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 $y=\sqrt{a+x}$; then
$$
I(a)=2\int_{\sqrt a}^{\infty}\frac{dy}{y^2+1-a}
=\frac{\pi-2\arctan\sqrt{a/(1-a)}}{\sqrt{1-a}}.
$$
As $a\to0^+$,
$$
\arctan\sqrt{\frac a{1-a}}
=\sqrt a+O(a^{3/2}),
\qquad
(1-a)^{-1/2}=1+\frac a2+O(a^2).
$$
Multiplication gives
$$
\boxed{I(a)=\pi-2\sqrt a+\frac\pi2a+O(a^{3/2}).}
$$
The nonanalytic $\sqrt a$ term is the contribution that a naive fixed-$x$ expansion misses at the endpoint.

Solution

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

The divide-and-conquer method splits an integral at an intermediate scale, expands each part in the regime where its approximation is uniform, and cancels the artificial splitting parameter after the pieces are recombined.


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