Any chances to find the following sum analytically?

Question

Consider the following sum:

\begin{equation} S(x, k, N, \alpha_1, \alpha_2) = \sum\limits_{m=1}^{N} \dfrac{ e^{i \phi_m k}}{x - \Omega(m, \alpha_1, \alpha_2, N)}, \end{equation}

where $x$ - complex valued parameter, $k, N$ - positive integers ($k<N$), $\phi_m = \dfrac{2 \pi (m-1)}{N}$, $\Omega (m, \alpha_1, \alpha_2, N) = i \alpha_1 \dfrac{\xi + \xi^{1/N} e^{-i \phi_m}}{1 + \xi^{1/N} e^{-i \phi_m}} - i \alpha_2$, and $\alpha_1, \alpha_2$ - real-valued constants, and $0<\xi<1$.

I really doubt that for any $N$ this can be written as a simple formula in terms of several known functions, but may there be any possible way to simplify it in certain limits? Like $N \to \infty$, or saying that $\alpha_2 \gg \alpha_1$.

Answer 1

This sum has a closed form. After some simplification (multiplying the numerator and the denominator of the summand by $1+\xi^{1/N}e^{-i\phi_m}$, replacing $m$ with $m+1$, etc.), it is a linear combination of sums of the form (here $z\in\mathbb{C}$) $$G(n,k,z)=\sum_{m=0}^{n-1}\frac{\omega^{km}}{1-z\omega^m},\qquad\omega=e^{2\pi i/n}.$$

This can be computed like I did here (or here), or as follows: if $|z|<1$ then $$G(n,k,z)=\sum_{m=0}^{n-1}\omega^{km}\sum_{j=0}^\infty z^j\omega^{mj}=\sum_{j=0}^\infty z^j\sum_{m=0}^{n-1}\omega^{m(k+j)},$$ and the inner sum is $0$ unless $k+j$ is a multiple of $n$; then, if $1\leqslant k\leqslant n$, $$G(n,k,z)=n\sum_{q=1}^\infty z^{nq-k}=\frac{nz^{n-k}}{1-z^n}.$$

Using $G(n,k,z)=G(n,n+k,z)$, and analytic continuation (out of $|z|<1$), $$G(n,k,z)=\frac{nz^{(-k)\bmod n}}{1-z^n}.$$