How do you integrate #ln(x) / x# from 4 to infinity?

1 Answer
Jul 30, 2016

the integral does not converge.

Explanation:

for the basic integration, spot the pattern:

#d/dx (ln^alpha x) = alpha ln^(alpha - 1) x * 1/x#

So

#d/dx (color{red}{1/2} ln^2 x) = 1/2 * 2 ln x * 1/x = ln x * 1/x#

So

#int_4^oo \ ln(x) / x \ dx = int_4^oo \ d/dx (1/2 ln^2 x) \ dx#

#= lim_(t to oo) (1/2 ln^2 x)_4^t #

the problem being, the integral does not converge.