Let $f\::\mathbb{R}\to\mathbb{R}$ be integrable function for all $[a,b],\hspace{0.2cm} (a<b, \hspace{0.5cm} a,b\in\mathbb{R}$).

and$\hspace{0.2cm}\forall c,x\in\mathbb{R} \hspace{0.2cm} f(x)\not=0 \hspace{0.2cm} and \hspace{0.2cm}{\displaystyle \int_{c}^{c+1}f(x)\,dx}=0$

Can I have an example for such function?

Thank you!

  • 1
    $\begingroup$ $f(x)=\{1 if x\in[\frac {2k} 2, \frac {2k+1} 2]; -1 if x\in[\frac {2k+1} 2, \frac {2k+2} 2]\}$ $\endgroup$ May 6, 2020 at 19:15
  • 2
    $\begingroup$ $sin 2\pi x$ should work $\endgroup$ May 6, 2020 at 19:15

1 Answer 1


You see according to what you say let us take

$f(x)=\sin 2\pi x, \text{ for } x\in\mathbb{R- Z}$

$f(x)=1, \text{ for } x\in\mathbb{ Z}$

$$\int_{a}^{b}\sin 2\pi x \mathrm dx <\infty, \forall a<b$$

And $$\int_{c}^{c+1}\sin 2\pi x \mathrm dx=\int_{0}^{1}\sin 2\pi x \mathrm dx=0, \forall c$$.

  • 1
    $\begingroup$ The question also asked for $\forall x \in \mathbb{R}, f(x) \neq 0$, but $\sin(2\pi 0) = \sin(0) = 0$. $\endgroup$ May 6, 2020 at 19:27
  • $\begingroup$ your answer works you just need to redefine sin where it is 0, to be something else, and it won't change the integral since it is a countable amount of points $\endgroup$
    – BinyaminR
    May 6, 2020 at 19:28
  • $\begingroup$ depends which kind of "integrable" he means but yeah $\endgroup$ May 6, 2020 at 19:34
  • $\begingroup$ @ChrisEagle I missed it sorry. $\endgroup$ May 6, 2020 at 19:37
  • $\begingroup$ @BinyaminR Thank you $\endgroup$ May 6, 2020 at 19:37

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