I have trouble to use Laplace method to find the asymptotic approximation as $x\rightarrow \infty$ of the following integral ($g(t)$ continuous and $g(0)\neq 0$):
$$\int_0^{\infty} e^{-xt^3}g(t)dt$$
If this is $xt^2$, then I could use Gaussian integral, which is related to the following nice discussion: Using Laplace’s method to find the leading-order of an integral with two variables in the exponent
But now it is $xt^3$, how do I solve this problem? please advice, thanks
I think it should go through just as in the linked answer, it's just that $$\int_0^{\infty}e^{-xt^3}dt=x^{-1/3}\int_0^{\infty}e^{-t^3}dt=\frac13x^{-1/3}\int_0^{\infty}e^{-t}t^{-2/3}dt=\frac13x^{-1/3}\Gamma\left(\frac13\right)$$ So you hope to arrive at the answer $$\int_0^{\infty}e^{-xt^3}g(t)dt\approx\frac13x^{-1/3}\Gamma\left(\frac13\right)g(0)$$ I.e. just the same as the previous problem with all the $2$'s changed to $3$'s.