= Solution
For small $x$, use $u=\cosh t$ to write the defining <oscillatory integral> as
$$
I(x)=\int_1^\infty\frac{\cos(xu)}{\sqrt{u^2-1}}\,du=\int_1^\infty\frac{\cos(xu)}u\,du+\int_1^\infty\cos(xu)\left(\frac1{\sqrt{u^2-1}}-\frac1u\right)du.
$$
The first integral is $-\operatorname{Ci}(x)$, where the <cosine integral> satisfies $\operatorname{Ci}(x)=\gamma_E+\log x+O(x^2)$. Its logarithm follows by integrating the $1/x$ singularity of its derivative; its conventional constant is the <Euler--Mascheroni constant>. The primary normalization is given by https://dlmf.nist.gov/6.6.E6 .
The second integrand has an integrable inverse-square-root endpoint singularity and an $O(u^{-3})$ tail. The <dominated convergence theorem> therefore gives its limit
$$
\int_1^\infty\left(\frac1{\sqrt{u^2-1}}-\frac1u\right)du=\lim_{L\to\infty}(\operatorname{arcosh}L-\log L)=\log2.
$$
Splitting at $u=1/x$ and using $|1-\cos(xu)|\le\min(x^2u^2/2,2)$ also bounds its error by $O(x^2|\log x|)$. Thus
$$
\boxed{g(x)=\frac2\pi\left(\log\frac x2+\gamma_E\right)+O(x^2|\log x|),\qquad x\downarrow0.}
$$
In particular, the leading term is $(2/\pi)\log x$.
For large $x$, the phase $\cosh t$ has its only stationary point at the endpoint $t=0$. Its expansion is $1+t^2/2+O(t^4)$, so the contributing width is $t=O(x^{-1/2})$. The endpoint <stationary phase method> and the <Fresnel integral> yield
$$
\int_0^\infty e^{ix\cosh t}dt\sim e^{ix}\int_0^\infty e^{ixt^2/2}dt=e^{i(x+\pi/4)}\sqrt{\frac\pi{2x}}.
$$
A smooth cutoff near the endpoint justifies this local replacement; the remaining nonstationary tail can be integrated by parts, with no stationary contribution at infinity. Taking the real part gives the <Bessel function of the second kind> asymptotic
$$
\boxed{g(x)=\sqrt{\frac2{\pi x}}\sin(x-\pi/4)+O(x^{-3/2}),\qquad x\to\infty.}
$$
The additive error formulation remains meaningful near zeros of the leading oscillation. Both limits were derived from the given integral, rather than identifying a special function and merely quoting its asymptotics.
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