When does the conclusion of the Cauchy integral theorem fail? I am interested in a condition  when following theorem of Cauchy integral fails:
 let $U$ be an open subset of $C$ which is simply connected, let $f : U → C$ be a holomorphic function, and let $\!\,\gamma$ be a rectifiable path in $U$ whose start point is equal to its end point. Then

As I understand, first condition of failing this statement should be that  function should not be  holomorphic or  function that is  not  complex differentiable in a neighborhood of every point in its domain.also maybe   also $U$ subset if it is not connected, then this theorem may fail, what is also other conditions? thanks in advance
 A: The conclusion of a theorem may fail if one of its hypotheses fails. In this case, if either 


*

*the domain is not simply-connected, or

*the function is not holomorphic, or

*the path is not closed  


the integral may be nonzero. However, there is nothing that says it has to be nonzero, even if all 1-2-3 fail together. This is what fedja commented on: 

If any condition or a set of conditions fail, it doesn't yet mean that the conclusion fails. It is the same as my phrase "If it is raining, I carry an umbrella with me" does not imply that if it is not raining, I don't (maybe it is sunny but I am just carrying the umbrella to a repair shop, or for whatever other reason). Most mathematical theorems are also one way (If ..., then...). When it is a two way theorem, it normally starts with "The following are equivalent:" or some other explicit phrase with the same meaning.

If you are looking for a set of conditions to guarantee that the integral is nonzero, here is one: $f$ is holomorphic except for one simple pole, and $\gamma$ is a simple closed curve that separates the pole from the boundary of the domain of $f$. This is just a consequence of the Residue theorem. 
But if the path $\gamma$ is not locally rectifiable, there is no obvious meaning of $\int_\gamma f$ at all. In this case the statement fails simply because we lack a definition of the integral over such path.
