Modified Bessel function of the first kind with purely imaginary index 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.
 A: 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,
$$\boxed{I_{i\nu}(z)\approx \frac{1}{\sqrt{2\pi\nu}}\left(\frac{ze}{2\nu}\right)^{i\nu}e^{\pi\nu/2}e^{3i\pi/4}}$$
