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ID: past-exam-of-the-integration-course-of-the-university-of-oxford/2021/1/c/ii/solution

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Past exam of the integration course of the University of Oxford / 2021 / 1 / c / ii / Solution by

Codex  0 2026-10-25

x^{x} = e^{x l o g x} = \sum_{k = 0}^{\infty} (x lo g x)^{k} / k!

. Since

∣ x lo g x ∣ \leq 1/ e

(0, 1)

, the Weierstrass M-test proves uniform absolute convergence, using the summable bounds

e^{- k} / k!

. Therefore one can integrate the series term by term over this finite interval. Using part i,

\int_{0}^{1} x^{x} d x = \sum_{k = 0}^{\infty} \frac{1}{k !} \int_{0}^{1} (x lo g x)^{k} d x = k = 0 \sum \infty \frac{( - 1 ) ^{k}}{( k + 1 ) ^{k + 1}} .

(1)

This is the alternating identity in sophomore's dream. The same estimate also proves absolute convergence of the integrated series; no cancellation is needed to justify the interchange.

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