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I'm looking for Riemann integrable function $f:\mathbb{R}\to\mathbb{R}$ with $\int^{a+1}_a f(x)dx=0$ for all $a\in\mathbb{R}$, but $ f(x)\neq 0$.

I suspect that floor function involves here, if so, then how?

Thank you all!

To clarify: $f$ must not be identically equal to $0$, and it should be integrable over any finite interval.

Sandra West
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  • It would be good if you clarified what you mean by $f(x)\ne0$. Does that mean that $f$ can never take the value $0$? Or rather that $f$ should not be identically equal to $0$? As you see, some of the examples below may or may not work, depending on what you are asking. – Andrés E. Caicedo May 11 '13 at 16:46
  • I meant that $f$ should not be identically equal to $0$. – Sandra West May 11 '13 at 16:49
  • Then Lana's (user:77181) example works fine, and then the question becomes whether a less "silly" example is possible. Anyway, could you clarify: By "Riemann integrable", do you mean that the improper integral $\int_{-\infty}^\infty f(x),dx$ exists, that the improper integral of $|f|$ exists, or that the integrals over finite intervals exist? – Andrés E. Caicedo May 11 '13 at 16:54
  • The integrals over finite intervals exist. – Sandra West May 11 '13 at 16:57
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    Ah, ok, thank you for the reply. Then the other two examples show a general approach. I would suggest to edit the question so these clarifications are not buried in the comments. – Andrés E. Caicedo May 11 '13 at 16:59

4 Answers4

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What about the characteristic function of a singleton?

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If you mean "Riemann integrable on every finite interval", try $f(x)=\sin{2\pi x}$. If it needs to be non-zero everywhere, you may redefine it to be $1$ for $2x\in\mathbb Z$.

Dejan Govc
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Consider any integrable periodic function with period $1$, which integrates to $0$, i.e., let $g(x)$ be any function defined on interval $[0,1]$. Then consider $$f(x) = \begin{cases} g(x) - \underbrace{\int_0^1 g(x) dx}_b & \text{if }x\in[0,1]\\ g(\{ x\}) - b & \text{else}\end{cases}$$

  • This is surely not improperly Riemann integrable over all of $\mathbb{R}$. – Martin May 11 '13 at 16:36
  • @Martin Oh ok. I interpreted the question as Riemann integrable over any unit interval, which I believe is what the OP is looking for. –  May 11 '13 at 16:38
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How about trigonometric functions?

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    This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. – vadim123 May 11 '13 at 23:01
  • @vadim123, it sure does... think $\sin x$ between 0 and $2 \pi$. – vonbrand May 11 '13 at 23:18
  • @vonbrand, in my understanding of Math.SE policy short answers like this should be comments instead. I did not mean to comment on its relevance or usefulness, unfortunately the "low quality post" selection menu inserted this text automatically on my behalf. – vadim123 May 12 '13 at 00:12
  • @vadim123, while answers which elaborate are usually desired, in this case being brief is pretty reasonable! – Mariano Suárez-Álvarez May 12 '13 at 05:41
  • @vadim123: a short answer that has gotten a fair number of votes. – robjohn May 12 '13 at 09:58
  • @vadim123, I wanted to leave a comment, but the 'Add Comment' button was somehow disabled. I did not comment to critique or request any clarification, but rather to give the OP a hint. – András Hummer May 12 '13 at 12:37