I'm interested in the following integral:
$$f(x) = \frac{x^6}{64} \int_0^\infty du \ u^5 J_0(x u) \frac{K_1(u)}{I_1(u)}$$
where $J_0$ is a Bessel function of the first kind and $K_1, I_1$ are modified Bessel functions. I want to understand the behaviour of this integral at large $x$. (This is a Hankel transform which arose in the context of a two-dimensional Fourier transform.)
I evaluated it numerically in Mathematica up to $x = 8$ and it seems that
$$\lim_{x \rightarrow \infty} f(x) = -1$$
with $1 + f(x)$ decaying exponentially in $x$. I'd like to prove this statement, but I don't know how.
I don't think the stationary phase approximation will work because for large $u$ the Bessel function oscillates like $\cos(xu - \pi/4)$, which doesn't have a point of stationary phase.
With the change of variable $v = x u$, the integral takes the form
$$ f(x) = \frac{1}{64} \int_0^\infty dv \ v^5 J_0(v) \frac{K_1(v/x)}{I_1(v/x)} . $$
I thought I could use the small-argument expansion of the modified Bessel function $K_1, I_1$, but they are not useful since the argument becomes arbitrarily large as $v \rightarrow \infty$.
How can I derive the large $x$ behaviour of this integral?