Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 68 / 1 / b

One convenient version of Watson's lemma is the following. Suppose $G(t)$ is integrable away from zero and has, as $t\downarrow 0$, an asymptotic expansion $G(t)\sim {\sum }_{j\ge 0}{c}_j{t}^{\beta +j-1}$, where $\beta >0$. On a finite interval, or on an infinite interval with suitable exponential growth control,To prove a finite truncation, split the integral at a fixed small $\delta >0$. The part away from zero is exponentially small. On $(0,\delta )$ subtract the first $N$ local terms and bound the remainder by $C{t}^{\beta +N-1}$; its integral is $O({\lambda }^{-\beta -N})$. Extending each polynomial integral to infinity changes it only exponentially, and the substitution $u=\lambda t$ produces the Gamma integral. A little-o local remainder gives the corresponding little-o asymptotic remainder. This is a proof by localization and remainder bounds, rather than merely formal integration of a series.

For the monotone-phase integral, assume the derivatives needed below exist and are bounded near $a$, and the rest of the integral is controlled. Put $t=\varphi (a)-\varphi (x)$. Since ${\varphi }^{\prime }(a)<0$, this is a regular endpoint coordinate andDifferentiating with $dx/dt=-1/{\varphi }^{\prime }$ gives $G(0)=-g(a)/{\varphi }^{\prime }(a)$ and $G^{\prime }(0)=g^{\prime }(a)/{\varphi }^{\prime }(a{)}^2-g(a){\varphi }^{\prime \prime }(a)/{\varphi }^{\prime }(a{)}^3$. Watson's lemma therefore gives the two-term monotone-endpoint Laplace expansionAn $O(e^{\lambda \varphi (a)}{\lambda }^{-3})$ remainder follows, for example, from a sufficiently smooth transformed amplitude with bounded second derivative near zero. The displayed second coefficient can vanish for special amplitudes without changing the expansion.

The unheaded transition request concerns a cubic endpoint-to-saddle transition. Near $x=0$ its exponent has the expansionAt the cubic endpoint $x$ has size ${\lambda }^{-1/3}$. Requiring the linear term to contribute on the same scale gives $1-\alpha =O({\lambda }^{-2/3})$. Thus the distinguished limit and its uniform leading integral areFor fixed $\nu$, the substitution $x={\lambda }^{-1/3}t$ makes the omitted exponent terms $O({\lambda }^{-2/3})$ on bounded $t$ intervals. Cubic decay controls the remaining tail, while the original interval away from zero is exponentially negligible. The estimate is uniform for $\nu$ in compact sets. In the printed range $\alpha \le 1$, $\nu \ge 0$; the same transition integral also makes sense for negative fixed $\nu$.

At $\nu =0$, the Gamma integral gives $h(0)=6^{1/3}\Gamma (1/3)/3$, recovering the cubic endpoint regime. For $\nu \to +\infty$, scale $t=u/\nu$ and expand $e^{-u^3/(6{\nu }^3)}$ to obtain $h(\nu )={\nu }^{-1}-{\nu }^{-4}+O({\nu }^{-7})$. Its leading term gives ${\lambda }^{-1/3}/\nu =1/[\lambda (1-\alpha )]$, recovering the ordinary endpoint regime in their overlap. Both matching limits are therefore checked explicitly.

For fixed $0\le \alpha <1$, the phase derivative is $-1+\alpha \cos x\le -(1-\alpha )<0$ on the integration interval. Its maximum is at $x=0$, with phase zero and derivative $-(1-\alpha )$. The Laplace endpoint estimate, or the substitution $x=t/[\lambda (1-\alpha )]$, givesThis fixed-parameter formula is not uniform as $\alpha \uparrow 1$; its diverging coefficient signals the distinguished limit derived in the parent 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: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 $1/6$.