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I want to show that for $F_2(t):=\exp\bigg(-\int_t^\infty (x-t)q(x)^2dx\bigg)$, where $q(t)$ satisfies the ODE and the asymptotic relation respectively: $q''=tq+2q^3$ $q(t)\sim Ai(t)$ as $t\to \infty$; and $Ai(x)$ is Airy function such that for $x>0$: $Ai(x)\sim\pi^{-1/2}x^{-1/4}e^{(-2/3)x^{3/2}}/2$;
$$\lim_{t\to \infty} \frac{1}{t^{3/2}}\log[1-F_2(t)]=-4/3$$
It seems the first step in calculating this limit is L'Hopital's rule, but I don't see how to proceed from there:
$$\lim_{t\to \infty}\frac{1}{t^{3/2}}\log[1-F_2(t)]=\lim_{t\to \infty} \frac{-F_2'(t)}{1-F_2(t)}1/((3/2)t^{1/2})=\lim_{t\to \infty}\frac{-\int_t^\infty q^2(x)dx F_2(t)}{F_2(t)-1}\frac{1}{(3/2)t^{1/2}}$$
How to continue from there? assuming I got the first step correctly.
P.S
I asked this question also in M.SE, in case someone watches over there.
Peace out!
$$\lim_{t\to \infty} \frac{1}{t^{3/2}}\log[1-F_2(t)]=-4/3$$
It seems the first step in calculating this limit is L'Hopital's rule, but I don't see how to proceed from there:
$$\lim_{t\to \infty}\frac{1}{t^{3/2}}\log[1-F_2(t)]=\lim_{t\to \infty} \frac{-F_2'(t)}{1-F_2(t)}1/((3/2)t^{1/2})=\lim_{t\to \infty}\frac{-\int_t^\infty q^2(x)dx F_2(t)}{F_2(t)-1}\frac{1}{(3/2)t^{1/2}}$$
How to continue from there? assuming I got the first step correctly.
P.S
I asked this question also in M.SE, in case someone watches over there.
Peace out!