Asymptotic expansion of integral involving modified Bessel-function I would like to obtain the asymptotic expression for $\alpha \to \infty$ of the following integral
$$I(\alpha)=\int_0^\infty\!dx\,x (1 - \cos[2\alpha K_0(x)]) = 
\int_0^\infty\!dx\, 2x \sin^2[\alpha K_0(x)]$$
where $K_0$ is the modified Besselfunktion of the second kind. I know that 
$$K_0(x) \sim  \begin{cases}
\log (2/x) -\gamma  & x\ll1, \\
\sqrt{\frac{\pi}{2x}}\,e^{-x} & x \gg1.
\end{cases}$$
The integral is finite for any $\alpha&lt\infty$: for small $x$ there is no problem because of the factor $x$. For large $x$ the Bessel-function approaches 0 exponentially fast and thus renders the integral finite. 
The problem I have is that the integral is both oscillatory and dominated by intermediate $x$. From numerics, I would believe that
$$ I(\alpha) \sim c_0 \log(\alpha) + c_1.$$
Do you have any ideas how to tackle integrals of this kind?
 A: I think I found a way to apply the method of stationary phase (please give feedback if you think that the answer is not sound).
As Jon noted via partial integration the integral can be brought into the form
$$I(\alpha) = \alpha \int_0^\infty\!dx\,x^2 K_1(x) \sin[2 \alpha K_0(x)].$$
Next I perform the substitution $y=2K_0(x)$ (note that $K_0$ is a monotonously falling function such that the inverse $K_0^{-1}$ is uniquely defined). I get
$$ I(\alpha) = \frac{\alpha}2 \int_0^\infty\!dy\,\sin(\alpha y)[K_0^{-1}(y/2)]^2; $$
note that $K_1$ cancelled with the derivative of $K_0^{-1}$ via inverse function theorem and the fact that $K_0'(x)=-K_1(x)$.
This integral is almost ready for a stationary phase analysis, we replace $\sin(\alpha y) = \operatorname{Im} e^{i\alpha y}$ and note that the path of stationary phase starting from $y=0$ is along the imaginary axis. Thus, we substitute $y=i \zeta$ and get
$$I(\alpha)=\frac{\alpha}2 \int_0^\infty \!d\zeta\, e^{-\alpha \zeta} \operatorname{Re}[K_0^{-1}(i\zeta/2)]^2 ;$$
here, we have analytically continued $K_0^{-1}$ into the complex plane.
The integral is dominated at $\zeta\approx 0$. Thus we can Taylor-expand $K_0^{-1}(i\zeta)$. We have $K_0(x) \sim e^{-x}$ thus $K_0^{-1}(y) \sim -\log y$. Using this asymptotic relation, we obtain
$$I(\alpha)\sim \frac{\alpha}2 \int_0^\infty \!d\zeta\, e^{-\alpha \zeta} \overbrace{\operatorname{Re}[\log(i\zeta/2)]^2}^{\sim \log^2 (\zeta/2)}
\sim \frac{\alpha}2 \int_0^\infty \!d\zeta\, e^{-\alpha \zeta}\log^2 (\zeta/2)
\sim \tfrac12 \log^2\alpha.  $$
I guess for the next term one has to work a bit harder and use $K_0^{-1}(y) \sim - \log y - \tfrac12 \log(-\log y)$. However, in principle it should be possible to extend this approach to next to leading terms.
A: Firstly, you need to integrate by parts obtaining
$$
I(\alpha)=
\int_0^\infty\!dx\, 2x \sin^2[\alpha K_0(x)]=
2\alpha\int_0^\infty\!dx\,x^2K_1(x) \cos[\alpha K_0(x)]\sin[\alpha K_0(x)].
$$
Now, change variable to $y=\alpha x$ so that
$$
I(\alpha)=\frac{1}{\alpha^2}\int_0^\infty\!dx\, y^2K_1\left(\frac{y}{\alpha}\right) \sin\left[2\alpha K_0\left(\frac{y}{\alpha}\right)\right].
$$
In the limit $y\rightarrow\infty$ one can take the asymptotic approximations $K_0(x)\sim\sqrt{\frac{\pi}{2x}}\,e^{-x} $ and $K_1(x)\sim\sqrt{\frac{\pi}{2x}}\,e^{-x} $ and you get
$$
I(\alpha)\sim\frac{1}{\alpha^\frac{3}{2}}\int_0^\infty\!dy\, y^\frac{3}{2}\sqrt{\frac{\pi}{2}}e^{-\frac{y}{\alpha}}\sin\left[\sqrt{2}\alpha^\frac{3}{2}\sqrt{\frac{\pi}{y}}e^{-\frac{y}{\alpha}}\right].
$$
This integral can be rewritten using Lambert W function with the change of variable $z=\frac{e^{-\frac{y}{\alpha}}}{\sqrt{y}}$ so that $y=\frac{\alpha}{2}W\left(\frac{2}{\alpha z^2}\right)$ that for $\alpha\rightarrow\infty$ gives $y\sim\frac{1}{z^2}-\frac{2}{\alpha z^4}$. This will turn the integral into
$$
I(\alpha)\sim\frac{\alpha^\frac{3}{2}}{4}\int_0^\infty\!dz\,\frac{\sin\left(\sqrt{2\pi}\alpha^\frac{3}{2}z\right)}{\frac{1}{W^2\left(\frac{2}{\alpha z^2}\right)}+\frac{1}{W^3\left(\frac{2}{\alpha z^2}\right)}}.
$$
Then, we put $w=\sqrt{2\pi}\alpha^\frac{3}{2}z$ and we arrive at
$$
 I(\alpha)\sim\frac{1}{4\sqrt{2\pi}}\int_0^\infty\!dw\,\frac{\sin w}{\frac{1}{W^2\left(\frac{4\pi\alpha^2}{w^2}\right)}+\frac{1}{W^3\left(\frac{4\pi\alpha^2}{w^2}\right)}}
$$
I checked the original integral proposed by OP and my approximation and I have got the following picture

so the approximation is really satisfactory and give a clear numerical proof of the finiteness of last approximate integral.
