Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 48 / 1

Put $L=\log \lambda$ and integrate by parts using $x^{-2}=-d(x^{-1})/dx$. The exponential cutoff has derivative $\lambda e^{-x}{e}^{-\lambda e^{-x}}$, so the endpoint at infinity vanishes and the lower endpoint givesWith $u=\lambda e^{-x}$ this becomesThe mass of the transformed integral lies at $u=O(1)$, whereTo control this asymptotic expansion, split at $u=e^{-L/2}$ and $u=e^{L/2}$. On the central interval the geometric-series remainder is bounded by $2(\log u{)}^2/L^3$, whose integral against $e^{-u}$ is finite. The lower tail has length $e^{-L/2}$ and denominator at least $L$; the upper tail is exponentially small in $e^{L/2}$, with the original denominator at least one. The omitted tails of the logarithmic moments are likewise negligible. Thus extending the moments to $(0,\infty )$ is justified.

The Gamma function gives ${\int }_0^{\infty }{e}^{-u}du=1$, and the first logarithmic moment is $-\gamma$, with $\gamma$ Euler's constant. ThereforeThis is the moving Gumbel cutoff in an inverse-power tail: replacing the original cutoff by a sharp step at $x=\log \lambda$ gives the first term, but its smooth transition supplies the Euler-constant correction.

The phase $f(x)=x(1-\cos x)$ is nonnegative on the interval and vanishes only at its two endpoints. An interval away from both endpoints contributes exponentially little, so both endpoint contributions must be expanded before their orders are ranked.

At the left endpoint,The Laplace's method scale is $x={\sigma }^{-1/3}z$. Hence its leading contribution isThe substitution $v=z^3/2$ evaluates the integral by the Gamma function. Since there is no quartic term in the phase, the next left-end correction is relatively $O({\sigma }^{-2/3})$, and thus absolutely $O({\sigma }^{-1})$.

At the right endpoint put $r=2\pi -x$. Thereso $r={\sigma }^{-1/2}z$ givesThe leading right-end integral is a Gaussian integral. Its first correction is also absolutely $O({\sigma }^{-1})$. The mixed cubic and quadratic Laplace endpoints therefore produce the first two global termsKeeping only the dominant cubic endpoint would miss the second term, which is larger than that endpoint's own first correction.