 Solution

For small $x$, use $u=\cosh t$ to write the defining oscillatory integral asThe first integral is $-\mathrm{Ci}(x)$, where the cosine integral satisfies $\mathrm{Ci}(x)={\gamma }_E+\log x+O(x^2)$. 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 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 limitSplitting at $u=1/x$ and using $|1-\cos (xu)|\le \min (x^2{u}^2/2,2)$ also bounds its error by $O(x^2|\log x|)$. ThusIn 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 yieldA 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 asymptoticThe 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.