 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-3/9b/c/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Regularity at zero gives

r^{2} ϕ^{'} (r) = \int_{0}^{r} \frac{s ^{2}}{1 + s ^{2}}, d s = r - arctan r .

Integrating once more,

ϕ (r) = \frac{1}{2} lo g (1 + r^{2}) + \frac{a r c t a n r}{r} + C,

where

arctan r / r

has removable limit one at zero. At

r = 1

, the nonconstant terms sum to

\frac{1}{2} lo g 2 + \frac{π}{4}

. Hence

ϕ (r) = \frac{1}{2} lo g (1 + r^{2}) + \frac{arctan r}{r} - 1 - \frac{1}{2} lo g 2 - \frac{π}{4} .

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