Sum of the series $\sum_{n\in \mathbb{Z}} \frac{e^{in x}}{(n+\alpha)^2}$

Question

Here $\alpha $ is a non-integer real number. The series is apparently uniformly convergent and periodic.

Is it possible to find a closed expression of it?

Answer 1

For $c \in (0,2\pi)$ then $$f(z) =e^{icz}\frac{\pi^2}{\sin^2(\pi z)}-i c e^{icz}\frac{2i\pi}{e^{2i\pi z}-1}- \sum_n \frac{e^{icn}}{(z-n)^2}$$ is entire and bounded thus constant. Since $\lim_{x\to \infty} f(-ix) = 0$ then $f(z)=0$.

Answer 2

The function is related to the Lerch zeta function. In particular, $$L\left(\frac{x}{2\pi}, \alpha,2\right) =\sum_{n=0}^\infty \frac{e^{in x}}{(n+\alpha)^2}.$$

Thus, we have that $$\sum_{n\in\mathbb{Z}} \frac{e^{in x}}{(n+\alpha)^2}=L\left(\frac{x}{2\pi}, \alpha,2\right) + L\left(-\frac{x}{2\pi}, \alpha,2\right)- \frac{1}{\alpha^2}\,.$$