$$\int_{0}^{\infty} e^{-t-x / t^{2}} d t$$ as $x \rightarrow 0$.
I understand that this is a Laplace integral with a moving maximum and that I will need to rescale to get a new variable $s=tx^{-1/3}$ to convert the integral into a standard Laplace integral. I know how to solve the problem as $x \rightarrow \infty$ (two bounds of $s$ is from 0 to $\infty$). However, when $x \rightarrow 0$ the two bounds for two bounds of $s$ should also be from $0$ to $\infty$. then what is the difference between these two limits?
Without Laplace method.
Using algebra, this integral involves the Meijer G function $$I=\int_0^\infty e^{-t-\frac{x}{t^2}}\,dt=\frac{1}{\sqrt{\pi }}\,G_{0,3}^{3,0}\left(\frac{x}{4}| \begin{array}{c} 0,\frac{1}{2},1 \end{array} \right)\qquad \text{if} \qquad \Re(x)>0$$
Expanded around $x=0$, $$I=1-\sqrt{\pi x} -\frac{1}{2} x \left(\log (x)+2 \gamma -1-2\log (2)-\psi \left(-\frac{1}{2}\right)\right)+O\left(x^{3/2}\right)$$