 Solution

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

Again the function is continuous on its domain and therefore a measurable function. For

0 < x < 1

∣ (lo g x) e^{- x} ∣ \leq - lo g x

, and

\int_{0}^{1} - lo g x d x = 1.

This standard improper integral follows from the antiderivative

x - x lo g x

and

x lo g x \to 0

at zero. For

x \geq 1

0 \leq lo g x \leq x

\int_{1}^{\infty} x e^{- x} d x = 2/ e

. Thus

\int_{0}^{\infty} ∣ (lo g x) e^{- x} ∣ d x \leq 1 + \frac{2}{e} < \infty.

The logarithmic singularity is integrable and the exponential factor controls the tail, so this too is a Lebesgue integrable function.

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