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 Incoming links: Past exam of the integration course of the University of Oxford / 2021 / 2 / c / i

Past exam of the integration course of the University of Oxford / 2021 / 2 / c / ii / Solution 2026-10-25

The function is Lebesgue integrable. The inequalities

∣ sin u ∣ \leq min {u, 1}

for

u > 0

give bounds that handle the two ends of the

x

-axis differently. For

0 < x < 1

,

\int_{0}^{\infty} e^{- 2 x y} ∣ sin x sin y ∣ d y \leq x \int_{0}^{\infty} e^{- 2 x y} d y = \frac{1}{2} .

For

x \geq 1

,

\int_{0}^{\infty} e^{- 2 x y} ∣ sin x sin y ∣ d y \leq \int_{0}^{\infty} y e^{- 2 x y} d y = \frac{1}{4 x ^{2}} .

Tonelli's theorem therefore proves

\iint_{(0, \infty)^{2}} e^{- 2 x y} ∣ sin x sin y ∣ d x d y \leq \frac{1}{2} + \int_{1}^{\infty} \frac{d x}{4 x ^{2}} = \frac{3}{4} < \infty.

The factors of sine suppress the near-axis contributions that caused divergence in part i.

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