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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-336/1/a/solution