= Solution
{c}
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: $\tan^2t=t^2+O(t^4)$. 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 $2\sin(\pi^2x/16)/(\pi x)$, of order $x^{-1}$.
Consequently \b[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>.
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