When I compute this contour integral around $s=1$ I get 0. $$\frac1{{2 \pi i}}\bigg(\int_{\frac{9}{10}-i}^{\frac{11}{10}-i} \frac{\zeta '(s)}{\zeta (s)} \, ds+\int_{\frac{11}{10}-i}^{\frac{11}{10}+i} \frac{\zeta '(s)}{\zeta (s)} \, ds+\int_{\frac{9}{10}+i}^{\frac{9}{10}-i} \frac{\zeta '(s)}{\zeta (s)} \, ds+\int_{\frac{11}{10}+i}^{\frac{9}{10}+i} \frac{\zeta '(s)}{\zeta (s)} \, ds\bigg)=0$$ But according to argument principle it should be equal to N-P where N is number of zeros and P number of poles. So the result should be -1 as there is one pole at s=1 inside contour and no zeros.
What am I doing wrong?
See any text on complex analysis. For $\gamma(t),t \in (0,1)$ a small enough closed-contour around $s=1$ in positive direction $$\int_\gamma \frac{\zeta'(s)}{\zeta(s)}ds \overset{def}=\int_0^1\frac{\zeta'(\gamma(t))}{\zeta(\gamma(t))}\gamma'(t)dt = -\int_\gamma \frac{1}{s-1}ds+\int_\gamma f(s)ds = -2i\pi$$ where $f(s) = \frac{\zeta'(s)}{\zeta(s)}+\frac{1}{s-1}$ is analytic on a simply connected open set containing the small enough contour, thus it has an analytic anti-derivative $F(s)= \int_1^s f(z)dz$ thus $\int_\gamma f(s)ds = F(s)|_{\gamma(0)}^{\gamma(1)} = 0$.