Antiderivitives of complex functions

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Discussion Overview

The discussion revolves around the relationship between the existence of antiderivatives and the analyticity of complex functions. Participants explore whether the existence of an antiderivative in a domain implies that the function itself is analytic within that domain, particularly focusing on the conditions of continuity and differentiability.

Discussion Character

  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant questions if the existence of an antiderivative in a domain implies that the function is analytic in that domain, given that the function is continuous.
  • Another participant suggests that if the antiderivative is analytic, then its derivative (the original function) must also be analytic.
  • A different participant seeks clarification, asking if a function can be analytic only if its antiderivative is also analytic, implying that the existence of an antiderivative alone does not guarantee the function's analyticity.
  • One participant challenges the presumption that an antiderivative is analytic, providing an example of a function that is infinitely differentiable but not analytic, specifically mentioning \( f(x) = e^{-\frac{1}{x^2}} \) for \( x \neq 0 \) and \( f(0) = 0 \).
  • Another participant asserts that for complex functions, having an antiderivative implies that the antiderivative is analytic, and thus the original function is also analytic.

Areas of Agreement / Disagreement

Participants express differing views on whether the existence of an antiderivative guarantees the analyticity of the original function. There is no consensus reached on this matter, as some participants argue for the implication while others provide counterexamples.

Contextual Notes

The discussion highlights the nuances of differentiability and analyticity in complex functions, particularly the conditions under which these properties hold. There are unresolved assumptions regarding the definitions of analyticity and the implications of differentiability.

fivestar
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My question is dose the existence of an antiderivitive in a domain imply that the function is anaylitic in that domain? (when f(x) is continuous on that domain.)
 
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Presumably this antiderivative is analytic on the domain, and hence so is its derivative, which is f.
 
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 existence of an antiderivitive alone dose not show that the function is analytic?


Thanks again.
 
" 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.
 
Last edited by a moderator:
HallsofIvy said:
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.
 
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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