Follow up question: asymptotics of a two dimensional integral This is a follow up question of Asymptotics of a two dimensional integral about asymptotics of integrals.
The problem is to find the leading order term of this integral.
$$\int_0^1d\epsilon\int_{-\epsilon}^\epsilon dt\,\sin[(\sqrt{1-(\rho+t)^2}-\sqrt{1-\rho^2})^{1/3}]e^{-N t^2},$$
where $\rho=1-\epsilon$, $N\rightarrow \infty$.
I feel the difficulty is that we cannot expand $\sin[(\sqrt{1-(\rho+t)^2}-\sqrt{1-\rho^2})^{1/3}]$ as a convergent series of $\epsilon$ and $t$ since they can be both small values.
I thank all the previous answers from Gary and DinosaurEgg. In the previous question Asymptotics of a two dimensional integral, switching the order of integral can significantly simplify the question and eventually gives a satisfying answer. I am sorry for being kind of picky about the method used there.  But I think the solution depends on the fact that we can get an analytical solution for the first layer of integral. The goal of making up these problems is to solve a more complex integral posted here Follow up question about Asymptotic behavior of integral with Laplace's method, where I have tried this method and failed to get an answer. I feel general method is indeed needed.
So, I try to make up another problem here, where the analytical solution is made no longer possible on purpose even after switching the order of integral. The main difficulty is to find the proper way of expanding the $\sin[(\sqrt{1-(\rho+t)^2}-\sqrt{1-\rho^2})^{1/3}]$.
 A: First, let's simplify the given integral using the same $\int_0^1 d\epsilon\int_{-\epsilon}^\epsilon dt=\int_{-1}^1 dt\int_{|t|}^1 d\epsilon$. Denote $$g(\rho,t)=\sin\big[({\textstyle\sqrt{1-(\rho+t)^2}-\sqrt{1-\rho^2}})^{1/3}\big],$$ then the integral is $\color{blue}{I_N}=\int_0^1 f(t)e^{-Nt^2}dt$ with $f(t)=\int_0^{1-t}[g(\rho,t)+g(\rho,-t)]\,d\rho$. Writing $$f(t)=\int_0^{1-t}g(\rho,t)\,d\rho+\int_0^t g(\rho,-t)\,d\rho+\int_t^{1-t}g(\rho,-t)\,d\rho,$$ we see that $\int_0^t g(\rho,-t)\,d\rho=0$ after $\rho\gets t-\rho$. Substitute $\rho\gets\rho+t$ in the last integral, and use $g(\rho+t,-t)=-g(\rho,t)$; we see that it equals $-\int_0^{1-2t}g(\rho,t)\,d\rho$, thus $$f(t)=\int_{1-2t}^{1-t}g(\rho,t)\,d\rho=t\int_0^1 g\big(1-t(1+x),t\big)\,dx.$$ Now the integrand can be expanded in powers of $t$, giving the asymptotics of $f(t)$ as $t\to 0^+$, needed for Watson's lemma. For the leading term, we get $\lim\limits_{t\to0^+}t^{-7/6}f(t)=-2^{1/6}L$, where $$L=\int_0^1\left(\sqrt{1+x}-\sqrt{x}\right)^{1/3}\,dx=\frac6{35}\Big((9+\sqrt2)(\sqrt2-1)^{1/3}-3\Big)$$ (evaluated using $\sqrt{1+x}-\sqrt{x}=y$). Hence $\color{blue}{I_N}\asymp-2^{-5/6}\Gamma(13/12)L\color{blue}{N^{-13/12}}$ as $N\to\infty$.
