Is there a method to find a tight upper bound on the given integral? Note that the integral is upper bounded by $\sqrt{\pi/2}$, and thus converges. I first thought about applying Laplace's method. However, the function $x^2/2a(1e^{x})$ is decreasing and achieves maximum at $x=0$ which is an endpoint of the domain of integration. As a result, I don't think using Laplace method is a good idea to find an upper bound. Numerical evaluation indicates an asymptotic of $\sim \frac{1}{a}$, but I am not sure how to proceed to say anything about an upper bound in terms of $a$. Any ideas? Thanks in advance.
The Laplace method is applicable to the integral $F(\lambda):=\int\limits_a^b f(x)e^{\lambda S(x)} $ in the case $\max\limits_{x\in [a,b]}S(x)=S(a)$ too. For example, this is described in V. Zorich, Mathematical Analysis II Chap. XIX, Par. 2.4, Theorem 1.
For the user's convenience here is a part of its statement
Then $$F(\lambda)=\frac {f(a)} {S'(a)} \frac {e^{\lambda S(a)}} \lambda (1+O(\lambda^{1})),\, \lambda \to \infty. $$
In the case under consideration $f(x):=e^{x^2/2},\,S(x):=e^{x}1,\,\lambda:=a.$ This gives the asymptotics $\frac 1 a \left(1+O( \frac 1 a) \right) ,\, a\to \infty.$

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$\begingroup$ @user64494 Do you have any advice on what to do when I have a function such that $S'(a)=\infty$, for example, if $a=0$, $S(x)=\sqrt{x}$. Though this case can be handled differently, I just want to see if the Laplace method can be applied here in any way. $\endgroup$ – Samrat Mukhopadhyay Feb 3 '18 at 10:48

$\begingroup$ @Samrat Mukhopadhyay: Can you present the whole integral you ask about? Did you try the change $t=\sqrt{x}?$ $\endgroup$ – user64494 Feb 3 '18 at 10:55

$\begingroup$ Actually, the integral I was asking about in the previous comment is $\int_0^\infty e^{a\sqrt{x}x^2/2} dx$. Although it can be easily upper bounded as $\int_0^\infty e^{a\sqrt{x}}dx$, which then is just $2/\lambda^2$, I want to see if Laplace's method can be applied to this case in any way so that I get a feel about applying this method to more general functions. $\endgroup$ – Samrat Mukhopadhyay Feb 3 '18 at 10:59

$\begingroup$ @Samrat Mukhopadhyay: I don't know whether the Laplace's method directly works for $S(x):=\sqrt{x}$. There are some generalisations of that method for $\int_a^b e^{S(\lambda,x)}\,dx$. $\endgroup$ – user64494 Feb 3 '18 at 11:26
The integral $$I(a)=\int_0^\infty e^{x^2/2a(1e^{x})}dx$$ approaches $1/a$ from above, for an upper bound I tried $$B(a)=a^{1/a1}$$ which seems to work, see this plot of $B(a)I(a)$ versus $a$:

$\begingroup$ Can you kindly elaborate how you obtained the bounds? $\endgroup$ – Samrat Mukhopadhyay Feb 3 '18 at 5:19