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AxiomOfChoice
Oct30-09, 09:18 PM
I'm trying to show that a function f(z) is analytic by showing f'(z) exists. But f(z) is defined in terms of a contour integral:

f(z) = \oint_{|\zeta - z_0| = r} g(z,\zeta) d\zeta.

Since the integral is being carried out with respect to \zeta and not z, am I allowed to bring the d/dz operator inside the integral? Or is it more complicated than that? Are there certain conditions that g(z,\zeta) must satisfy? If so, what are they?

THANKS!

Gib Z
Oct30-09, 09:23 PM
I'm not sure if it will work the same for Contour Integrals, but the conditions for bringing a derivative inside are given nicely here: http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign

Phyisab****
Nov6-09, 08:18 AM
I think you might want to use leibnitz rule.