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),
$$\hat{U}_s(x)=i^sU_s(-ix),\quad \hat{p}=(z^2-1)^{-1/2}.$$
Best Answer
I begin with a discussion of the uniform asymptotic expansion of $K_\nu(\nu z)$. We start with the integral representation $$\tag{1} K_\nu (z) = \int_0^{ + \infty } {{\rm e}^{ - z\cosh t} \cosh (\nu t){\rm d}t} = \frac{1}{2}\int_{ - \infty }^{ + \infty } {{\rm e}^{ - z\cosh t} {\rm e}^{ - \nu t} {\rm d}t} ,$$ which is valid when $\nu>0$ and $|\arg z|\leq \frac{\pi}{2}$. It is convenient to work with the variable $\zeta \in \mathcal{D}: = \left\{ {w:\Re w \geq 0,\; \left| \Im w \right| \leq \frac{\pi }{2}},\; w\neq 0 \right\}$. Note that the function $\operatorname{csch}$ is a bijection between $\mathcal{D}$ and the sector $|\arg z|\leq \frac{\pi}{2}$. Thus, by $(1)$, $$ K_\nu (\nu \operatorname{csch} \zeta ) = \frac{1}{2}\int_{ - \infty }^{ + \infty } {\exp ( - \nu f(t,\zeta )){\rm d}t} , $$ where $$ f(t,\zeta ) = \operatorname{csch} \zeta \cosh t + t, $$ provided $\nu>0$. The saddle points of $f(t,\zeta )$ are $$ t_e^{(n)} := - \zeta + 2n\pi {\rm i}, \quad t_o^{(n)} := \zeta + (2n + 1)\pi {\rm i}, $$ for any $n\in \mathbb{Z}$. Hence, by deforming the path of integration, $$\tag{2} K_\nu (\nu \operatorname{csch} \zeta ) = \frac{{{\rm e}^{ - \nu (\coth \zeta - \zeta )} }}{2}\int_{\mathscr{C}_e^{(0)} } {\exp ( - \nu (f(t,\zeta ) - f(t_e^{(0)} ,\zeta ))){\rm d}t} , $$ where $\mathscr{C}_e^{(0)}$ is the doubly infinite steepest descent path through the saddle $t_e^{(0)}$. We now parametrise the path of steepest descent $\mathscr{C}_e^{(0)}$ by introducing the new variable $s$: $$\tag{3} s := s(t,\zeta ) := f(t,\zeta ) - f(t_e^{(0)} ,\zeta ). $$ Note that, since $\nu>0$ and the definiton of $\mathscr{C}_e^{(0)}$, $s$ is real and positive on each half of the contour $\mathscr{C}_e^{(0)}$ divided by $t_e^{(0)}$. Thus, corresponding to each $s>0$, there are two values of $t$, which we denote by $t^+$ and $t^-$ that satisfy $(3)$. With this parametrisation, $(2)$ becomes $$\tag{4} K_\nu (\nu \operatorname{csch} \zeta ) = \frac{{{\rm e}^{ - \nu (\coth \zeta - \zeta )} }}{2}\int_0^{ + \infty } {\rm e}^{-\nu s}F(s,\zeta ){\rm d}s , $$ where $$ F(s,\zeta ) := \frac{{{\rm d}t^+(s,\zeta )}}{{{\rm d}s}}-\frac{{{\rm d}t^-(s,\zeta )}}{{{\rm d}s}}. $$ Since $s(t,\zeta)$ is entire and has no finite asymptotic values, the only singularities of $t(s,\zeta)$ (and hence $F(s,\zeta )$) are branch points located at $$ s_e^{(n)} := f(t_e^{(n)} ,\zeta ) - f(t_e^{(0)} ,\zeta ) = 2\pi {\rm i}n, $$ $$ s_o^{(n)} := f(t_o^{(n)} ,\zeta ) - f(t_e^{(n)} ,\zeta ) = - 2(\coth \zeta - \zeta ) + (2n + 1)\pi {\rm i}, $$ for any $n\in \mathbb{Z}$. Note that $F(s,\zeta )$ is the difference of the values of $t(s,\zeta)$ on the two sides of the branch cut running from $s_e^{(0)}=0$. Now we can complexify $\nu$ in $(4)$ by rotating the path of integration so that $\nu s$ remains real and positive. This process works until the path runs into one of the singularities of $F(s,\zeta)$. There are two cases that have to be considered.
