This is my first post about this topic and now I am trying to evaluate the integral, $$N=\frac{1}{2\pi i}\oint_{|z-1|=1}\frac{\zeta'(z)-1}{\zeta(z)-z}dz+1$$ $\zeta-$is the Riemann Zeta function. I am expecting that $N=1$ i.e. the integration will be zero but I am unable to do so. Further, I have tried to check for $|z-1|=r$ for any suitable $r>0$ but nothing meaningful happened.
If my assumption is wrong or somehow there is any mistake in the integral [something with well definedness] kindly let me know.
Any help/guide/complete derivation will be highly appreciated. Thanks in advance!
Let $f(z) =\frac{\zeta'(z)-1}{\zeta(z)-z}$. It is the derivative of $\log (\zeta(z)-z)$ so you can approximate it numerically to find the branch of $\log (\zeta(z)-z)$ continuous on $1+e^{it},t\in (0,2\pi)$.
Otherwise with the residue theorem, if $\zeta(z) -z$ doesn't have any zero on $|z-1|\le 1$ then $z=1$ is the only pole of inside $|z-1|<1$ and $$\int_{|z-1|=1} f(z)dz = 2i\pi Res(f(z),z=1)=-2i\pi$$ The residue is $-1$ because $\zeta(z)$ has a simple pole at $z=1$ so $\zeta(z) = \frac{C}{z-1}+O(1)$ and $$ f(z) = \frac{\frac{-C}{(z-1)^2}+O(1)}{\frac{C}{z-1}+O(1)} =\frac{-1}{z-1}+O(1)$$
Also you can do it without the residue theorem using that $\int_{|z-1|=r} f(z)dz$ doesn't depend on $r\le 1$ (Cauchy integral theorem) ie. $$\int_{|z-1|=1} f(z)dz = \lim_{r \to 0}\int_{|z-1|=r} f(z)dz= \lim_{r \to 0}\int_{|z-1|=r} \frac{-1}{z-1}dz+\int_{|z-1|=r}O(1)dz\\=\lim_{r \to 0} -2i\pi+O(r)=-2i\pi$$