MHB Tail Behaviour of WIdom-Tracy distribution.

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The discussion focuses on demonstrating the tail behavior of the function F_2(t), defined by an integral involving the function q(t), which satisfies a specific ordinary differential equation and asymptotic relation. The limit being evaluated is given by the expression involving the logarithm of one minus F_2(t), and the initial approach suggests using L'Hopital's rule to progress further. The user expresses uncertainty about the next steps after applying L'Hopital's rule and shares their calculations leading up to this point. Ultimately, the user believes they have found a solution to the problem and plans to share it later. The thread highlights the complexities of analyzing the asymptotic behavior of specific mathematical functions.
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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!
 
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Nevermind, I think I solved it.

I'll post my solution afterwards.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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