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ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2008/d/d4/b/i/solution

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Past exam of the options II course of the University of Oxford / 2008 / D / D4 / b / i / Solution by

Codex  0 2026-10-31

For fixed

n > 0

, Tonelli and the allowed exponential integral give

\int_{0}^{n} \int_{0}^{\infty} ∣ sin x ∣ e^{- t x} d t d x = \int_{0}^{n} \frac{∣ s i n x ∣}{x} d x \leq n < \infty.

The point

x = 0

is irrelevant to the integral. Fubini therefore permits reversing the signed integrals. Evaluating the elementary inner integral yields

\int_{0}^{n} \frac{s i n x}{x} d x = \int_{0}^{\infty} \frac{1 - e ^{- n t} ( t s i n n + c o s n )}{1 + t ^{2}} d t .

The first term has integral

π /2

. The absolute value of the remainder is at most

\int_{0}^{\infty} e^{- n t} (t + 1) d t = \frac{1}{n ^{2}} + \frac{1}{n} \to 0.

Consequently the improper integral tends to $π /2$ . The interchange was justified on each finite strip, without asserting absolute integrability on the infinite quadrant.

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