I'm just learning in my class at college about the cauchy integral theorem. We learned is that $\oint_c \frac{dz}{z} = 2\pi i$ where c is the unit circle oriented counter clockwise. I'm wondering what happens if it is not a closed loop. As far as I understand it, if you want to integrate from one point to another on the complex plane, you simply need to draw a simply connected region that encompasses the whole path of the integral and show that it is analytic everywhere inside. If this were true I would expect the integral the above function around the unit circle from say $\theta = 3\pi /4$ to $5\pi /4$ to be $\frac{3\pi}{2} i$. See below:
However if I integrate along the path which I know always works regardless of analyticity, I get the parameterized version: $\oint_c \frac{dz}{z} =\int_{\frac{3\pi}{4}}^{\frac{5\pi}{4}}\frac{ie^{it}}{e^{it}}$ and this evaluates to $\frac{\pi}{2} i$
I understand that there is something of a discontinuity of the graph of ln(z) at the line along the real axis extending to the left of the origin. Essentially, values of ln(z). Essentially ln(z) jumps by $-2\pi i$ when it crosses this axis in the counterclockwise orientation.
Is there a singularity along the whole line of the real axis starting at the origina and moving to the left? Or do I not understand the conditions for indefinite integration with relation to the function 1/z?
$\ln(z)$ doesn't have a discontinuity. However, it is multivalued, so if you want to make an actual function out of it on some domain, you have to pick one of the multiple possible values for each point. This cannot be done in a continuous manner on all of the complex plane, so a standard "maximal domain" to use is the complex plane minus the non-positive reals (this is called doing a "branch cut"). One can also include the negative reals in the domain, but then you do necessarily get a discontinuity.
You're specifically interested in a domain which crosses the non-positive reals, so you have to see that as a subset of some other "maximal" domain.