thus, indeed \begin{equation} I(\ep)\sim\frac\pi{\ep^2}. \end{equation}

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  • $\begingroup$ It seems like your claim $J_t(\varepsilon) \sim \int_0^{r_\ast} \frac{r dr}{(\varepsilon^2 + r^2)^2}$ follows solely from the assumption that $r_\ast(\varepsilon) << \varepsilon^{1/2}$ (and that $h$ has a vanishing linearization at the origin.) I could not follow the argument, but now also think that it cannot be true: if all I have is an upper bound on $r_\ast$ , then I can make it zero and then $J_t$ is also zero. (Put differently, the $1/(2\varepsilon^2)$ asymptotic cannot follow just from an upper bound for $r_\ast$.) Am I missing anything? $\endgroup$ – TYp Feb 29 at 17:40
  • $\begingroup$ p.s. The proof for $K_t$ growing strictly slower than $1/\varepsilon^2$ is very clear, and uses only the lower bound for $r_\ast$. $\endgroup$ – TYp Feb 29 at 17:53
  • $\begingroup$ @TYp : I have added details on the first asymptotic equivalence, $\sim$, in (1). In that equivalence, you cannot set $r_*=0$, because then the equivalence would become the nonsensical "equivalence" $0\sim0$. More importantly, the lower bound on $r_*$ given by $\varepsilon<<r_*$ is needed for the second asymptotic equivalence in (1). $\endgroup$ – Iosif Pinelis Mar 1 at 1:55
  • $\begingroup$ Now I see the lower bound is used and the doubt is gone. Thanks. The technical issue I have is how that "interchange of integral and asymptotic" in the first equivalence can be justified. For simplicity, if $h(u,v)$ is the radially symmetric $1/2(u^2+v^2)$. Then $J_t(\varepsilon)$ is exactly $\int_0^{\varepsilon^{3/4}} \frac{\sqrt{1+r^2} r dr}{[(r^2+\varepsilon)^2+r^2]^2}$. And the claim is that for small $\varepsilon$ this is equivalent to $\int_0^{\varepsilon^{3/4}} \frac{r dr}{[\varepsilon^2+r^2]^2}$. I am trying to figure out the (elementary?) missing lemma. $\endgroup$ – TYp Mar 2 at 23:09
  • $\begingroup$ @TYp : When people write (as I did) "for $r\le r_*$ as $\epsilon\to0$", they of course mean "uniformly in $r\in[0,r_*]$ as $\epsilon\to0$" (which is how I have now rewritten that). So, the first equivalence in (1) follows by the general fact that, if $f_\epsilon(r)\sim g_\epsilon(r)>0$ uniformly in $r\in[0,r_*]$ (as $\epsilon\to0$), then $\int_0^{r_*}f_\epsilon(r)\,dr\sim \int_0^{r_*}g_\epsilon(r)\,dr$. This fact should be an easy exercise for you. $\endgroup$ – Iosif Pinelis Mar 3 at 3:25

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