We notice in advance that $\tilde{\alpha}_{1}$ and $\bar{\alpha}_{1}$ are periodic in $\eta$, so it suffices to assume $|x|\leq 1/2$. The main idea of a variant of Cauchy’s integral test for convergence.

Using the fact that $|\frac{\partial^{2}\alpha_{1}}{\partial x^{2}}|\leq C\frac{a_{22}}{(\mu_{1}^{2}+a_{22}|\eta|^{2})^{3/2}}$, and the mean value type inequality

$$f(0)-\int_{-1/2}^{1/2}f(s)ds\leq C\sup_{|s|\leq 1/2}|f^{\prime\prime}(s)|,$$

we deduce that for $|\eta|^{2}+\frac{\mu_{1}^{2}}{a_{22}}\gtrsim 1$,

$$|\alpha_{1}(\mu_{1},\mu_{2},\eta)-\int_{x-1/2}^{x+1/2}\alpha_{1}(\mu_{1},\mu_{2},s+\sqrt{-1}y)ds|\leq CA^{-1/4}(|\eta|^{2}+\frac{\mu_{1}^{2}}{a_{22}})^{-3/2}$$

Thus for $|x|\leq 1/2$,

$$\begin{split}&|\sum_{n\in\mathbb{Z}\setminus\{0\}}\{\alpha_{1}(\mu_{1},\mu_{2},\eta+n)-\int_{x-1/2}^{x+1/2}\alpha_{1}(\mu_{1},\mu_{2},s+n+\sqrt{-1}y)ds\}|\\ &\leq CA^{-1/4}\sum_{|n|\neq 0}(|\eta+n|^{2}+\frac{\mu_{1}^{2}}{a_{22}})^{-3/2}\leq CA^{-1/4},\end{split}$$

and the sum converges to zero as $\mu_{1}^{2}+|\eta|^{2}\to\infty$.

In particular if $y^{2}+a_{22}^{-1}\mu_{1}^{2}\gtrsim 1$, then adding the above two inequalities already implies the bound

$$\begin{split}|\tilde{\alpha}_{1}-\bar{\alpha}_{1}|&=|\sum_{n\in\mathbb{Z}}\{\alpha_{1}(\mu_{1},\mu_{2},\eta+n)-\int_{x-1/2}^{x+1/2}\alpha_{1}(\mu_{1},\mu_{2},s+n+\sqrt{-1}y)ds\}|\leq CA^{-1/4},\end{split}$$

and that $|\tilde{\alpha}_{1}-\bar{\alpha}_{1}|$ converges to zero as $\mu_{1}^{2}+|\eta|^{2}\to\infty$.

If however $y^{2}+a_{22}^{-1}\mu_{1}^{2}\ll 1$ but $\mu_{2}\lesssim-A^{1/4}$, then we can make

$$\frac{a_{22}\mu_{2}+a_{12}\mu_{1}}{A^{1/2}\sqrt{\mu_{1}^{2}+a_{22}|\eta|^{2}}}\lesssim-1,$$

and the Taylor expansion of $\arctan$ will ensure $|\alpha_{1}(\mu_{1},\mu_{2},\eta)|\leq\frac{C}{-\mu_{2}}$, so

$$|\alpha_{1}(\mu_{1},\mu_{2},\eta)-\int_{x-1/2}^{x+1/2}\alpha_{1}(\mu_{1},\mu_{2},s+\sqrt{-1}y)ds|\leq C\mu_{2}^{-1}\leq CA^{-1/4},$$

from which we again deduce $|\tilde{\alpha}_{1}-\bar{\alpha}_{1}|\leq CA^{-1/4}$. ∎