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

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 $\mathbb{R}$. 

