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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: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 $1/6$.

At $a=1$, the endpoint is degenerate: $S(t)=\sinh t-t=t^3/6+O(t^5)$. The contributing width is therefore $t=O({\nu }^{-1/3})$, rather than the width in part (i). With $v=\nu t^3/6$, the formula for a degenerate endpoint in Laplace's method givesThe coefficient is a Gamma function value obtained by the Gamma integral.