Antiderivitives of complex functions

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My question is dose the existance of an antiderivitive in a domain imply that the function is anaylitic in that domain? (when f(x) is continous on that domain.)
 

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morphism
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Presumably this antiderivative is analytic on the domain, and hence so is its derivative, which is f.
 
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Thanks for the reply. Let me make sure i have this right. A function is only analytic if its antiderivitive is analytic? That means that just the existance of an antiderivitive alone dose not show that the function is analytic?


Thanks again.
 
HallsofIvy
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" Presumably this antiderivative is analytic on the domain, and hence so is its derivative, which is f."

Why would that be presumed? The question was whether or not the fact tha a function is once differentiable is enough to conclude that is is analytic on an interval. The function [itex]f(x)= e^{-\FRAC{1}{x^2}}[/itex] if x is not 0. f(0)= 0 is infinitely differentiable but not analytic.
 
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morphism
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Why would that be presumed?
Because I was under the impression that this question was complex analytic in flavor, i.e. that we're talking about complex derivatives.
 
mathwonk
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for complex functions, having an antiderivative implies that the antiderivative has one derivative and hence is analytic, thus so is the derivative, i.e. the original function.

simply out, for complex functions, the answer is yes.
 
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