Cauchy's integral theorem says that the integral around a closed contour of a function which is analytic on and inside of the contour is zero. If there is a region within the contour on which the function is undefined am I right in saying that the integral around this region is still path independent but just no longer equal to zero?
Is contour integration still path independent around a non-analytic region
complex integrationcomplex-analysis
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Best Answer
If you change the path by "sliding" it continuously over a region where the function is analytic, this will not change the result of the integral over the path. So, for instance, integrating $\frac1z$ over a path that goes one time counterclockwise around the origin is $2\pi i$, regardless of how that path goes one time counterclockwise around the origin. That's because any two such paths may be continuously deformed to one another without touching the origin.
But that's not true path independence. The value of the integral still depends on whether the path goes around the origin. If you want the value of the integral to be completely independent of what path you integrate over (which is to say, the value of the integral only depends on what the end points of the path is), then you need your region to be simply connected (i.e. not have any holes), and you need the function to be analytic on the whole region.