So I have to calculate the following Integral: $\int_{0}^{\infty} \frac{|\sin(\pi*x)|} {\lfloor x \rfloor} dx$

I know how to calculate improper Integrals, but I really do have a problem with the $\lfloor x \rfloor$ (everytime it comes to any analysis excercise i can't solve it) even though i have read it's definition a lot of times and as far as i understand it it would be the greatest integer less than or equal to x. I also have a problem with he absolute value when it comes to integrals.

I would like to add up here my part of the solution but in this case i really am stuck and have no clue what to do. Therefore i would really be greatful for any sort of help.

Thanks in advance

  • 1
    $\begingroup$ The integrand function is not defined on the interval $(0, + \infty)$. If $0 < x < 1$ you have $\frac{0}{0}$. $\endgroup$
    – Crostul
    Oct 4, 2020 at 9:57
  • $\begingroup$ I see but that's exactly how it is given in the excercise $\endgroup$
    – Annalisa
    Oct 4, 2020 at 10:02
  • $\begingroup$ I would say "the integral is not defined", and conclude the exercize in this way. $\endgroup$
    – Crostul
    Oct 4, 2020 at 10:04
  • $\begingroup$ What's $\Huge *$ ?. $\endgroup$ Oct 4, 2020 at 15:01

1 Answer 1


I will solve the integral $$\int_1^{+ \infty} \frac{|\sin ( \pi x)|}{\lfloor x \rfloor} \mathrm d x$$ since for $0 < x < 1$ you have something divided by zero.

Split the integral into a series: $$\int_1^{+ \infty} \frac{|\sin ( \pi x)|}{\lfloor x \rfloor} \mathrm d x = \sum_{n=1}^{+ \infty} \int_n^{n+1} \frac{|\sin ( \pi x)|}{\lfloor x \rfloor} \mathrm d x $$ In the interval $(n,n+1)$ the floor function is constant and evaluates $n$.

Moreover $|\sin ( \pi x)|$ is a periodic function with period $1$. In particular $$\int_n^{n+1} |\sin ( \pi x)| \mathrm d x = \int_0^1 |\sin ( \pi x)| \mathrm d x = C$$ where $C$ is a positive constant I don't want to compute.

Hence $$\sum_{n=1}^{+ \infty} \int_n^{n+1} \frac{|\sin ( \pi x)|}{\lfloor x \rfloor} \mathrm d x= \sum_{n=1}^{+ \infty} \int_n^{n+1} \frac{|\sin ( \pi x)|}{n} \mathrm d x= \sum_{n=1}^{+ \infty} \frac{C}{n} = + \infty $$ So the integral is divergent.

  • $\begingroup$ ..Bravo A.! :-) $\endgroup$
    – Joe
    Oct 4, 2020 at 11:07
  • $\begingroup$ @Joe <3 this was easy... $\endgroup$
    – Crostul
    Oct 4, 2020 at 12:15
  • $\begingroup$ $\displaystyle +1$. Good job. I like your "I don't want to compute" statement. $\endgroup$ Oct 4, 2020 at 15:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.