Large $t$ asymptotics of $\int_0^{\infty}\exp(-tx)\exp(-\frac{1}{x^2})dx$

Question

How do I find the asymptotic behavior of $$\int_0^{\infty}\exp(-tx)\exp\left(-\frac{1}{x^2}\right)dx$$ as $t\to\infty$? The Laplace method apparently doesn't work since $\exp(-\frac{1}{x^2})$ isn't analytic at $0$, the point where $-x$ has a maximum on $[0,\infty)$.

Answer 1

Since the function $g(x)=\frac{1}{x^2}+tx $ has a minimum in $x=\left(\frac{2}{t}\right)^{1/3}$ and in such a point we have: $$ g(x) = \left(2^{-2/3}+2^{1/3}\right)t^{2/3},\qquad g''(x) = 6\cdot 2^{-4/3} t^{4/3}, $$ by the saddle point method it is expected that the integral for large values of $t$ behaves like: $$ \sqrt{\frac{\pi}{3}}\,\left(\frac{2}{t}\right)^{2/3}\exp\left(-\left(2^{-2/3}+2^{1/3}\right)t^{2/3}\right).$$