 Solution

ID: past-exam-of-the-integration-course-of-the-university-of-oxford/2021/1/c/i/solution

For each nonnegative integer

k

x = e^{- t}

gives

\int_{0}^{1} ∣ (x lo g x)^{k} ∣ d x = \int_{0}^{\infty} t^{k} e^{- (k + 1) t} d t = \frac{Γ ( k + 1 )}{( k + 1 ) ^{k + 1}} < \infty.

Γ (k + 1) = k!

follows by repeated integration by parts. Since

lo g x < 0

(0, 1)

, the sign of the original integrand is

(- 1)^{k}

, so

\int_{0}^{1} (x lo g x)^{k} d x = \frac{( - 1 ) ^{k} k !}{( k + 1 ) ^{k + 1}} .

For

k = 0

, the integral is one, consistent with the formula.

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