 Solution

ID: past-exam-of-the-probability-statistics-and-options-ii-course-of-the-university-of-oxford/2008/d/3/c/i/solution

Substitute

u = lo g x

d x = e^{u} d u

. The nonnegative integral becomes

\int_{0}^{\infty} u e^{- (α - 1) u} d u .

For

α > 1

1/ (α - 1)^{2}

; its boundary term vanishes because exponential decay dominates

u

. Hence it is finite, proving integrability and

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

 New to topics? Read the docs here!