Let $g$ be a holomorphic function on $C$ and its interior, where $C$ is a circle with a radius $r$ around the origin. Furthermore let's assume that $g$ doesn't assume a zero value on $C$ and its interior. Evaluate the integral:
$$\oint_C \log z \frac{g'(z)}{g(z)} dz$$
I tried parametrizing the circle as $re^{it}$ and using the line $t = \pi$ as a branch cut for the logarithm. The most I could get is:
$$\oint_C \log z \frac{g'(z)}{g(z)} dz = -r \int_{-\pi}^{\pi}te^{it}\frac{g'(re^{it})}{g(re^{it})} dt = 2\pi i \log(g(-r)) - i\int_{-\pi}^{\pi}\log(g(re^{it}))dt$$
where in the last part I used integration by parts. However I can't evaluate the last integral. If I apply integral by parts again to it I return back to the initial problem.
On the other side the integral resembles the Argument Principle, but this time $g$ is holomorphic on the region, so I don't how we can take advantage of it.
Additionally if I'm not mistaken we have that:
$$\lim_{r \to 0}\oint_C \log z \frac{g'(z)}{g(z)} dz = 0$$
I proved this by using the fact that $\left|\frac{g'(z)}{g(z)}\right|$ can be bounded by some $M$, not dependent on $r$ and then:
$$\left|\oint_C \log z \frac{g'(z)}{g(z)} dz\right|\le\oint_C \left|\log z \frac{g'(z)}{g(z)}\right| dz \le 2\pi r (\ln r + \pi)M$$ Upon taking limit from both sides we get the answer.
This somehow does the job for me, but I'm still interested in whether the integral can be evaluated explicitly for any $r>0$
Let $f(z)=g’/g$.
Your aim is to find $$\int_{\gamma_1}f(z)\log z~dz$$
By Cauchy’s integral theorem, $$\int_{\gamma_1}=-\left(\int_{\gamma_2}+ \int_{\gamma_3}+ \int_{\gamma_4}\right)$$
You have proved $\int_{\gamma_3}=0$.
By the theorem I wrote here, $$\left(\int_{\gamma_2}+ \int_{\gamma_4}\right)f(z)\log z~dz=-2\pi i\int^{re^{ia}}_0f(z)dz$$
So, indeed, $$\int_{\gamma_1}\frac{g’(z)}{g(z)}\log z~dz=2\pi i[\ln g(re^{ia})-\ln g(0)]$$