Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2010 / A / 2

For an asymptotic sequence $a_{n+1}(ϵ)=o(a_n(ϵ))$, the asymptotic expansion condition is that every fixed truncation has remainder smaller than its last retained term:This limit is taken as $ϵ\to 0$ in the specified direction or sector. The number $N$ stays fixed while taking that limit. The infinite series need not converge for any nonzero parameter.

Let $g_x(t)=(1+xt{)}^{-1}$. Repeated integration by parts using the factor $e^{-t}$ givesBoundary terms at infinity vanish, and $g_x^{(n)}(t)=(-1{)}^{n}n!x^n(1+xt{)}^{-n-1}$. ThereforeFor every fixed $N$, this is $o(x^N)$ as $x\downarrow 0$, provingFor each fixed $x>0$ the displayed infinite series diverges, since successive term magnitudes have ratio $(n+1)x\to \infty$. Its usefulness is precisely that of an asymptotic expansion, not a convergent power series.

Fix $0<ϵ<\pi /2$. On $[ϵ,\pi /2]$, ${\sin }^{2}t\ge {\sin }^2ϵ>0$, so that part of the integral is bounded by $(\pi /2)e^{-x{\sin }^2ϵ}$, exponentially smaller than $x^{-1/2}$.

By the assumed local equivalence, the remaining part is asymptotic to ${\int }_0^{ϵ}{e}^{-x{t}^2}dt$. Substitute $s=\sqrt{x}t$:The last step uses the Gaussian integral and monotone convergence of the upper limit to infinity. Hence

A fixed small-angle approximation alone is insufficient because its error is multiplied by $x$. Introduce the shrinking cutoff $\eta =x^{-1/3}$, which satisfies $\eta \ll ϵ$ but $\sqrt{x}\eta \to \infty$.

On $[0,\eta ]$, the Taylor expansion ${\sin }^{2}t=t^2+O(t^4)$ gives $x({\sin }^{2}t-t^2)=O(x{\eta }^4)=O(x^{-1/3})$ uniformly. Thusuniformly there, and the two central integrals are asymptotically equal. On $[\eta ,ϵ]$, choose a fixed $c>0$ with $\sin t\ge ct$ for $0\le t\le ϵ$. The sine-integral tail is at most $ϵe^{-c^{2}x{\eta }^2}=ϵe^{-c^2{x}^{1/3}}$; the Gaussian tail has the same exponential bound with $c=1$. Both are $o(x^{-1/2})$, while the central Gaussian integral is asymptotic to $\sqrt{\pi }/(2\sqrt{x})$. HenceThis justifies the localized Laplace method rather than interchanging a nonuniform approximation with integration.