 Solution

ID: past-exam-of-the-integration-course-of-the-university-of-oxford/2023/2/a/i/solution

The function is continuous on

(0, \infty)

(0, 1)

its absolute value is at most one, while on

[1, \infty)

the elementary estimate

∣ sin u ∣ \leq ∣ u ∣

gives

∣ sin (1/ x^{3}) ∣ \leq x^{- 3}

. By comparison for the Lebesgue integral,

\int_{0}^{\infty} ∣ sin (1/ x^{3}) ∣ d x \leq 1 + \int_{1}^{\infty} x^{- 3} d x = \frac{3}{2} < \infty.

It is therefore a Lebesgue integrable function. The oscillations near zero do not obstruct absolute integrability because their amplitude stays bounded on an interval of finite measure.

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