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 ?

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

1 Answer 1


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.

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

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