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

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  • $\begingroup$ 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 $\frac1{27}t^3$? $\endgroup$ Commented May 19 at 15:45
  • $\begingroup$ @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 $e^{\frac{5}{27}t^3 - \frac{4}{27}t^3}=e^{\frac{1}{27}t^3}$ $\endgroup$
    – random0620
    Commented May 19 at 15:52

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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.

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  • $\begingroup$ Good insight! But it seems they make this substitution before applying the saddle point method $\endgroup$
    – random0620
    Commented May 19 at 16:24
  • $\begingroup$ @SamKirkiles Right, but they know in advance that they're planning to use the saddle point method, and therefore only need to care about the form of the integrand in a region where it's OK to approximate the Bessel function in this way. $\endgroup$ Commented May 19 at 16:31
  • $\begingroup$ What about sliding the contour sufficiently far to the right so that every point is big enough, and then using the fact that the integral over the translated contour equals that of the original? $\endgroup$
    – random0620
    Commented May 19 at 16:37
  • $\begingroup$ @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). $\endgroup$ Commented May 19 at 16:44
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    $\begingroup$ Ohh!! I see now! So as we make $x$ large it makes t large and t is the thing that's inside the $K_{1/6}(T)$? $\endgroup$
    – random0620
    Commented May 19 at 17:18

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