In the quoted section from this paper, why is the author able to "substitute this result into Eq. (2.1)"? This should hold for $z$ large. But not everything on the contour is large. Why can the author make this substitution?
Let $T=\frac{4}{27} t^3$ and $K_{1 / 6}(T)$ be the modified Bessel functions of order $\frac{1}{6}$. Now previously the paper showed$$\operatorname{Ai}^3(x)=C_1 \int_{\mathcal{L}_1} t^{1 / 2} K_{1 / 6}(T) \exp \left(\frac{5}{27} t^3-x t\right) d t \qquad \qquad (2.1)$$
(If you see a rectangle, the symbol is mathcal{L}_1). Next we need to show that the constant $C_1$ can be chosen so that both sides of this equation have the same asymptotic behavior as $x \rightarrow \infty$. For that purpose we first recall that in the complete sense of Watson [3]$$K_{1 / 6}(T) \sim\left(\frac{3}{2}\right)^{3 / 2} \pi^{1 / 2} t^{-3 / 2} e^{-T} \quad\left(|\operatorname{ph} t|<\frac{1}{3} \pi\right)$$Substituting this result into Eq. (2.1) Why? then gives$$\operatorname{Ai}^3(x) \sim C_1\left(\frac{3}{2}\right)^{3 / 2} \pi^{1 / 2} \int_{\mathcal{L}_1} t^{-1} \exp \left(\frac{1}{27} t^3-x t\right) d t$$and an application of the saddle-point method to this integral gives$$\operatorname{Ai}^3(x) \sim i C_1 2^{3 / 2} 3 \pi x^{-3 / 4} e^{-3 \xi}$$
