Is there any difference between a function being integrable (since there are many ways to define whether a function is integrable, let's stick to Riemann integration.) and it's antiderivative existing? The antiderivative need not be any simple/analytic function.
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Yes. For instance, the function $f(0) = 1$, $f(x) = 0$ for $x \ne 0$ is integrable in any interval, but has no antiderivative.
Yoni Rozenshein
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it's antiderivative is any constant – ABC Apr 05 '13 at 17:02
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1@exploringnet, the derivative of a constant is $0$, so that can't be an antiderivative of $f$. – Yoni Rozenshein Apr 05 '13 at 23:11
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Adding to the discussion, if a function has an antiderivative, is it integrable? This depends on how you define the integration
http://en.wikipedia.org/wiki/Volterras_function
This function is not Riemannian integrable even though it is bounded and antiderivative exists everywhere. I wonder, does existence of an antiderivative imply Lebesgue integrability?
