Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 13 / c / i

Table of contents
- Solution i

Solution

Use integration by substitution with the reciprocal variable

t = 1/ x

, so

d x = - d t / t^{2}

and

lo g x = - lo g t

. The endpoints

x = 1, \infty

become

t = 1, 0

. Thus

\int_{1}^{\infty} x^{- 3} lo g x d x = \int_{1}^{0} t lo g t d t = - \int_{0}^{1} t lo g t d t .

Renaming the dummy integration variable proves the identity without evaluating either side.

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