Evaluate $\int_{\Gamma_{r}}\frac{1}{z}\,dz$ 
Let ${\Gamma_{r}}$ be a curve disk around $(0,0)$ evaluate :$$\int_{\Gamma}\frac{1}{z}dz$$

Using ${\Gamma_{r}}(t)=R(\cos(t)+i\sin(t))=Re^{it}$ we get:
$$\int_{\Gamma}\frac{dz}{z}=\int_{0}^{2\pi}\frac{dt}{Re^{it}}iRe^{it}=\int_{0}^{2\pi}idt=2\pi i$$
Now there is a theorem that if a function have an antiderivative on a closed curve so the integral is $0$
In this case $\log(z)=\ln|z|+i\text{Arg}(z)$ when $-\pi<\text{Arg}(z)\leq \pi$ is defined on our curve $Re^{it}$ except for one point, namely $t=-\pi$ so that theorem can not be used?
Or it fails to fill the requirements of Cauchy's integral theorem? Because $Re^{it}$ close a domain which include $(0,0)$ where $\frac{1}{z}$ is not defined?
 A: Indeed $\frac 1z$ fails the requirements for the Cauchy-Goursat integral theorem, but that doesn't guarantee that the integral should be non-zero. Also it is impossible to use the antiderivative theorem directly since it is impossible to define an antiderivative which is differentiable over the entirety of the contour; the fact that we must exclude some point (such as that at $t = \pi$) is essential to the fact that this integral is non-zero.
In fact: we could have found the integral using that antiderivative as follows: break $\Gamma$ up into $\Gamma_0$ which traces the curve from angles $-\pi + \epsilon$ to $\pi - \epsilon$, and the curve $\Gamma_\epsilon$ which traces the rest of the circle, i.e. the angles from $\pi - \epsilon$ to $\pi + \epsilon$.  We note that
$$
\int_\Gamma \frac 1z \,dz = \int_{\Gamma_0} \frac 1z \,dz + \int_{\Gamma_\epsilon} \frac 1z \,dz
$$
With the antiderivative theorem, we have
$$
\int_{\Gamma_0} \frac 1z \,dz = i \operatorname{Arg}(\pi - \epsilon) - i \operatorname{Arg}(-\pi + \epsilon)
$$
We can also show that $\lim_{\epsilon \to 0} \int_{\Gamma_\epsilon} \frac 1z \,dz = 0$.
Thus, we have
$$
\int_\Gamma \frac 1z \,dz = 
\lim_{\epsilon \to 0^+} \left[ \int_{\Gamma_0} \frac 1z \,dz + \int_{\Gamma_\epsilon} \frac 1z \,dz\right]
= 
\lim_{\epsilon \to 0^+}\int_{\Gamma_0} \frac 1z \,dz + 
\lim_{\epsilon \to 0^+}\int_{\Gamma_\epsilon} \frac 1z \,dz\\ = 
\lim_{\epsilon \to 0^+} \left[i \operatorname{Arg}(\pi - \epsilon) - i \operatorname{Arg}(-\pi + \epsilon)\right] = 2 \pi i
$$
