Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 3 / 30E (Asymptotic Methods)

PutOn $[0,2]$, its global maximum is the endpoint $t=2$, whereand the amplitude $\log t$ has value $\log 2$. The endpoint form of Laplace's method therefore givesIndeed, with $u=2-t$, Taylor expansion gives $\varphi (2-u)=2-5u+O(u^2)$ and $\log (2-u)=\log 2+O(u)$; the scale contributing to the integral is $u=O(x^{-1})$. Outside any fixed neighbourhood of $2$, $\varphi$ is bounded above by $2-\delta$, so that part is exponentially smaller. This also controls the integrable logarithmic singularity at zero.

Near zero, the absolute value of the integrand is comparable towhich is integrable for $z>0$. At infinity, the factor $e^{-t}$ dominates every power of $t$ and $\log t$, so the integral also converges there.

Direct use of Laplace's method is inconvenient because the phasedoes not have the form $z\varphi (t)$ with a fixed phase: its stationary point $t=z$ moves to infinity as $z\to \infty$. Rescaling is needed to fix the maximum.

Set $t=zs$. ThenThe fixed phasehas its unique maximum at $s=1$, withThe amplitude at the maximum is $\log z$. The interior-maximum form of Laplace's method givesThusso $\boxed{a=2\pi }$. This is the Laplace asymptotic for the derivative of the Gamma function.