Let
$$
D(x)=x(x+\varepsilon)+\varepsilon^3e^{-x^2},
\qquad
D_0(x)=x(x+\varepsilon)+\varepsilon^3.
$$
The inequality $|1-e^{-x^2}|\leq x^2$ and a split into $0<x<\varepsilon^2$, $\varepsilon^2<x<\varepsilon$, and $\varepsilon<x<1$ show that
$$
\int_0^1\left|\frac1D-\frac1{D_0}\right|dx
\leq C\int_0^1\frac{\varepsilon^3x^2}{D_0(x)^2}\,dx
=O(\varepsilon^2).
$$
Thus replacing the exponential by one does not affect any term through $O(1)$.

Factor
$$
D_0(x)=(x+a)(x+b),
$$
where
$$
a=\frac{\varepsilon}{2}
\left(1-\sqrt{1-4\varepsilon}\right),
\qquad
b=\frac{\varepsilon}{2}
\left(1+\sqrt{1-4\varepsilon}\right).
$$
<Partial fraction decomposition> gives
$$
\int_0^1\frac{dx}{D_0(x)}
=\frac{1}{\varepsilon\sqrt{1-4\varepsilon}}
\left[
\log\frac ba+
\log\frac{1+a}{1+b}
\right].
$$
The required <asymptotic expansion> follows from
$$
\frac1{\sqrt{1-4\varepsilon}}
=1+2\varepsilon+O(\varepsilon^2),
$$
$$
\log\frac ba=-\log\varepsilon-2\varepsilon
+O(\varepsilon^2),
\qquad
\log\frac{1+a}{1+b}=-\varepsilon+O(\varepsilon^2).
$$
Therefore
$$
\boxed{
I(\varepsilon)
=\frac{\log(1/\varepsilon)}{\varepsilon}
+2\log(1/\varepsilon)-3+o(1).}
$$
The two small roots reveal the same nested scales $x=O(\varepsilon^2)$ and $x=O(\varepsilon)$ that a <divide-and-conquer asymptotic expansion> would match explicitly.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 336 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-336/2/a/solution