I want to evaluate the series
\begin{equation} S(\alpha,\omega)=\sum_{k=-\infty}^{\infty}\frac{|\Psi(2k\pi-\omega)|^2}{|2k\pi-\omega|^\alpha} \end{equation} where $0\le\omega<2\pi$, $0<\alpha<1$ and $\Psi(\omega)$ is the Fourier transform of the Daubechies wavelet.
From the basics of Daubechies wavelet, we know $\Psi(\omega)$ decays very fast thus the above series converges. But without a closed-form expression for $\Psi(\omega)$, this series is hard to evaluate numerically to a tolenrated precision, in my opinon. I guess by calculating the first $M$ terms of the series and using an asymptotics for the remaining would work, i.e.,
\begin{equation} S(\alpha,\lambda)=\sum_{k=-M}^{M}\frac{|\Psi(2k\pi-\omega)|^2}{|2k\pi-\omega|^\alpha} + 2\sum_{k=M+1}^{\infty}\frac{|\Psi(2k\pi-\omega)|^2}{|2k\pi-\omega|^\alpha} \end{equation}
but it requires an asymptotic for the second term of the right side of the equation, or more specifically, an asymptotic for $\Psi(\omega)$.
Is there an asymptotic for $\Psi(\omega)$, or is there a better way to evaluate the series?
Thank you for your answer