The modified Bessel function of the second kind $K_{\nu} (x)$ should have an exponential – decreasing – behaviour with respect to its variable $x$, as shown in this document (page 19, fig. 4.4).
As stated here (page 14), the Bessel functions $J_{\nu}(iz)$ and $N_{\nu}(iz)$ of the first and the second kind with a complex argument $iz$ have the following asymptotic behaviour
$$J_{\nu}(iz) = \sqrt{\frac{2}{\pi iz}} \cos \left( iz – \frac{\nu \pi}{2} – \frac{\pi}{4} \right)$$
$$N_{\nu}(iz) = \sqrt{\frac{2}{\pi iz}} \sin \left( iz – \frac{\nu \pi}{2} – \frac{\pi}{4} \right)$$
How to prove from them that $K_{\nu} (x)$ has an exponential behaviour? "Google" provides so many definitions of $K_{\nu} (x)$, but not this proof.
Best Answer
The modified Bessel Function of the second kind $K_{\nu}(x)$ may be expressed as $$K_{\nu}(x) = \frac{\pi}{2}i^{\nu + 1}(J_{\nu}(ix) + i N_{\nu}(ix))$$ With this it follows that for $x\rightarrow\infty$ $$K_{\nu}(x) \sim \frac{\pi}{2}i^{\nu + 1}\sqrt{\frac{2}{\pi ix}}\Big[\cos \big(ix+\alpha\big)+i \sin \big(ix + \alpha\big)\Big]$$ $$\implies K_{\nu}(x)\sim \frac{\exp (-x)}{\sqrt x}$$ by Euler's formula.