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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.

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  • $\begingroup$ 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. $\endgroup$ May 11, 2013 at 16:46
  • $\begingroup$ I meant that $f$ should not be identically equal to $0$. $\endgroup$ May 11, 2013 at 16:49
  • $\begingroup$ 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? $\endgroup$ May 11, 2013 at 16:54
  • $\begingroup$ The integrals over finite intervals exist. $\endgroup$ May 11, 2013 at 16:57
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    $\begingroup$ 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. $\endgroup$ May 11, 2013 at 16:59

4 Answers 4

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What about the characteristic function of a singleton?

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  • $\begingroup$ How I do that...? $\endgroup$ May 11, 2013 at 17:16
  • $\begingroup$ What is there to do? $\endgroup$ May 11, 2013 at 17:18
  • $\begingroup$ Ok..... got it! $\endgroup$ May 11, 2013 at 17:29
  • $\begingroup$ what is characteristics function of singletons? explain for me please. I do not understand your one line answer. $\endgroup$
    – Marso
    May 17, 2013 at 8:14
  • $\begingroup$ Pick $x \in \mathbb{R}$, the characteristic function $ \chi: \mathbb{R} \to \mathbb{R}$ of its singleton $ \{ x \}$ it's defined as follows: $\chi(z):=1$ iff $z=x$, otherwise $\chi(z):=0$. $\endgroup$ May 17, 2013 at 12:23
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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 $2x\in\mathbb Z$.

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    $\begingroup$ If you want it to be integrable as an improper riemann integral over the whole real axis, simply damp it with $e^{-x^2}$ $\endgroup$
    – fgp
    May 11, 2013 at 16:37
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    $\begingroup$ @fgp You wont then have $\int_{a}^{a+1}$ to be invariant for all $a$. $\endgroup$
    – user17762
    May 11, 2013 at 16:39
  • $\begingroup$ Hm, true. I missed that. $\endgroup$
    – fgp
    May 11, 2013 at 16:47
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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 $$f(x) = \begin{cases} g(x) - \underbrace{\int_0^1 g(x) dx}_b & \text{if }x\in[0,1]\\ g(\{ x\}) - b & \text{else}\end{cases}$$

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  • $\begingroup$ This is surely not improperly Riemann integrable over all of $\mathbb{R}$. $\endgroup$
    – Martin
    May 11, 2013 at 16:36
  • $\begingroup$ @Martin Oh ok. I interpreted the question as Riemann integrable over any unit interval, which I believe is what the OP is looking for. $\endgroup$
    – user17762
    May 11, 2013 at 16:38
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How about trigonometric functions?

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    $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. $\endgroup$
    – vadim123
    May 11, 2013 at 23:01
  • $\begingroup$ @vadim123, it sure does... think $\sin x$ between 0 and $2 \pi$. $\endgroup$
    – vonbrand
    May 11, 2013 at 23:18
  • $\begingroup$ @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. $\endgroup$
    – vadim123
    May 12, 2013 at 0:12
  • $\begingroup$ @vadim123, while answers which elaborate are usually desired, in this case being brief is pretty reasonable! $\endgroup$ May 12, 2013 at 5:41
  • $\begingroup$ @vadim123: a short answer that has gotten a fair number of votes. $\endgroup$
    – robjohn
    May 12, 2013 at 9:58

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