 Solution

Use a counterclockwise keyhole contour around the positive real branch cut, with small radius $\epsilon$ and large radius $R$. On the upper bank $\arg z=0$, so the integral tends to $I$. On the lower bank $\arg z=2\pi$, $z^{1/2}=-\sqrt{x}$, giving the negative of the upper-bank integrand. That bank is traversed from $R$ to $\epsilon$, so its contribution also tends to $I$.

The outer circle contributes $O(R^{-3/2})$, since $f=O(R^{-5/2})$ and its length is $O(R)$. The small circle contributes $O({\epsilon }^{1/2})$, since $f=O({\epsilon }^{-1/2})$. Both vanish. HenceThe contour encloses both double poles $\pm i$ but excludes the branch point.