 Solution

The only nondegenerate stationary point of a phase of the phase $t^2$ is the endpoint zero. Its usual leading stationary phase method term is proportional to the amplitude there, but ${\tan }^2(0)=0$. Thus the nominal $x^{-1/2}$ stationary contribution vanishes.

Expand one order further in the local amplitude: ${\tan }^{2}t=t^2+O(t^4)$. The rescaling $t=u/\sqrt{x}$ then gives a first nonzero stationary contribution of order $x^{-3/2}$, with coefficient $-\sqrt{2\pi }/8$ in the real integral, as in part (i). Meanwhile the upper endpoint is nonstationary and has nonzero amplitude; its integration by parts contribution is $2\sin ({\pi }^{2}x/16)/(\pi x)$, of order $x^{-1}$.

Consequently the nonstationary endpoint dominates the first nonzero saddle contribution. A stationary phase method formula retaining only the usual leading saddle term would miss the answer. The full stationary phase method, with the vanishing-amplitude correction and the endpoint term, agrees with method of steepest descent. This is quadratic stationary endpoint with quadratically vanishing amplitude.