Approximation of an integral with exponentials Given a positive real $a$, is it possible to evaluate a good approximation of the following integral
$$
\int_{\mathbf{R}}\mathrm{exp}(-ae^{-\frac{x}{\sqrt{2}}}-x^2)\,\mathrm{d}x\,\,\,\,\,\,?
$$
 A: The integrand has its maximum at $x = \sqrt{2} W(a/4)$, where $W$ is the Lambert $W$ function.  You should be able to use Laplace's method when $a$ is large.
EDIT: That is, you approximate the integrand by 
$\exp(f(x_0) + f''(x0) (x-x_0)^2/2))$, where $f(x) = -a \exp(-x/\sqrt{2} - x^2$
and $x_0 = \sqrt{2} W(a/4)$.  The result is that the integral is approximately
$$ {\frac {{{\rm e}^{-2\,{\rm W} \left(a/4\right) \left( {\rm W} \left(a/
4\right)+2 \right) }}\sqrt {\pi}}{\sqrt {{\rm W} \left(a/4\right)+1}}}
$$
Thus for $a=10$, the numerical value of the integral as found by Maple is 
$0.004351294730$ while the approximation above is $0.004357452968$.
A: By setting $x=\sqrt{2}\log t$ and exploiting parity, the given integral equals
$$ \sqrt{2}\int_{0}^{+\infty}\frac{e^{-2\log^2 t}}{t}\,e^{-at}\,dt $$
that is the Laplace transform of $\frac{e^{-2\log^2 t}}{t}$. On the interval $(1,+\infty)$ such function can be approximated with a Padé approximant at $t=1$
$$ \frac{e^{-2\log^2 t}}{t}\approx \frac{1}{t+2(t-1)^2} $$
where the Laplace transform of the RHS is given by an exponential integral $\text{Ei}$ evaluated at complex arguments.
