# Calculate the following Integral $\int_{0}^{\infty} \frac{|\sin(\pi*x)|}{\lfloor x \rfloor} dx$

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.

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

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.

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