I know that integral of removable discontinuous functions are differentiable but jump discontinuous aren't. However, When 2xsin(1/x)-cos(1/x) is integrand, which is derivative of x^2sin(1/x) has no limit defined at x=0(oscillating discontinuity) and its primitive(integral) exist as x^sin(1/x), but not differentiable. 2xsin(1/x)-cos(1/x) is Riemann integrable function(bounded) but it's integral is not differentiable. Can I say all integral of functions that have oscillating discontinuous points(not finite) aren't differentiable? If it's true, how can I prove it?
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$\begingroup$ But, $x^2 \sin(1/x)$ is differentiable everywhere! It's the classic example of a differentiable, but not continuously differentiable function. $\endgroup$– Theo BenditSep 17, 2022 at 6:43
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$\begingroup$ Well at points of oscillating discontinuity the integral may or may not be differentiable. If $f(x) =\int_0^x\cos(1/t)\,dt,g(x)=\int_0^x|\cos(1/t)|\,dt$ then both $f, g$ are differentiable everywhere. The case of $f$ is easier to handle than that of $g$. $\endgroup$– Paramanand Singh ♦Nov 10, 2022 at 7:26
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$\begingroup$ See math.stackexchange.com/q/1551332/72031 $\endgroup$– Paramanand Singh ♦Nov 10, 2022 at 7:28
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