Summation of a quotient with a square root: $ \sum\limits_{n=-\infty}^{\infty}\frac{e^{inx}}{\sqrt{k^{2}-(n+\alpha)^{2}}(n+\beta)} $ Is anyone aware of an expression for the following series, presumably involving a special function?
$$
\sum_{n=-\infty}^{\infty}\frac{e^{inx}}{\sqrt{k^{2}-(n+\alpha)^{2}}(n+\beta)}
$$
Mathematica won't evaluate the sum and I can't find it in Gradshteyn and Ryzhik. It would be helpful even to have an expression for the case $x=0$ (i.e. $1$ on the numerator). Thanks in advance for any help.
 A: I tried this, but it clearly fails. I learned a technique which uses a floor function like for this case. You can find the sum over the negative reals from the positive reals sum. Note that the $\lfloor x+a\rfloor$ may take on different “a” values.
$$
\sum_{n=0}^{\infty}\frac{e^{inx}}{\sqrt{k^{2}-(n+a)^{2}}(n+b)}=\int_0^\infty \lfloor x+a\rfloor \frac{d}{dx}\left(\frac{e^{inx}}{\sqrt{k^{2}-(n+a)^{2}}(n+b)}\right)dx=\int_0^\infty\frac{\lfloor x+1\rfloor(a + n) e^{i n x}}{(b + n) (k^2 - (a + n)^2)^\frac32} + \frac{\lfloor x+1\rfloor i x e^{i n x}}{(b + n) \sqrt{k^2 - (a + n)^2)}} - \frac{\lfloor x+1\rfloor e^{i n x}}{(b + n)^2 \sqrt{k^2 - (a + n)^2)}}dx=\int_0^\infty\frac{\lfloor x\rfloor(a + n) e^{i n x}}{(b + n) (k^2 - (a + n)^2)^\frac32} + \frac{\lfloor x\rfloor i x e^{i n x}}{(b + n) \sqrt{k^2 - (a + n)^2)}} - \frac{\lfloor x\rfloor e^{i n x}}{(b + n)^2 \sqrt{k^2 - (a + n)^2)}}dx+ \frac{e^{i\infty x}}{\sqrt{k^{2}-(\infty+a)^{2}}(\infty+b)}-\frac{1}{b\sqrt{k^2-a^2}} $$
Also see alternate forms of the Floor function. This should work as the bolded link link demonstrates so. There are other options we have, but I may not find a function which “fits” the summation. Please correct me and give me feedback!
