Evaluate $$ \oint_{\left\vert z\right\vert\ =\ 1} \log\left(\left\vert \zeta\left(\frac{z}{z - 1}\right)\right\vert\right) \,{\mathrm{d}z \over z} $$ where $\zeta(s)$ is the analytic continuation of Riemann zeta function which is analytic except for a simple pole at $s = 1$.
MY TRY- Define $f(z)=\frac{\log \left| \zeta(\frac{z}{z-1})\right|}{z}$
Poles of $f(z)$ are $z=0$.
R=Residue of $f(z)$ at $z=0$.
R$=\lim_{z\rightarrow 0} (z-0) f(z)$
R= $\lim_{z\rightarrow 0} \log \left|\zeta(\frac{z}{z-1})\right| $
$\zeta(0)=-1/2 $ gives,
R$=\log\left(\left| \frac{-1}{2}\right|\right)=-\log \ 2$
By Cauchy's Residue Theorem,
$$\oint_{|z|=1}\frac{\log \left| \zeta(\frac{z}{z-1})\right|}{z}dz=-2\pi i \log \ 2$$
Is it correct?
$g(z)=\frac{z}{z-1}$ is biholomorphic $\Bbb{C}-1$ to $\Bbb{C}-1$ and $|z|<1$ to $\Re(s)<1/2$.
$g(g(z))=z$ thus $\zeta( g(z))$ is analytic on $|z|<1$ with some zeros at $g(-2n)$ and at the $g(\rho),\Re(\rho)<1/2$.
$\log \zeta(g(z))$ has some logarithmic branch points there.
Then apply Jensen's formula.
For $r<1$
$$\frac1{2i\pi}\int_{|z|=r} \log |\zeta(g(z))| \frac{dz}z= \log |\zeta(0)|-\sum_{g(-2n)<r} \log \frac{g(-2n)}{r} - \sum_{g(\rho)<r} \log \frac{|g(\rho)|}{r}$$ so that $$\lim_{r\to 1^-}\frac1{2i\pi}\int_{|z|=r} \log |\zeta(g(z))| \frac{dz}z= \infty$$