 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-3/10b/b/solution

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

For fixed

x > 0

x \neq = 1

\int_{0}^{1} x^{y} d y = \frac{x - 1}{l o g x},

with limiting value one at

x = 1

. Integrating over

0 < x < 1

in the opposite order gives

\int_{0}^{1} \int_{0}^{1} x^{y} d y d x = \int_{0}^{1} \frac{d y}{y + 1} = lo g 2.

On the other hand, after setting

x = e^{- u}

\int_{0}^{1} \frac{x - 1}{l o g x} d x = \int_{0}^{\infty} \frac{e ^{- u} - e ^{- 2 u}}{u} d u .

Consequently

\int_{0}^{\infty} \frac{e ^{- u} - e ^{- 2 u}}{u} d u = lo g 2 .

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