Asymptotic expansion of $\int_0^1 e^{\lambda(-x+\alpha \sin x)} \ \mathrm{d}x$ My question concerns the final part of the following question from this document. For parts (i) and (ii) I obtain $-\frac{1}{\lambda(\alpha-1)} $ and $\frac{1}{3}6^{\frac{1}{3}}\frac{\Gamma(1/3)}{\lambda^{1/3}}$, respectively. To find these, I used a previous result in the same question and Watson's Lemma. However, I have no idea how to 'deduce' from these results that there is a distinguished limit, and indeed where the expressions in terms of $\nu$ come from. I am new to perturbation methods, so I'd appreciate it if someone could explain how this problem should be tackled. Many thanks.
 A: The results you have already suggest that, if $\alpha=\alpha(\lambda)$ is allowed to vary, something "breaks down" when $\lambda(1-\alpha)$ and $\lambda^{1/3}$ are of the same order (which happens at $p=2/3$), so that neither of the two leading asymptotes is dominant. Explicitly, let $g_p(\lambda,\nu)=f(\lambda,1-\lambda^{-p}\nu)$. Then, if $0<p<2/3$, we substitute $x=\lambda^{p-1}t$ in the integral: $$\lim_{\lambda\to\infty}\lambda\big((1-\lambda^{-p}\nu)\sin(\lambda^{p-1}t)-\lambda^{p-1}t\big)=-\nu t;$$ after taking limit under integral sign, which is easy to justify, this gives $\lim\limits_{\lambda\to\infty}\lambda^{1-p}g_p(\lambda,\nu)=1/\nu$. Thus, (i) dominates. Similarly, if $p>2/3$, we use $x=\lambda^{-1/3}t$ and get $\lim\limits_{\lambda\to\infty}\lambda^{1/3}g_p(\lambda,\nu)=\int_0^\infty e^{-t^3/6}\,dt$ much like you did; (ii) is the "leader". Finally, if $p=2/3$, the substitutions coincide, and we obtain $$\lim_{\lambda\to\infty}\lambda^{1/3}g_{2/3}(\lambda,\nu)=\int_0^\infty e^{-\nu t-t^3/6}\,dt.$$ Perhaps this is what is asked for: $p=2/3$, $q=1/3$, and $h(\nu)$ is the integral above.
