Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 336 / 1

Factor out the elementary exponential and writeUse a matched asymptotic expansion with $0<\lambda \ll \delta \ll 1$. In the outer part,The given identity for the Euler--Mascheroni constant, integrated by parts, impliesHenceapart from pure higher matching powers in $\delta$.

In the inner part, use $t={\lambda }^2/s$:Since $s={\lambda }^2/t\le {\delta }^2$, expand the last exponential. At the required order,With $a={\lambda }^2/{\delta }^2$, this givesThe overlap terms ${\delta }^2$ and $\log \delta$ cancel, leaving the logarithmic switchback termFinally,so

The exponential phase isIts saddle points satisfyFor positive real $\lambda$, the original positive-real contour passes through $s=\lambda$; the singularity at $s=0$ separates it from the other saddle. At the contributing saddle,Set $s=\lambda +{\lambda }^{1/2}t$. ThenThe method of steepest descent therefore gives

WriteThe saddle equation is unchanged, soThe contour connected to the original positive-real integration path passes through $s_{+}$. Since $\Phi (s_{+})=0$, its steepest-descent branch hasFor $s=r{e}^{i\psi }$ and ${\lambda }^2=i{L}^2$,Thus the contour equation isIt approaches the origin along the positive imaginary axis, passes through $r=L$, $\psi =\pi /4$, and tends to infinity along the positive real axis. Along both ends the real part of the phase tends to $-\infty$, so the contour rises from the origin nearly vertically, bends through the first-quadrant saddle, and leaves nearly horizontally. The other saddle lies on a different constant-imaginary-phase contour and does not connect to this deformation.