 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2005/d/2/i/solution

Since

lo g x \leq 0

(0, 1]

, substitute

x = e^{- s}

to obtain

\int_{0}^{1} ∣ x^{t - 1} lo g x ∣ d x = \int_{0}^{\infty} s e^{- t s} d s = \frac{1}{t ^{2}} < \infty (t > 0) .

The functions are measurable, so this proves Lebesgue integrability. Also

0 < e^{- x} \leq 1

gives

∣ h_{t} (x) ∣ \leq ∣ g_{t} (x) ∣

. Hence $h_{t}$ is integrable on $(0, 1]$ as well.

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