The modified Bessel function (Macdonald function) $K_\alpha(z)$ is known to have the following asymptotic expansion for large positive $z$: $$ K_\alpha(z)=\sqrt{\frac{\pi}{2z}}e^{z}\sum_{k=0}^\infty \frac{b_k(\alpha)}{z^ k} $$ where $b_1(\alpha)=1$, $b_2(\alpha)=\frac{\alpha^21^2}{1!8}$, $b_3(\alpha)=\frac{(\alpha^21^2)(\alpha^23^2)}{2!(8)^2}$ and so on. Is there any simple integral representation, for which it would be a perturbative expansion such that $$ K_\alpha(z)=h(z) \int_C \exp\left(\frac{f(y)}{z}\right) g(y)^\alpha d\mu(y) $$ where $f(x)$, $g(x)$, $h(x)$ and $d \mu(x)$ are $\alpha$independent?
Integral representation of the modified Bessel functions of the second kind and asymptotic expansion
1 Answer
The DLMF lists multiple integral representations of $K_\nu(z)$. Here's one that fits your bill:
$$\mathop{K_{{\nu}}}\nolimits\!\left(z\right)=\frac{\pi^{{\frac{1}{2}}}(\frac{1}{2}z)^{\nu}}{\mathop{\Gamma}\nolimits\!\left(\nu+\frac{1}{2}\right)}\int _{0}^{\infty}e^{{z\mathop{\cosh}\nolimits t}}(\mathop{\sinh}\nolimits t)^{{2\nu}}dt .$$
For integer $\nu$, the contour could be extended by symmetry to all of $\mathbb{R}$.

$\begingroup$ Thank you, I know this nice integral representation, but I want something (maybe) more complicated  with $\frac{1}{z}$ as I wrote. $\endgroup$– SashaJan 31, 2013 at 15:54

$\begingroup$ Sorry, I don't follow your logic about $1/z$ in $g(f(y)/z)$. What kind of expansion would you expect to perform? The integral formula I suggested should be amenable to a Laplacetype (or steepest descent) expansion for large $z$. In any case, most known integral representations are in fact listed on the DLMF page. $\endgroup$ Jan 31, 2013 at 20:53

$\begingroup$ I just want to have simple $1/z$ expansion for large $z$ (and not a usual steepest descent expansion). $\endgroup$– SashaFeb 1, 2013 at 13:06