Determine the asymptotic behavior of the modified bessel-function of second kind $K_\nu(x)$ for $x\to\infty$, where $\nu$ is a constant$. It has the following integral form:
$K_\nu(x)=\frac{1}{2}\int_0^\infty \frac{ds}{s^{1-\nu}}\exp{[(-x/2)(s+1/s)]}$
Here's my approach to this:
First I deformed the integral to pass through a saddle point and use the steepest descents method then. Taking $w=-(x/2)(s+1/s)$, there is a saddle point at $s=1$, where $w=-x$ and $w"=-x/s^3=-x$. Assuming x to be positive, $arg(x)=\pi$, and the angle $\theta$ needed for the steepest-descents method is therefore 0. The slowly varying quantity is $s^{1-\nu}$, which is $1$ at $s=1$. Inserting all of that into the steepest descents formulat, I get
$K_\nu(x)\sim \frac{1}{2}e^{-x}\sqrt{\frac{2\pi}{|-x|}}=\sqrt{\frac{\pi}{2x}}e^{-x}$
I solved this by following a similar example in a workbook but I'm not quite understanding the approach.
And, what do I get then? For $x\to\infty$ it's simply $0$?
