Upper bound for ratio of modified Bessel functions of second kind

Question

I was wondering if someone has an idea if for $0 < x < y$, and $0< \nu \leq \frac{1}{2}$, one can obtain an upper bound for the ratio

$$ \frac{K_{\nu}(x)}{K_{\nu}(y)} $$

Thanks.

Answer 1

$$K_\nu(x)=\int_{0}^{+\infty}\cosh(\nu t)\exp(-x\cosh t)\,dt = \frac{\pi}{2\sin(\nu \pi)}\left(I_{-\nu}(x)-I_{\nu}(x)\right)$$ together with the $\phantom{}_2 F_1$ representations: $$ I_\nu(x) = \frac{1}{\Gamma(1+\nu)}\left(\frac{x}{2}\right)^{\nu}\left\{1+\frac{(x/2)^2}{1(1+\nu)}\left(1+\frac{(x/2)^2}{2(2+\nu)}\left(1+\frac{(x/2)^2}{3(3+\nu)}\left(1+\ldots\right)\right)\right)\right\}$$ $$ I_{-\nu}(x) = \frac{1}{\Gamma(1-\nu)}\left(\frac{x}{2}\right)^{-\nu}\left\{1+\frac{(x/2)^2}{1(1-\nu)}\left(1+\frac{(x/2)^2}{2(2-\nu)}\left(1+\frac{(x/2)^2}{3(3-\nu)}\left(1+\ldots\right)\right)\right)\right\}$$ should give you all the approximation accuracy you need.

Probably its is also useful to recall that $K_{1/2}(x)=e^{-x}\sqrt{\frac{\pi}{2x}}$ and that $K_\nu(x)$ is a logarithmically convex function with respect to $\nu$ (for a fixed $x$) due to Cauchy-Schwarz inequality, hence if $0<\nu\le \frac{1}{2}$ and $x,y>0$,

$$\frac{K_\nu(x)}{K_{\nu}(y)}\leq \frac{K_{1/2}(x)}{K_{1/2}(y)} = e^{y-x}\sqrt{\frac{y}{x}}.$$