why is Cauchy's integral formula not 0

Question

why is the Integral in Cauchy's integral formula not equal to 0 ? Im asking this because the Cauchy's integral theorem says that the integral of a holomorphic function over a closed curve is =0. But the in the Cauchy's integral formula ($f(z)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(\phi)}{\phi-z}d\phi$) you integrate over a closed curve and the function you integrate over($\frac{f(\phi)}{\phi-z}$) is only not holomorphic at z. But $\gamma$ never crosses z. I hope someone can see my mistake and correct me. Thank you :)

Answer 1

The theorem says the integral is $0$ if the function is holomorphic in an open set that includes all points in the region surrounded by the curve.

With $\displaystyle \int_\gamma \frac{f(\varphi)}{\varphi-z} \, d\varphi, $ the function is not holomorphic in an open set that contains $\varphi=z$ and $z$ is within the region surrounded by $\gamma.$