# Strange substitution made in a paper to find asymptotics

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 $$\operatorname{Ai}^3(x) \sim i C_1 2^{3 / 2} 3 \pi x^{-3 / 4} e^{-3 \xi}$$

• Even before that question: is $T$ in equation (2.1) supposed to be $t$? What caused the exponential term in the integral to go from $\frac5{27}t^3$ to $\frac1{27}t^3$? Commented May 19 at 15:45
• @GregMartin The reason is because we get an $e^{-T}=e^{-\frac{4}{27}t^3}$ from the asymptotics then when plugged into 2.1 we get $e^{\frac{5}{27}t^3 - \frac{4}{27}t^3}=e^{\frac{1}{27}t^3}$ Commented May 19 at 15:52

To put this in a form readily amenable to use of the saddle-point method (i.e. where everything inside the argument of the exponential contains a factor of the same power of $$x$$), one has to do a substitution $$t = x^{1/2}u$$. Hence, if the saddle point is at a value of $$u$$ that's of the order of $$1$$ (as it happens, I think it's at $$u = \sqrt{27/2}$$), and $$x$$ is our large parameter, that guarantees that $$t$$ in the region of the saddle point is large enough to justify using the asymptotic series for the Bessel function. Since the whole basis of the saddle point method is that only the behaviour of the integrand near the saddle point matters, that means that $$t$$ is large enough to justify using the asymptotic series for the Bessel function everywhere we care about.
• @SamKirkiles IIRC, the (proof of large-$x$ asymptotic validity of the) saddle point method needs the contour to be shifted to a particular path anyway (the path along which the imaginary part of $u^3/27-u$ is constant). Commented May 19 at 16:44
• Ohh!! I see now! So as we make $x$ large it makes t large and t is the thing that's inside the $K_{1/6}(T)$? Commented May 19 at 17:18