= Solution
Evaluate the supplied real <logarithmic derivative> at $2+it$. Since $1<2-\beta<2$, its positive summand is bounded above and below by constant multiples of $(1+|t-\gamma|^2)^{-1}$. On the other hand, differentiating the defining xi expression gives
$$
\frac{\xi'}{\xi}(s)=\frac1s+\frac1{s-1}-\frac12\log\pi
+\frac12\frac{\Gamma'}{\Gamma}(s/2)+\frac{\zeta'}{\zeta}(s).
$$
At real part two the last term is bounded by the <absolutely convergent> series $\sum\Lambda(n)n^{-2}$; the gamma <logarithmic derivative> is $O(\log t)$. Therefore
$$
\boxed{\sum_\rho\frac1{1+|t-\gamma|^2}=O(\log t).}
$$
For $\gamma\in[T,T+1]$, use $t=T$: each term in the displayed sum is at least $1/2$. Hence the number of zeros in that unit ordinate interval, counted with multiplicities, is $O(\log T)$. These are the <local zeta zero-count bound> and the corresponding smoothed bound.
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