# 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 I don't understand why the set of discontinuity points of $$f(x)$$ is infinite and has an infinite number of limit points. If so, it's 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
Commented 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. Commented 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 Commented 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
Commented Apr 7, 2017 at 9:38
• @Did i have copied it line to line from the book..... Commented Apr 7, 2017 at 13:10

We aren't integrating f(x), so we don't have to worry about f(x) being continuous. Rather, we are integrating g(x) = (1/x)^3, which is continuous.

Since |f(x)| = g(x), we can say that Since we know g(x) > f(x) and -g(x) < f(x). If g(x) approaches C, then f(x) is between C and -C. Therefore, it approaches a finite value.

Basically, we avoid having to worry about integrating f(x) by integrating a different function, g(x), instead and comparing the two.

Edit because my answer was unsatisfactory: Basically squeeze theorem is what proves the correctness of this method. Because squeeze theorem does not require a function to be continuous, this method works.

• I understand the method being followed ..... just concerned about its correctness Commented Apr 7, 2017 at 5:20