 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2008/d/d3/c/i/solution

Substitute

t = lo g x

, with

d x = e^{t} d t

. The nonnegative integral becomes

\int_{0}^{\infty} t e^{- (α - 1) t} d t = \frac{1}{( α - 1 ) ^{2}},

α > 1

\int_{1}^{\infty} x^{- α} lo g x d x = (α - 1)^{- 2} .

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