Is this closed-form summation a special case of known Lerch zeta function formulas? With some Poisson summation manipulations (credit: Michał Pacholski) I have convinced myself of a closed form expression for this conditionally convergent series:
$$\sum_{n=-\infty}^\infty \frac{e^{in\alpha}}{z+n}=\frac{2\pi  i}{e^{i\alpha z}-e^{i(\alpha-2\pi) z}},\;\;\alpha\in(0,2 \pi),\;\;z\in\mathbb{C}\backslash\mathbb{Z}.$$
Mathematica returns a Lerch zeta function (more precisely: a sum of two Hurwitz-Lerch transcendents), without further simplification. The above formula implies for the Lerch zeta function $L(a,b,s)=\sum_{n=0}^\infty \frac{e^{2\pi ina}}{(n+b)^s}$ the reflection formula
$$L(a,b,1)-L(-a,-b,1)=\frac{1}{b}+\frac{2\pi  i}{e^{2\pi iab}-e^{2\pi i(a-1) b}},\;\;a\in(0,1),\;\;b\in\mathbb{C}\backslash\mathbb{Z},$$
or more generally for $s\in\mathbb{N}$
$$L(a,b,s)+(-1)^{s}L(-a,-b,s)=\frac{1}{b^{s}}+\frac{(-1)^{s-1}}{(s-1)!}\frac{d^{s-1}}{db^{s-1}}\left(\frac{2\pi i}{e^{2\pi iab}-e^{2\pi i(a-1) b}}\right).$$
I would be pleased to learn if this is a known result for this special function, or a reference to a derivation in the literature.
 A: The $z$-derivative of your sum is given in M.Engelhardt and B.Schreiber, Z.Phys.A 351 (1995) 71, cf. eqs. (6) and (7) therein (sorry, I don't have a freely accessible online link). As already mentioned as a comment to Christian Remling's answer, these sums can be evaluated by pulling out a factor $e^{-iz\alpha }$ and replacing factors $1/(z+n)$ with integrations, upon which one can perform the sum.
A: This is the Fourier series for the RHS, as a function of $\alpha\in (0,2\pi)$,
$$
f(\alpha)=\frac{2\pi i}{1-e^{-2\pi iz}}\, e^{-iz\alpha} .
$$
The series representation follows by computing the Fourier coefficients and noting that $f$ (as a function on the circle) is smooth away from $\alpha\equiv 0\bmod 2\pi$, so the Fourier series converges to the function.
A: I eventually found the reflection formula for the Lerch zeta function in the literature, it's theorem 2 in New properties of the Lerch's transcendent (2016), by E. M. Ferreira, A. K. Kohara, J. Sesma. They give a generalized expression for this relation, which also holds (by analytical continuation) when $z\equiv e^{2\pi ia}$ is not on the unit circle. I find it quite remarkable that it's such a recent result.
when comparing formulas, note that $\frac{2\pi i}{e^{2\pi iab}-e^{2\pi i(a-1) b}}=\pi  (\cot \pi  b+i) e^{-2 i \pi  a b}$.
A: Let me suggest a more pedestrian approach than the elegant solution of Christian Remling but which is arguably easier to come up with: contour integration.
To get the sum over $\mathbb{Z}$ from a contour integral we need a function which has poles at all integers. And thankfully we know such a function - it is $\frac{1}{\sin(\pi z)}$ ! More precisely, the function
$$f(z)=\frac{\pi e^{iz(\alpha - \pi)}}{(z+w)\sin(\pi z)}$$
has residue $\frac{e^{in\alpha}}{w+n}$ at the point $n$ (I switched $z$ from the OP to $w$ because I want $z$ be a free complex variable). Note that $-\pi$ in the exponent is because residues of $\frac{1}{\sin(\pi z)}$ alternate between $\frac{1}{\pi}$ and $\frac{-1}{\pi}$.
Now let us consider a circle of radius $R$ centred at the origin with, say, $R = N+\frac{1}{2}$ for integer $N$ just to be far from the poles. Since $-\pi < \alpha - \pi < \pi$ we can see that the integral tends to $0$ (exponent $e^{iz(\alpha - \pi)}$ is being killed by the growth of $\sin (\pi z)$) and therefore sum of the residues is zero. On the other hand there's only one more residue that we have: $z = -w$ and the value at it gives us precisely the result.
