 Solution

ID: past-exam-of-the-probability-statistics-and-options-ii-course-of-the-university-of-oxford/2008/d/4/b/ii/solution

No. On each interval

[kπ, (k + 1) π]

, for

k \geq 1

\int_{kπ}^{(k + 1) π} \frac{∣ s i n x ∣}{x} d x \geq \frac{1}{( k + 1 ) π} \int_{kπ}^{(k + 1) π} ∣ sin x ∣ d x = \frac{2}{( k + 1 ) π} .

The resulting harmonic lower bound diverges. Therefore

sin x / x \in / L^{1} (0, \infty)

. Its convergent improper integral does not make it Lebesgue integrable; its positive and negative parts both have infinite integral.

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