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A divide-and-conquer asymptotic expansion splits an integral at an intermediate scale, expands each region where its approximation is uniform, and recombines the pieces. Terms depending on the arbitrary splitting point must cancel to the claimed error order. This is useful when an endpoint and the rest of the domain require different approximations.

For $0<ϵ\ll 1$, considerChoose $ϵ\ll \delta \ll 1$. On $[0,\delta ]$, put $x=ϵX$ and integrate the inner form exactly:On $[\delta ,1]$, replacing $(x+ϵ{)}^{-1}$ by $x^{-1}$ givesTheir sum is $-\log ϵ+O(ϵ/\delta )$, independent of $\delta$ at leading order. Taking $\delta =\sqrt{ϵ}$ gives a vanishing error. Naively expanding the entire original integrand as $x^{-1}-ϵx^{-2}+\cdots$ fails at zero; the split makes the domain of each approximation explicit.

LetThe inequality $|1-e^{-x^2}|\le x^2$ and a split into $0<x<{\epsilon }^2$, ${\epsilon }^2<x<\epsilon$, and $\epsilon <x<1$ show thatThus replacing the exponential by one does not affect any term through $O(1)$.

FactorwherePartial fraction decomposition givesThe required asymptotic expansion follows fromThereforeThe two small roots reveal the same nested scales $x=O({\epsilon }^2)$ and $x=O(\epsilon )$ that a divide-and-conquer asymptotic expansion would match explicitly.

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.