I need the following sum ( in the sense of principal value):
$$\sum\_{s=-\infty}^{\infty}\frac{(-1)^{s}e^{-2\pi isy}}{x+s}$$


It is possible to show that
$$\sum\_{s=-\infty}^{\infty}\frac{e^{-2\pi isy}}{x+s}=\pi\frac{e^{\pi ix(2FractionalPart[y]-1)}}{\sin(\pi x)}$$ 


Hence the sum I need is
$$f(y)=\sum\_{s=-\infty}^{\infty}\frac{e^{-2\pi is(y+0.5)}}{x+s}=\pi\frac{e^{\pi ix(2FractionalPart[y+0.5]-1)}}{\sin(\pi x)}$$
But now it is difficult to see that the function f is continuous on $(0,1)$.
How to obtain expression without 0.5? 


The second question is what are this series in both cases? 
Maybe I wrong, but it seems that it is not Fourier series of $\pi\frac{e^{\pi ix(y-1)}}{\sin(\pi x)}$



 series    Sum[(-1)^n/(x+n)]
This is in a typical complex analysis text
$$
\frac{1}{x} + \sum\_{s = 1}^{\infty}\left( \frac{1}{x + s} + \frac{1}{x - s}\right) = \pi \cot (\pi x)
$$
When grouped this way, it converges...


Maple says
$$
\frac{1}{x} + \sum\_{s = 1}^{\infty} \left(\frac{\operatorname{e} ^{(-2 i\pi s y)}}{x + s} + \frac{\operatorname{e} ^{2 i \pi s y}}{x - s}\right) = 
-\frac{1 - LerchPhi \biggl(\frac{1}{\operatorname{e} ^{2 i \pi y}},1,x\biggr) x + LerchPhi \bigl(\operatorname{e} ^{2 i \pi y},1,-x\bigr) x}{x}
$$
and numerically this seems to agree with what you said.


So your formula looks correct. Why do you say it is not continuous? The complex exponential has period $2\pi i$.


