In the series (3.4) defining $\tilde{\alpha}_{1}$, we can separate the sum into two ranges $|n|\gtrsim A^{-1/2}|\vec{\mu}|_{a}+1$ and $1\leq|n|\lesssim A^{-1/2}|\vec{\mu}|_{a}$. In the first range, using elementary Taylor expansion of arctan,

$$|\alpha_{1}(\mu_{1},\mu_{2},\eta+n)-\frac{1}{4|n|\sqrt{a_{22}}}|\leq\frac{C|\vec{\mu}|_{a}}{A^{3/4}|n|^{2}},$$

which implies after summation

$$\begin{split}\sum_{|n|\gtrsim A^{-1/2}|\vec{\mu}|_{a}+1}|\alpha_{1}(\mu_{1},\mu_{2},\eta+n)-\frac{1}{4|n|\sqrt{a_{22}}}|\leq CA^{-1/4}\min\{1,A^{-3/4}|\vec{\mu}|_{a}\}.\end{split}$$

The second range only appears if $1\lesssim A^{-1/2}|\vec{\mu}|_{a}$. This sum is crudely estimated by

$$\begin{split}&\sum_{1\leq|n|\lesssim A^{-1/2}|\vec{\mu}|_{a}}|\alpha_{1}(\mu_{1},\mu_{2},\eta+n)-\frac{1}{4|n|\sqrt{a_{22}}}|\leq CA^{-1/4}\sum_{1\leq|n|\lesssim A^{-1/2}|\vec{\mu}|_{a}}\frac{1}{n}\\ &\leq CA^{-1/4}\log(A^{-3/4}|\vec{\mu}|_{a}).\end{split}$$

Combining the discussions gives the result. ∎