Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 76 / 1 / a / Solution

For $0<\lambda <1$ fixed, the denominator stays away from zero, so dominated convergence givesThis approximation remains appropriate on the left of the transition when $1-\lambda \gg ϵ$. To resolve the Lorentzian peak approaching an integration endpoint, let $\lambda -1=ϵ\tau$ with bounded $\tau$, and put $x=ϵX$. The cosine is $1+O({ϵ}^2{X}^2)$ in the contributing region. ThereforeThus the distinguished transition has width $|\lambda -1|=O(ϵ)$. For large negative $\tau$ this expression matches $1/(1-\lambda )$ to leading order; for large positive $\tau$ it gives the full interior-peak contribution.

For $\lambda >1$ away from the transition, the denominator has a narrow minimum at $x_{\ast }=1-1/\lambda$. Set $c_{\ast }=\cos (\pi x_{\ast }/2)=\sin (\pi /(2\lambda ))$ and $x-x_{\ast }=(ϵc_{\ast }/\lambda )s$. The slowly varying cosine may be frozen over the peak and the integration limits extended, givingThe small parameter controlling the left-end truncation near $\lambda =1$ is $ϵ/(\lambda -1)$, so this formula requires $\lambda -1\gg ϵ$ there.

For $\lambda \gg 1$, the peak approaches $x=1$, but its width is $ϵ\pi /(2{\lambda }^2)$ whereas its distance from that endpoint is $1/\lambda$. Their ratio is $O(ϵ/\lambda )$, and the relative variation of the cosine across the peak is also $O(ϵ/\lambda )$. Hence proximity to the endpoint does not destroy the approximation: the peak becomes narrower still. Equivalently, $s=\lambda (1-x)$ gives $J={\lambda }^{-1}{\int }_0^{\lambda }[(1-s{)}^2+{ϵ}^2{\sin }^2(\pi s/(2\lambda )){]}^{-1}ds$, with its peak at $s=1$. The large-$\lambda$ limit of the leading result is $J\sim 2/ϵ$.