A question about Yitang Zhang's paper "On the zeros of ζ’(s) near the critical line"

Question

We conclude that, in the case $\sigma = 1/2$ and $\zeta’\left(s\right) \neq 0$,

$$\mathrm{Re}\frac{\eta’}{\eta}\left(s\right) = \sum_{\beta’ \gt 0}{\mathrm{Re}\frac{1}{s - \rho’}} + O(1)$$

It is easy to see that, for $s = 1/2 + it$, the error term in (2.9) is continuous in $t$. If $\eta(1/2 + it) \neq 0$, then

$$-\sum_{\beta’ \gt 0}{\mathrm{Re}\frac{1}{1/2 + it - \rho’}} = F_{1}(t) - F_{2}(t)$$

where

$$F_{1}(t) = -\sum_{\beta’ \gt 1/2}{\mathrm{Re}\frac{1}{1/2 + it - \rho’}}$$ $$F_{2}(t) = -\sum_{0 \lt \beta’ \lt 1/2}{\mathrm{Re}\frac{1}{1/2 + it - \rho’}}$$

I am wondering where the terms with $\beta’=1/2$ are.

Answer 1

As Carlo Beenakker mentions in the comments,

\begin{equation*} \begin{split} -\sum_{\beta'=1/2} \mathrm{Re} \left(\frac{1}{1/2+it-\rho'}\right)&=- \mathrm{Re}\left( \frac{1}{1/2+it-(1/2+it')}\right)\\ &=- \mathrm{Re}\left( \frac{1}{(t-t')i}\right)=0 \end{split} \end{equation*}

I suggest that you run this type of question by math.stackexchange first.