The integral converges as it is easily seen to be upper bounded by $\sqrt{\pi/2}$.
 However, [Laplace's method](https://en.wikipedia.org/wiki/Laplace%27s_method) does not seem to work out as the maxima of the function $S(x) = -a\sqrt{1-e^{-x}}-x^2/2$ is located at the end point $0$. [This question](https://mathoverflow.net/questions/291978/finding-a-tight-upper-bound-of-int-0-infty-e-x2-2-a1-e-xdx-a0-a) enquires about a similar problem, however, with the major difference that the $S(x)$ function there is given by $-a(1-e^{-x})-x^2/2$. The second answer to that problem suggests using a modified form of Laplace's method as given by V. Zorich, Mathematical Analysis II Chap. XIX, Par. 2.4, Theorem 1, to tackle the issue of maxima at an endpoint. However, for the problem at hand, this method breaks down as the function $S(x)$ is not differentiable at $0$. So, Laplace's method cannot be applied here. 


I tried using the transformation $1-e^{-x}\to x^2$ to obtain the integral $$\int\_0^1 \frac{\exp(-a x-(\ln(1-x^2))^2/2)}{1-x^2}2xdx$$ From this, intuitively, it seems to me that, at least for large $a$, the integrand is concentrated highly around $0$, and there it seems to be approximated ``well'' by $2xe^{-ax}$, which produces a $\sim\frac{1}{a^2}$ trend. However, all this is very intuitive and I don't know how to transform this intuition into rigorous statements. Also, this intuition seems to serve well for getting asymptotics, but my true intention is to obtain tight upper bounds. As Laplace's method seems not to be a suitable choice, I do not have much idea about how to proceed to say anything about an upper bound. Please help.



Finding a tight upper bound of $\int_0^\infty e^{-a\sqrt{1-e^{-x}}-x^2/2} dx$ as a function of $a$, $a>0$
Here again, like in the previous question the Laplace method [DLMF](https://dlmf.nist.gov/2.3#E15) applies. With $p(x)=\sqrt{1-\exp(-x)}$ and $q(x)=\exp(-x^2/2)$, we have the expansion
$$ p(x)=x^{1/2}-\frac1 4 x^{3/2}+\frac 5{96}x^{5/2}+\cdots$$
which is of the form
$$p(x)\sim p(0)+\sum\_{s=0}^{\infty}p\_{s}t^{s+\mu}$$ with $\mu=\frac1 2$ and $p(0)=0$.
One may calculate the coefficients for the expansion using the residue formula in the link above to obtain
$$I(a)\sim \frac 2{a^2}+\frac {12}{a^4}+\frac {120}{a^6}+\cdots$$


