I'm looking for an example of a function (if such a function exists) that cannot be integrated, but its derivative can.

Also, Does such a positive function exist, such that its co-domain is always positive? If so, do you have an example of one?

Thank you

  • 3
    $\begingroup$ No. The existence of derivatives implies continuity. Continuity implies integrability. $\endgroup$
    – A.S
    Apr 13, 2013 at 6:58
  • 1
    $\begingroup$ Lebesgue integral always improves the regularity of an integrand. Thus such thing cannot happen. $\endgroup$ Apr 13, 2013 at 6:58

1 Answer 1


If a function is differentiable on its entire domain, then the set of discontinuities is empty, therefore it is Riemann integrable. There do exist integrable functions that are not differentiable however.

  • $\begingroup$ I can see what you've said is true if the domain is a closed interval, but what if $f$ is defined on an open interval? $\endgroup$
    – user1551
    Apr 13, 2013 at 7:26
  • $\begingroup$ Even if it is closed you have to be pricese: Take the function $f : \mathbb{R} \to \mathbb{R} : x \mapsto 1$. It is not even Lebesgue integrable on $\mathbb{R}$. The theorem you state is true for bounded intervals. Then again Riemann integration only works for functions defined on a bounded domain. $\endgroup$
    – MrOperator
    Apr 13, 2013 at 7:34
  • $\begingroup$ For open domains this is not true. Take for example $f: ]0,1] \to \mathbb{R}: x \mapsto \dfrac{1}{x}$. $\endgroup$
    – MrOperator
    Apr 13, 2013 at 7:35

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