Write $e^x=1+xg(x)$, where $g=(e^x-1)/x$ is smooth at zero. Then
$$
\int_0^1\frac{dx}{\sqrt{x+\epsilon}}
=2-2\sqrt\epsilon+\epsilon+O(\epsilon^2),
$$
and a uniformly integrable expansion gives
$$
\int_0^1\frac{xg(x)}{\sqrt{x+\epsilon}}dx
=\int_0^1\sqrt x,g(x)dx
-\frac\epsilon2\int_0^1\frac{g(x)}{\sqrt x}dx
+O(\epsilon^{3/2}).
$$
By <integration by parts>,
$$
\int_0^1\frac{e^x-1}{x^{3/2}}dx=-2(e-1)+2I(0).
$$
Therefore
$$
\boxed{I(\epsilon)=I(0)-2\sqrt\epsilon+[e-I(0)]\epsilon+O(\epsilon^{3/2}).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 336 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-336/1/b/solution