Asymptotics of second kind Bessel function I would like to obtain asymptotic of $K_{i\nu}(\nu z)$, which the modified Bessel function of the second kind of order $i\nu$ for large $\nu$. The parameter $\nu$ is positive, $\nu>0$ and $z >0$, too.
I have read the corresponding chapter from the book by Olver ("Asymptotics and special functions") (according to DLMF, pages 378-382 from Chapter 8 are relevant). However, my understanding is still poor. Olver suggests as an exercise to derive this asymptotics and states the final answer,
$$K_{i\nu}(\nu z)\approx\sqrt{\frac{\pi}{2\nu}}\frac{\exp(-\pi\nu/2-\nu\zeta)}{(z^2-1)^{1/4}}\left(\sum_{s=0}^{n-1}\frac{(-1)^s\hat{U}_s(\hat{p})}{\nu^s}+\phi_n(\nu,z)\right), \tag{*}$$
where $\zeta=\sqrt{z^2-1}-\text{Arcsec}(z)$. I have tried to rederive this expression and in general I understand how it appears.
However Olver refers to original work by Balogh (link). I have read this paper and my understanding disappeared. One of the final results in this paper is the following expression,
$$K_{i\nu}(\nu z) = \frac{\pi\sqrt{2}e^{-\pi\nu/2}}{\nu^{1/3}}\left(\frac{\eta}{z^2-1}\right)^{1/4}\left(\text{Ai}(\beta)+\frac{\text{Ai}'(\beta)}{\nu^{4/3}}+...\right),(**)$$
where quantities $\eta$ and $\beta$ are given by
$$\frac{2}{3}\eta^{3/2}=\sqrt{z^2-1}-\text{Arcsec}(z),\quad \beta=-\nu^{2/3}\eta.$$
Finally, there is one more paper by Dunster (which is the most of interest for me, link), where the quite similar (up to some redefinitions) expression appears.
In addition, Dunster states that $(**)$ and $(*)$ coincides (cf. (4.8) with (4.25)). I tried my best but still cannot connect $(*)$ and $(**)$ to each other. I do not understand what should I do (I tried to perform large-$\nu$ expansion in Airy functions).
My final goal is to find large $\nu$ asymptotics for $I_{i\nu}(\nu z)$ and $K_{i\nu}(\nu z)$. Can anyone clarify how $(**)$ becomes $(*)$ or simply give more refences with more details?
Further investigation is performed around well-known uniform expansion,
$$K_{\nu}(\nu z)=\sqrt{\frac{\pi}{2\nu}}\frac{e^{-\nu\gamma}}{(1+z^2)^{1/4}}\sum_{s=0}^{\infty}(-1)^s\frac{U_s(p)}{\nu^s}, \tag{***}$$
where $U_s(p)$ is some function and $p=(1+z^2)^{-1/2}$. I am not interested in this term). Naively, I can just replace $\nu\rightarrow i\nu$ and $z\rightarrow -iz$ in order to deal with $K_{i\nu}(\nu z)$. Surprisingly for me, it works: I reproduce the result in Olver book (the very first formula). The most tricky moment is to remind that
$$\text{Arcsec}(z)=i\ln\frac{z}{1+\sqrt{1-z^2}},$$
which seems correct. Finally, there is the following mapping (based on Olver book, Ch. 8),
$$\hat{U}_s(x)=i^sU_s(-ix),\quad \hat{p}=(z^2-1)^{-1/2}.$$
 A: First of all, careful check with replacements $\nu\rightarrow i\nu$ and $z\rightarrow -iz$ for $K_{\nu}(\nu z)$ expansion has shown that such naive replacement does not work.
Next, Airy functions in $(**)$ should be expanded with positive argument (i.e. one should write down expansions for $\text{Ai}(x)$ not for $\text{Ai}(-x)$). In large $\nu$ limit, the second term in $(**)$ does not contribute because it contains greater negative power of $\nu$ in compare with the first term with $\text{Ai}(\beta)$ function ($7/6$ vs $1/2$). For Airy function, the following expansion holds,
$$Ai(x)\sim \frac{e^{-\gamma}}{2\sqrt{\pi}x^{1/4}},\quad \gamma=\frac{2}{3}x^{3/2}.$$
Accurate substitution of all the quantities gives the also called Debye expansion (expansion in terms of elementary functions, not in special functions), which is $(*)$.
The same procedure can be performed for $I_{i\nu}(z)$. The uniform expansion of such function in terms of Airy functions was derived by Dunster, 1990. However, in this paper there is a typo (it seems): the argument of Airy functions should be $(-\nu^{2/3}\xi)$ (cf. with expansion for $L_{i\nu}(z)$ from this paper).
