9
$\begingroup$

Okay so I'm trying to determine whether $\int_{0}^{1}\left(\left\lceil \frac{1}{x} \right\rceil-\left\lfloor \frac{1}{x} \right\rfloor\right) \, dx$ converges and if so, to what value? So the function $\left\lceil \frac{1}{x} \right\rceil-\left\lfloor \frac{1}{x} \right\rfloor = 0$ when $x=\frac{1}{k}$ where $k$ is a natural number, and everywhere else $\left\lceil \frac{1}{x} \right\rceil-\left\lfloor \frac{1}{x} \right\rfloor = 1$ From there I split the integral:

\begin{align*} \int_0^1 \left(\left\lceil \frac{1}{x} \right\rceil-\left\lfloor \frac{1}{x} \right\rfloor\right) \, dx &= \int_{1/2}^1 1 \, dx +\int_{1/3}^{1/2} 1 \, dx + \int_{1/3}^{1/4} 1 \, dx + \cdots \\[10pt] &= (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + \cdots \\[10pt] &= \lim_{m \to \infty} 1-1/m = 1\;. \end{align*}

Am I correct?

$\endgroup$
1
  • 2
    $\begingroup$ Very nice solution indeed...I could learn some thing new from it. $\endgroup$
    – DeepSea
    Jul 16, 2015 at 0:02

1 Answer 1

4
$\begingroup$

Yes, you are correct. But it is much faster if you say that $\lceil \frac{1}{x} \rceil-\lfloor \frac{1}{x} \rfloor$ is $1$ except in a set of measure zero. If you don't know what the meausre of a set is, just say that the set is countable. Then $$\int_0^1\left(\left\lceil \frac{1}{x}\right \rceil-\left\lfloor \frac{1}{x} \right\rfloor\right)dx=\int_0^11dx=1 $$

Your solution is still very nice.

$\endgroup$
5
  • $\begingroup$ Thank you. I haven't taken any measure theory yet, and I wasn't sure of it being correct because of the function taking on a value of 0 at all of the integrals bounds. I guess that's a nice explanation as to why my first equality holds. $\endgroup$
    – lamyvista
    Jul 16, 2015 at 0:07
  • $\begingroup$ The problem then is that this function might not be integrable in Riemann's sense. I think it is, but it is not obviuos, that is, if you are using Riemann's integrals, you should show that it is integrable. $\endgroup$
    – ajotatxe
    Jul 16, 2015 at 0:17
  • $\begingroup$ @ajotatxe: One can just use the theorem that says that a function is Riemann integrable on an interval iff its set of discontinuities has measure zero. I think most introductory analysis classes cover that theorem. $\endgroup$
    – shalop
    Jul 16, 2015 at 1:07
  • $\begingroup$ Dirichlet's function (rationals map to $0$ and irrationals to $1$) is not integrable in Riemann's sense. $\endgroup$
    – ajotatxe
    Jul 16, 2015 at 3:09
  • 1
    $\begingroup$ @ajotatxe: That is not a counterexample to my statement. Think about it: where is that function discontinuous? $\endgroup$
    – shalop
    Jul 16, 2015 at 3:29

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.