I deal with the modified Bessel function of the firts kind with purely imaginary index, $I_{i\nu}(z)$, where $\nu\in\mathbb{R}$. I am interested in large $\nu$ expansion of this function. In order to find this expansion, I use the following integral representation, which is valid (see 10.32.2),
$$I_{i\nu}(z)=\frac{1}{\sqrt{\pi}\Gamma(i\nu+1/2)}\left(\frac{z}{2}\right)^{i\nu}\int_{-1}^{+1}dt\,\exp\left\lbrace-zt+\left(i\nu-\frac{1}{2}\right)\ln(1-t^2)\right\rbrace.$$
In order to investigate large $\nu$ behavior, I try to evaluate this integral (I denote it as $J$) with help of steepest descent method. I rewrite the integral as
$$J=\int_{-1}^{+1}dt\,e^{-zt}\exp\left\lbrace\nu\left(i-\frac{1}{2\nu}\right)f(t)\right\rbrace\approx \int_{-1}^{+1}dt\,e^{-zt}\exp\left\lbrace i\nu f(t)\right\rbrace,\quad f(t)=\ln(1-t^2).$$
So, I apply steepest descent for the function $F(t)=if(t)$. The integral saturates near the point $t_0$, which corresponds to a solution of $F'(t)=0$. This point is $t_0=0$. Next, I extend limits of integration to infinity and write
$$J\approx e^{-zt_0}e^{\nu t_0}\sqrt{\frac{2\pi}{\nu|F''(t_0)|}}=\sqrt{\frac{2\pi}{2\nu}}.$$
Then, I expand the function $\Gamma(i\nu+1/2)$ and obtain
$$\frac{1}{\Gamma(i\nu+1/2)}\approx \frac{1}{\sqrt{2\pi}}\exp\left\lbrace i\nu+\frac{\pi\nu}{2}-i\nu\ln\nu\right\rbrace.$$
Combining approximated $J$ and this expansion, I write
$$\boxed{I_{i\nu}(z)\approx \frac{1}{\sqrt{2\pi\nu}}\left(\frac{ze}{2\nu}\right)^{i\nu}e^{\pi\nu/2}} \tag{*}$$
Obtained expression $(*)$ seems right but I would like to be completely sure that my derivations are correct.
Modified Bessel function of the first kind with purely imaginary index
bessel functionsdefinite integralsspecial functions
Best Answer
Thanks for Max & Gary comments, I assume that below derivation is correct. I start from $$I_{i\nu}(z)=\frac{1}{\sqrt{\pi}\Gamma(i\nu+1/2)}\left(\frac{z}{2}\right)^{i\nu}\int_{-1}^{+1}dt\,\left(1-t^2\right)^{i\nu-1/2}e^{-zt}.$$ Then, denote the appeared integral as $J$, so $$ J = \int_{-1}^{+1}dt\,\left(1-t^2\right)^{i\nu-1/2}e^{-zt}.$$ The integrand can be represented as $$g(t)e^{\nu f(t)}, \quad g(t)=\exp\left(-zt-\frac{1}{2}\ln(1-t^2)\right), f(t)=i\ln(1-t^2).$$ The steepest descent method gives $$J\approx g(t_0)e^{\nu f(t_0)}e^{i\phi}\left(\frac{2\pi}{\nu|f''(t_0)|}\right)^{1/2},$$ where $f''(t_0)=|f''(t_0)|\exp(i\theta)$, $\phi=(\pi-\theta)/2$ and $t_0$ is a solution of $f'(t)=0$. It is easy to see that $t_0=0$. Then, $$f''(t_0)=-2i\rightarrow |f''(t_0)|=2e^{-i\pi/2}\rightarrow \phi=\frac{3\pi i}{4},\quad g(t_0)=1, f(t_0)=1.$$ Considering all the above, $$J\approx \sqrt{\frac{\pi}{\nu}}e^{3\pi i/4}.$$ Next, I expand $1/\Gamma(i\nu+1/2)$ for large $\nu$ and obtain $$\frac{1}{\Gamma(i\nu+1/2)}\approx \frac{1}{\sqrt{2\pi}}\exp\left\lbrace i\nu+\frac{\pi\nu}{2}-i\nu\ln\nu\right\rbrace.$$ So, finally, $$\boxed{I_{i\nu}(z)\approx \frac{1}{\sqrt{2\pi\nu}}\left(\frac{ze}{2\nu}\right)^{i\nu}e^{\pi\nu/2}e^{3i\pi/4}}$$