Asymptotic behavior of integral with Laplace's method I am working on the following integral
$\int_0^1 dx\int_0^1 dT \sqrt{1-(1-\sqrt{x}+\sqrt{xT})^2} e^{-n xT},$
as $n\rightarrow \infty$. The goal is to find the asymptotic behavior of the integral to the leading order of $n$.
Obviously, there is a saddle point at $(x,T)=(0,0)$, which is the main difficulty of this calculation. It seems this is related to the Laplace's method introduced in particular in Chapter VIII of [Wong, R. (2001). Asymptotic approximations of integral]. But when I try to change the variable $y_1=(x+T)/2$,  $y_2=(x-T)/2$,  $y_1=\sqrt{\xi}\cosh{\mu}$,  $y_2=\sqrt{\xi}\cosh{\mu}$. It seems all the higher order terms of $\xi$ in the expansion of $\sqrt{1-(1-\sqrt{x}+\sqrt{xT})^2}$ will contribute to the results.
Thanks for the satisfying solutions. I think the question is already addressed. This question is made up by myself with the goal of addressing a more complex problem posted here
Follow up question about Asymptotic behavior of integral with Laplace's method
This may need some more efforts and possibly a more general method.
 A: Elaborating on Maxim's comment. If $v=xT$, then
\begin{align*}
I(n)&=\int_0^1\!\! {\int_0^1 {\mathrm{e}^{ - nxT} \sqrt {1 - (1 - \sqrt x  + \sqrt {xT} )^2 } \,\mathrm{d}T}\, \mathrm{d}x} \\& = \int_0^1\!\! {\int_0^x {\mathrm{e}^{ - nv} \frac{{\sqrt {1 - (1 - \sqrt x  + \sqrt v )^2 } }}{x}\,\mathrm{d}v} \,\mathrm{d}x} \\ & = \int_0^1 {\mathrm{e}^{ - nv} \int_v^1 {\frac{{\sqrt {1 - (1 - \sqrt x  + \sqrt v )^2 } }}{x}\,\mathrm{d}x} \,\mathrm{d}v} .
\end{align*}
Now,
\begin{align*}
& \int_v^1 {\frac{{\sqrt {1 - (1 - \sqrt x  + \sqrt v )^2 } }}{x}\,\mathrm{d}x} 
\\ & = 2(\sqrt {1 - v}  + (1 + \sqrt v )\arccos (\sqrt v ) - v^{1/4} \sqrt {2 + \sqrt v } \arccos (v + \sqrt v  - 1))
\\ & = (\pi  + 2) - 2\sqrt 2 \pi v^{1/4}  + (\pi  + 2)v^{1/2}  - \frac{{\sqrt 2 }}{2}\pi v^{3/4}  + \frac{1}{3}v +  \ldots 
\end{align*}
as $v\to 0^+$. Thus, by Watson's lemma,
$$
I(n) \sim \frac{{\pi  + 2}}{n} - \frac{{\sqrt 2 \pi \Gamma (1/4)}}{{2n^{5/4} }} + \frac{{(\pi  + 2)\sqrt \pi  }}{{2n^{3/2} }} - \frac{{3\sqrt 2 \pi \Gamma (3/4)}}{{8n^{7/4} }} + \frac{1}{{3n^2 }} +  \ldots 
$$
as $n\to +\infty$.
A: As suggested in the comments by Ian, make the change of variables $u=xT$, $v=x$, after which your integral takes the form
$$\int_0^1\int_0^v \left(1-(1-\sqrt{v}-\sqrt{u})^2\right)^{1/2}e^{-nu}\frac{1}{v}du\,dv.$$
We Taylor expand the square root term to obtain that the integral equals
$$\int_0^1\int_0^v \left(\left(1-(1-\sqrt{v})^2\right)^{1/2}+O(u)\right)e^{-nu}\frac{1}{v}du\,dv=\int_0^1 \frac{\left(1-(1-\sqrt{v})^2\right)^{1/2}}{n}dv+O\left(\frac{1}{n^2}\right).$$
We compute the value of the integral over $v$ to be $\pi/2-2/3$, giving the following final result for the initial expression:
$$\frac{\pi/2-2/3}{n}+O\left(\frac{1}{n^2}\right).$$
