Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 336 / 1

The divide-and-conquer asymptotic expansion separates the $x=O(a)$ endpoint region from the $x=O(1)$ bulk, whose expansions individually contain terms that are nonuniform in the other region. Here the recombined answer can also be checked exactly. Put $y=\sqrt{a+x}$; thenAs $a\to 0^{+}$,Multiplication givesThe nonanalytic $\sqrt{a}$ term is the contribution that a naive fixed-$x$ expansion misses at the endpoint.

Write $w=z-1=X+iY$. The phase and its derivative areso the saddle points arewith $\varphi (z_{+})=2i$ and $\varphi (z_{-})=-2i$. The contour geometry can be drawn fromThe stationary-phase level $\mathrm{Re}\varphi =0$ consists of $X=0$ and $X^2-3Y^2+3=0$, meeting at the saddles. The steepest curves through $z_{\pm }$ are the levels $\mathrm{Im}\varphi =\pm 2$. Far away, sectors with $\cos (3\arg w)>0$ are exponential hills and those with $\cos (3\arg w)<0$ are valleys.

The stated contour deforms through the upper saddle. If $z=z_{+}+s$, thenThe descent tangent has $s=e^{i\pi /4}t$, because $3i{s}^2=-3t^2$. The simple-saddle contribution in steepest descent is therefore

When the contour begins at $z=1$, deform it first from the endpoint into the decaying negative-real direction and then onto the same upper-saddle descent path. Near the endpoint, $s=z-1$ and $\varphi =3s+O(s^3)$, so the endpoint contribution in steepest descent isAdding the saddle and endpoint pieces gives