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Why do analytic functions always have an antiderivative on a simply connected region?

Thank you for your time,

Chris

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Well, authors are not in universal agreement about the definition of analytic, so let us know where you're starting from. –  user17794 Jun 17 '12 at 2:35
    
$f$ is analytic on a set $\Omega \subseteq \mathbb{C}$, if it is complex differentiable in every point in $\Omega$. –  Chris Jun 17 '12 at 2:37
    
I'd imagine $\Omega $ is also required to be open. See my answer. –  user17794 Jun 17 '12 at 3:01
    
Yes, you are right. –  Chris Jun 17 '12 at 3:03
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1 Answer 1

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Cauchy's Theorem tells us that $\displaystyle\int_{\gamma_1} f(x) dz = \displaystyle\int_{\gamma_2} f(x) dz$ whenever $\gamma_1 \gamma_2$ are homotopic, simple curves sharing the same endpoints in $f$'s domain of analyticity $\Omega .$ We can use this fact to define the antiderivative of $f$ in $\Omega $ as follows: $F(\omega ) = \displaystyle\int_{\gamma } f(z) dz,$ where $\gamma $ is a contour connecting a fixed point $z_0\in \Omega $ to $\omega.$ Analyticity is required simply so that $F$ is well-defined (in general it is not.) Using a parametrization $\rho (t) = \omega_0 + (\omega - \omega_0)t, 0\le t \le 1,$ you can show that the difference quotient satisfies $\frac{F(\omega ) - F(\omega_0)}{\omega - \omega_0} = \displaystyle\int_0^1 f(\omega_0 + t(\omega - \omega_0)) dt.$

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I omitted some detail here but hope it makes sense - let me know. –  user17794 Jun 17 '12 at 21:49
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