Actually, I don't get from where the big-$O$ comes from. Let $ \chi $ be a primitive Dirichlet character of modulus $q$, $L(s,\chi)$ the $L$-function and $\Lambda(s,\chi)$ the completed $L$-function. Let $ N(T,\chi) $ let be denote the number non-trivial zeros $\rho=\beta+i\gamma$ of $L(s,\chi)$ in the rectangle $\left| \gamma \right| \leq T$ and $N'(T,\chi)$ the number of zeros of $ \Lambda(s,\chi) $ with $0 \leq \beta \leq 1 $ and $ 0 < \gamma \leq T $. Prove that
- $N(T,\chi)= N'(T,\chi) + N'(T,\overline{\chi}) + O(\log q T)$
- $ N'(T,\chi) = I(T) + O(\log q T)$ where $$ I(t) = \frac{1}{2\pi i} \oint_{C} \frac{\Lambda'}{\Lambda}(s,\chi)ds $$ here $C$ denote the recatngle of verticies $3-i\delta, 3+iT, -2+iT, -2-i\delta$ and where $\delta>0$ is choosen small such that $\Lambda(s,\chi)\neq 0 $ in $ - \delta \leq \Im s < 0 $.
Actually for 1. by the functional equation of $\Lambda(s,\chi)$, $\rho=\beta+ i \gamma$ is zero with $0 < \gamma \leq T $ of $\Lambda(s,\overline{\chi})$ if and only if $1-\rho = \beta' + i \gamma' $ is a zero with $-T \leq \gamma' < 0 $ of $ \Lambda(s,\chi) $. Hence I would say that $$N(T,\chi)= N'(T,\chi) + N'(T,\overline{\chi}) $$ For 2. I would say that by the choice of $\delta$ and by the argument principle we have that $$ I(T) = N'(T,\chi)$$ since $\Lambda(s,\chi)$ is an entire function hence it has no poles inside $C$.
Where is my mistake?