Asymptotic of an area integral 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?
 A: $\newcommand{\ep}{\varepsilon}$
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}
