Asymptotic expansions (large argument) for Bessel K function The $K$ Bessel function is defined as (let's assume $x$ is real):
$$
K_{\nu}=\frac{\pi}{2}\frac{I_{-\nu}(x)-I_{\nu}(x)}{\sin(\nu\pi)},
$$
where
$$
I_{\nu}(x)=\sum_{n=0}^{\infty}\frac{1}{n!\Gamma(1+n+\nu)}\Big(\frac{x}{2}\Big)^{2n+\nu}.
$$
I would to know how it is possible to deduce the asymptotic expansions  for large $x$:
$$
I_{\nu}(x)\sim \frac{e^x}{\sqrt{2\pi x}}\Big(1-\frac{4\nu^2-1}{8x}+\frac{(4\nu^2-1)(4\nu^2-9)}{2!(8x)^2}+\ldots\Big)
$$
and
$$
K_{\nu}(x)\sim \sqrt{\frac{\pi}{2x}}e^{-x}\Big(1+\frac{4\nu^2-1}{8x}+\frac{(4\nu^2-1)(4\nu^2-9)}{2!(8x)^2}+\ldots\Big).
$$

You can establish them by applying Watson's lemma to the integral representations
$$
I_\nu  (z) = \frac{1}{{\sqrt \pi  \Gamma \left( {\nu  + \frac{1}{2}} \right)}}(2z)^\nu  e^z \int_0^1 {e^{ - 2zt} t^{\nu  - 1/2} (1 - t)^{\nu  - 1/2} dt} 
$$
and
$$
K_\nu  (z) = \frac{{\sqrt \pi  }}{{\Gamma \left( {\nu  + \frac{1}{2}} \right)}}\left( {\frac{2}{z}} \right)^\nu  e^{ - z} \int_0^{ + \infty } {e^{ - 2zt} t^{\nu  - 1/2} (1 + t)^{\nu  - 1/2} dt} 
$$
where $\Re z>0$ and $\Re \nu >- \frac{1}{2}$. See https://en.wikipedia.org/wiki/Watson%27s_lemma
