Complex integration theorem validation

If $f$ is analytic on a path from $a$ to $b$, then the integral on that path is $\int_{a}^{b} f(z)dz = F(b) - F(a)$, where $F' = f$.

However, if the path is a circuit, then for the above formula to work (meaning the integral is $0$), we have the additional requirement that $f$ must be analytic inside the region enclosed by the circuit.

Are the statements I mentioned correct?

Solution 1:

The first statement needs a qualification: "if there exists $F$ in a neighbourhood of the path such that $F' = f$ there."

Analyticity of $f$ in a region containing the circuit and the region enclosed by it is sufficient (but not necessary) for such $F$ to exist.