$(i)$ If $\Re (\coth \zeta - \zeta ) \geq 0$, $\zeta \in \mathcal{D}$, then all the singularities lie in the closed left half-plane $\Re s \leq 0$. This region corresponds to the grey region in the closed right half-plane in the variable $z=\operatorname{csch} \zeta$ depicted on Figure 8.3 on page 380 of Olver's book Asymptotics and Special Functions. Since $F(s,\zeta)$ remains bounded for $\Re s>0$, an application of the extended Watson lemma (Theorem 3.3 on page 114 of Olver's book) yields $$\tag{5} K_\nu (\nu \operatorname{csch} \zeta ) \sim \sqrt {\frac{\pi }{{2\nu \coth \zeta }}} {\rm e}^{ - \nu (\coth \zeta - \zeta )} \sum\limits_{n = 0}^\infty ( - 1)^n \frac{U_n (\tanh \zeta )}{\nu ^n} $$ as $\nu \to \infty$ in the sector $|\arg \nu|\leq\pi-\delta\,(<\pi)$ provided that $\Re (\coth \zeta - \zeta ) \geq 0$, $\zeta \in \mathcal{D}$. Here, we follow the standard notation.
$(ii)$ If $\Re (\coth \zeta - \zeta )< 0$, $\zeta \in \mathcal{D}$, then the singularities $s_o^{(n)}$ move into the right half-plane. This region corresponds to the white region in the closed right half-plane in the variable $z=\operatorname{csch} \zeta$ depicted on Figure 8.3 on page 380 of Olver's book. It is not hard to show that in this case $2n\pi \leq \Im s_o^{(n)} \leq (2n + 2)\pi$. In particular $0 \leq \Im s_o^{(0)} \leq 2\pi$ and $- 2\pi \leq \Im s_o^{( - 1)} \leq 0$. Since $\Re s_o^{(n)}$ is the same for all $n$, the path rotation argument works until we hit one of the two singularities $s_o^{(0)}$ or $s_o^{(-1)}$. Thus, the $\nu$-sector of validity of $(5)$ will depend on the phases of these singularities in this case. Nevertheless, we can say that $(5)$ will definitely be valid as long as $|\arg \nu|\leq\frac{\pi}{2}-\delta\,(<\frac{\pi}{2})$. Furthermore, if $z=\operatorname{csch} \zeta$ is bounded away from the imaginary axis inside the white domain, $(5)$ remains valid for $|\arg \nu|\leq\frac{\pi}{2}$.
Now, let me turn to your problem. We see from the above analysis that in order to derive the asymptotics of $K_{\mathrm{i}\nu}(\nu z)$ from that of $K_{\nu}(\nu z)$ via the substitutions $\nu \to \mathrm{i}\nu$ and $z \to \mathrm{e}^{-\frac{\pi}{2}\mathrm{i}}z$ ($\nu$ and $z$ are real and positive), we have to assume $z>1$ (a special case of case $(i)$ above). This is because if $0<z\leq 1$, we are on the imaginary axis inside the white region and the asymptotic expansion of $K_{\nu}(\nu z)$ is not valid for $|\arg \nu| =\frac{\pi}{2}$. Indeed, Olver says in Exercise 8.3 on page 382 that, in particular, $0< z\leq 1$ has to be excluded. The expansion for the case $0<z<1$ is \begin{align*}\tag{6} K_{\mathrm{i}\nu } (\nu \operatorname{sech}\alpha ) \sim \sqrt {\frac{{2\pi}}{\nu\tanh \alpha}} {\mathrm e}^{ - \pi \nu /2} & \left( \cos \omega \sum\limits_{n = 0}^\infty {( - 1)^n \frac{{U_{2n} (\coth \alpha )}}{{\nu ^{2n} }}} \right. \\ & \left. \qquad - \sin \omega \sum\limits_{n = 0}^\infty {( - 1)^n \frac{{U_{2n + 1} (\coth \alpha )}}{{\nu ^{2n} }}} \right) \end{align*} as $\nu \to +\infty$ with $\alpha>0$ and $\omega = \nu (\alpha - \tanh \alpha ) - \frac{\pi }{4}$. Note that Balogh mistakenly writes $\sec \alpha$ in place of $\operatorname{sech}\alpha$. For the case $z>1$, the substitutions $\nu \to \mathrm{i}\nu$ and $\zeta \to \mathrm{i}\left(\frac{\pi}{2}-\beta\right)$, $0 < \beta < \frac{\pi}{2}$, in $(5)$ give $$\tag{7} K_{\mathrm{i}\nu } (\nu \sec \beta ) \sim \sqrt {\frac{\pi }{{2\nu \tan \beta }}} {\mathrm e}^{ - \pi \nu /2}\mathrm{e}^{ - \nu (\tan \beta - \beta )} \sum\limits_{n = 0}^\infty {\mathrm{i}^n \frac{{U_n (\mathrm{i}\cot \beta )}}{{\nu ^n }}} $$ as $\nu \to +\infty$. Note that due to the properties of $U_n$, the coefficients of this expansion are real. Both of these expansions break down near the turning point $z=1$ (i.e., when $\alpha=0$ or $\beta=0$). In fact, both of these expansions are useful only if $|\nu-z| \gg \nu^{1/3}$.
