 # Antiderivative exists but not integrable In the sense of Riemann, could Maliyah Robles 2022-07-05 Answered
Antiderivative exists but not integrable
In 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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Explanation:
It is possible. Let $f\left(x\right)={x}^{1.2}\mathrm{sin}\left(\frac{1}{x}\right)$ when $x\in \left(0,1\right]$ and $f\left(0\right)=0$. It is a differentiable function (check it) in the interval [0,1] and its derivative equals to ${f}^{\prime }\left(x\right)=1.2{x}^{0.2}\mathrm{sin}\left(\frac{1}{x}\right)-\frac{1}{{x}^{0.8}}\mathrm{cos}\left(\frac{1}{x}\right)$ when $x\ne 0$ and ${f}^{\prime }\left(0\right)=0$. So obviously f′ has an antiderivative in [0,1], but it isn't even bounded, hence not Riemann integrable.