# Contour integral of complex differentiable function

From my response to a recent post, I began wondering what theorems don't hold for complex differentiable, non-holomorphic functions on $\mathbb{C}$ that hold for holomorphic functions. In particular suppose we know that a function is only complex differentiable on some connected path $X \subseteq \mathbb{C}$ (but is not holomorphic). Then if $\gamma$ is a parameterization of $X$ (i.e. $\gamma:[0,1]\rightarrow X$), can we say that $\int_{\gamma} f'(z)dz = f(\gamma(1))-f(\gamma(0))$? What if $X$ is a closed contour?

If a meromorphic function has a pole of order 1 inside $\gamma$, this would show up in the contour integral around the pole. Could we conclude anything about the nature of our complex differentiable, non-holomorphic function inside $\gamma$ if the above relationship is ture, i.e could we conclude it has no pole of order 1 inside $\gamma$ since the contour integral would be zero? I'm sure I'm missing something fundamental but I can't see where my flawed logic is when reading the proofs in my graduate complex analysis text.

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You first ask about a function that is differentiable on some closed, connected path $X$... Later, you ask, "What if $X$ is closed?" Is there something else you mean? –  Clayton Apr 25 '13 at 18:23
That was my mistake. I changed the sentence while I was writing the post but forgot to omit "closed" when I did. I edited the post to reflect what I meant. –  Cameron Williams Apr 25 '13 at 18:30
What do you mean by "differentiable" function here, if not holomorphic? Do you only want $f$ to be real differentiable as a function $f: R^2 \to R^2$? If so, then $\int_\gamma f'(z) dz$ does not make much sense, because $f'(z)$ is a linear map $R^2 \to R^2$ and not a number. –  xyzzyz Apr 25 '13 at 18:57
I mean complex differentiable, i.e. satisfying the Cauchy-Riemann equations, but not on an open set in this case. I'll edit my post to reflect that. –  Cameron Williams Apr 25 '13 at 18:58