# Examine the convergence of the improper integral $\int_1^{\infty} f(x) dx$

Examine the convergence of the improper integral $\int_1^{\infty} f(x) dx$ where

$f(x) = \frac{1}{x^3}$ if $x$ be rational $\geq 1$ and

$f(x) = \frac{-1}{x^3}$ if $x$ be irrational $>1$.

The solution given is straight forward $\int_{1}^{\infty}|f(x)|dx= \int_{1}^{\infty}\frac{1}{x^3}dx$ so its convergent, since is $f(x)$ absolutely convergent so is $\int_1^{\infty} f(x) dx$.

But what I don't understand is, the set of points of discontinuity of $f(x)$ is infinite and has infinite number of limit points. If so its not even riemann integrable in the first place. How then can we comment on its convergence ?

• Not Riemann integrable but Lebesgue integrable since $f(x)=-1/x^3$ almost everywhere.
– Did
Apr 6, 2017 at 19:11
• Just to be sure, does the book handle only Riemann integrability? Or does it discuss Lebesgue integrals as well? The comment of Did applies. Apr 6, 2017 at 19:12
• @mickep just reimann integrability .....its mostly undergraduate math...so would help if the answer is from reimann integration point of view Apr 6, 2017 at 21:10
• If the book writes the formula $\int_1^\infty f(x)\,dx$ and means it in the Riemann sense, then the book is wrong. Does the book actually write this formula, or only the perfectly correct formula $\int_1^\infty|f(x)|\,dx$
– Did
Apr 7, 2017 at 9:38
• @Did i have copied it line to line from the book..... Apr 7, 2017 at 13:10