Antiderivative exists but not integrableIn the sense of Riemann, could an integral (if there exists could you give the less pathological counter-example possible) have an antiderivative but not be integrable on a compact subset?

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Kayley Jackson · Accepted Answer

Explanation:It is possible. Let   f  (  x  )  =      x          1.2        sin  ⁡  (      1    x    ) when   x  ∈  (  0  ,  1  ] and   f  (  0  )  =  0. It is a differentiable function (check it) in the interval [0,1] and its derivative equals to       f    ′    (  x  )  =  1.2      x          0.2        sin  ⁡  (      1    x    )  −      1          x              0.8              cos  ⁡  (      1    x    ) when   x  ≠  0 and       f    ′    (  0  )  =  0. So obviously f′ has an antiderivative in [0,1], but it isn&#039;t even bounded, hence not Riemann integrable.

Antiderivative exists but not integrable In the sense of Riemann, could

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