Now, there are two ways to overcome the above problem.
$(i)$ For fixed (complex) $\tau$, the following transitional expansion holds: $$ \frac{2}{\pi }{\rm e}^{\pi \nu /2} K_{\mathrm{i}\nu } (\nu + \tau \nu ^{1/3} ) \sim \frac{{2^{4/3} }}{{\nu ^{1/3} }}\operatorname{Ai}(2^{1/3} \tau )\sum\limits_{n = 0}^\infty {\frac{{P_n (\tau )}}{{\nu ^{2n/3} }}} + \frac{{2^{5/3} }}{\nu }\operatorname{Ai}'(2^{1/3} \tau )\sum\limits_{n = 0}^\infty {\frac{{Q_n (\tau )}}{{\nu ^{2n/3} }}} $$ as $\nu \to +\infty$, where $P_n$ and $Q_n$ are polynomials in $\tau$. In particular, $$ P_0 (\tau ) = 1,\quad Q_0 (\tau ) = - \frac{3}{{10}}\tau ^2 . $$ In fact, the above expansion remains valid as long as $|\tau| =o(\nu^{4/15})$. Note that the large parameter $\nu$ is not present in the argument of the Airy function and its derivative.
$(ii)$ The second option is to use the expansion in terms of the Airy function and its derivative obtained by Balogh. This expansion is uniformly valid for $z>0$ as $\nu \to +\infty$. One may separate Balogh's result into two cases: \begin{align*} \frac{{{\rm e}^{\pi \nu /2} }}{\pi }K_{\mathrm{i}\nu } (\nu \operatorname{sech}\alpha ) \sim & \; \sqrt {\frac{{12^{1/3} (\alpha - \tanh \alpha )^{1/3} }}{{\tanh \alpha }}} \\ & \times \left( {\frac{{\operatorname{Ai}(\nu ^{2/3} \zeta )}}{{\nu ^{1/3} }}\sum\limits_{n = 0}^\infty {\frac{{A_n (\zeta )}}{{\nu ^{2n} }}} + \frac{{\operatorname{Ai}'(\nu ^{2/3} \zeta )}}{{\nu ^{5/3} }}\sum\limits_{n = 0}^\infty {\frac{{B_n (\zeta )}}{{\nu ^{2n} }}} } \right) \end{align*} and \begin{align*} \frac{{{\rm e}^{\pi \nu /2} }}{\pi }K_{\mathrm{i}\nu } (\nu \sec \beta ) \sim & \;\sqrt {\frac{{12^{1/3} (\tan \beta - \beta )^{1/3} }}{{\tan \beta }}} \\ & \times \left( \frac{{\operatorname{Ai}(\nu ^{2/3} \zeta )}}{{\nu ^{1/3} }}\sum\limits_{n = 0}^\infty {\frac{{A_n (\zeta )}}{{\nu ^{2n} }}} + \frac{{\operatorname{Ai}'(\nu ^{2/3} \zeta )}}{{\nu ^{5/3} }}\sum\limits_{n = 0}^\infty {\frac{{B_n (\zeta )}}{{\nu ^{2n} }}} \right) \end{align*} as $\nu \to +\infty$, with $\zeta = - (3/2)^{2/3} (\alpha - \tanh \alpha )^{2/3}$ in the first- and $\zeta = (3/2)^{2/3} (\tan \beta - \beta )^{2/3}$ in the second expansion, respectively. These expansions have no issue at $\alpha=\beta=0$ (the apparent singularity is removable) and they coincide in this special case. So we can assume $0 \leq \beta < \frac{\pi}{2}$ and $\alpha \geq 0$. The price we have to pay for uniformity is that the coefficients $A_n$ and $B_n$ are complicated functions of their argument and possess removable singularities at the turning point (when $\alpha=\beta=0$). To recover the expansions $(6)$ and $(7)$, one can replace the Airy function and its derivative by their standard asymptotic expansions but this can be done only if $\alpha$ and $\beta$ are bounded away from $0$.