I am working on the following integral
$$\int_0^1d\epsilon\int_{-\epsilon}^\epsilon dt\left(\sqrt{1-(\rho+t)^2}-\sqrt{1-\rho^2}\right)e^{-N t^2},$$
where $\rho=1-\epsilon$, $N\rightarrow \infty$. The problem is to find the leading order term of this integral.
I feel the difficulty is that we cannot expand $\sqrt{1-(\rho+t)^2}-\sqrt{1-\rho^2}$ as a convergent series of $\epsilon$ and $t$ since they can be both small values.
It turns out that @Gary 's suggestion to reverse the order of integration is a fruitful first step because the integral over $\epsilon$ is analytically tractable. Exchanging the order results in the region
$$\int_{0}^1d\epsilon\int_{-\epsilon}^\epsilon dt=\int_{-1}^1 dt\int_{|t|}^1 d\epsilon$$ and then we can rewrite the desired integral, after changing the dummy variable to $\rho=1-\epsilon$
$$I(N)=\int_{-1}^1dt~ e^{-Nt^2}\int_{0}^{1-|t|}d\rho\left(\sqrt{1-(\rho+t)^2}-\sqrt{1-\rho^2}\right)$$
The integral over $\rho$ is elementary since
$$\int\sqrt{1-x^2}\,dx=\frac{1}{2}\left(x\sqrt{1-x^2}+\arcsin x\right)+C$$
After some algebra, the integral can be reduced to the sum of several integrals on the interval $(0,1)$:
$$I(N)=\frac{1}{2}\int_{0}^1dt ~e^{-Nt^2}\left[\pi/2+2(1-2t)\sqrt{t(1-t)}+\text{arcsin}(1-2t)\\-2(1-t)\sqrt{t(2-t)}-2\arcsin(1-t)\right]$$
Note here that most of the summands have square-root singularities at $t=0.$ This poses no problem for applying Watson's lemma (and the fact that the range of integration is finite can be handled easily, just assume that the integrand is bounded from above). To finish this off, expand
$$[t^n]\sqrt{1-t}=(-1)^n{1/2\choose n}:=c_n$$
$$[t^{n}]\frac{\pi/2-\text{arcsin}(1-t)}{\sqrt{t}}=\frac{\sqrt{2}}{8^n(2n+1)} {2n\choose n}:=d_n$$
and hence the integral can be analytically expanded in an asymptotic nonconvergent series of the form (Noting that $\int_{0}^\infty t^{\alpha}e^{-Nt^2}dt=\Gamma\left(\frac{\alpha+1}{2}\right)/N^{(\alpha+1)/2})$
$$I(N)\sim \sum_{n=1}^{\infty}\left[\left(1-\frac{1}{2^{n-1/2}}\right)c_n+\left(\frac{1}{2^{n-3/2}}-2\right)c_{n-1}-(2^{n-1/2}-1)d_n\right]\frac{\Gamma(n/2+3/4)}{N^{n/2+3/4}}$$