What is $\int_{\lvert z\rvert=1} \mathrm{Log}(z)\,dz$?
By letting $z = \mathrm{e}^{it}$, we get
$$\int_{\lvert z\rvert=1} \mathrm{Log}(z)\,dz = \int_0^{2\pi} \mathrm{Log}(\mathrm{e}^{it}) i\mathrm{e}^{it} dt = \int_0^{2\pi} (it) i\mathrm{e}^{it}\, dt = -\int_0^{2\pi}t\mathrm{e}^{it}\, dt $$
Does this make sense? But since the function $\,\mathrm{Log}(z)$ is not analytic on $\{(x,0) : x\leq 0\}$, therefore it is not analytic at the point $(-1,0)$ which is on the contour. And I have read that contour integral does not make sense when there is a singularity on the contour.
Thank you very much!
It does make sense.
Although $\log z$ is a multi-valued function and one might think that the value of the integral depends of the choice of the branch, in reality IT DOES NOT depend, as we simply and a constant, the contour integral of which if zero.