Let $h$ be the function $$ h(x) = \sum_{k\geq0} \frac{(ix)^k}{k!}\zeta(2k), $$ with the Laplace transform $$ \tilde h(s) = -\frac{\pi}{2s}\sqrt{i/s}\cot\left(\pi\sqrt{i/s}\right), $$ which has an essential singularity at $s=0$.
How can I derive an asymptotic form for $h$ in the limit $x\to\infty$ ($x$ real)? I would like to understand how to do it with the inverse Laplace transform. The inversion formula gives $$ -\frac{\sqrt{x}}{4i} \int \sqrt{i/s} \cot\big(\pi\sqrt{ix/s}\big) e^s\frac{ds}{s}, $$ with the integral taken along a vertical line in the right half-plane.
But the essential singularity at $s=0$ looks rather tricky to deal with and I am not sure how to approximate the integral. Heuristic semi-rigorous reasoning is welcome.