= Solution
At $\alpha=1$, the phase is $-x+\sin x=-x^3/6+O(x^5)$, so the ordinary nonstationary endpoint formula fails. In the <degenerate endpoint in Laplace's method>, take $x=\lambda^{-1/3}t$ and extend the localized <integral> to infinity:
$$
f(\lambda,1)\sim\lambda^{-1/3}\int_0^\infty e^{-t^3/6}\,dt=\boxed{\frac{6^{1/3}\Gamma(1/3)}{3\lambda^{1/3}}.}
$$
The last equality follows by $u=t^3/6$. It is the $\nu=0$ value of the <cubic Laplace transition integral> after its normalization is rescaled to the coefficient $1/6$.
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