Challenging series of a fraction with a cosine and a variable Is it possible to express the following series explicitly (e.g. as a polynomial in $\alpha$):
$$f(\alpha )=\sum_{k=1}^{\infty} \frac{\cos(\alpha2\pi k)}{(2\pi k)^2((2\alpha k)^2-1)^2} , $$
where $0\leq\alpha\leq1$ ?
Do you have any idea on how to approach this? Thank you so much!
 A: $$S_0(a)=\sum_{k=1}^{\infty} \frac{\cos(2\pi ka)}{(2\pi k)^2((2ka)^2-1)^2}=\frac{1}{4\pi^2}\frac{1}{16a^4}\Re \frac{1}{2}S(a)$$
where
$$S(a)=\sum_{\binom{k=-\infty}{k\neq0}}^{\infty}\frac{e^{2\pi ika}}{k^2(k-1/2a)^2(k+1/2a)^2} ,\,\,a\in(0;1), \,\,a\neq\frac{1}{2n} \text(n - integer)$$
To evaluates $S(a)$ we go in the complex plane and integrate the function $f(z)=\frac{2\pi i \,e^{2\pi iza}}{(e^{2\pi iz}-1)z^2(z-1/2a)^2(z+1/2a)^2} $ along a big circle with the radius $R\to\infty$.
The integrand declines fast enough at $|z|\to \infty $, so
$$\oint f(z)dz\to 0$$
On the other hand,
$$\oint f(z)dz =2\pi i\sum Res $$
We have inside the cirlcle the simple poles in the points $z=\pm1, \pm2,...$ , the pole of the third degree at $z=0$, and a couple of simple poles at $z=\pm1/2a$

The residues at $z=\pm1, \pm2,...$ give us the sum $S(a)$, therefore
$$S(a)=-Res_{z=0}f(z)-Res_{z=\pm1/2a}f(z)$$
The easiest way to evaluate $Res_{z=0}$ is to decompose the integrand at $z=0$ and evaluate the coefficient at $z^2$
$$f(z)=-2\pi i \,16a^4\frac{1+2\pi iaz-2\pi^2a^2z^2+...}{(1-1-2\pi iz+2\pi^2z^2-(2\pi iz)^3/3!+...)z^2(1-4a^2z^2)^2}$$
$$-Res_{z=0}f(z)=-(2a)^4(8a^2-2\pi^2a^2-\pi^2/3+2\pi^2a)$$
The residues evaluation at $z=\pm1/2a$ is straightforward:
$$-Res_{z=1/2a}f(z)=-\frac{8\pi ia^4}{e^{\pi i/a}-1}\Big(6a-2\pi ia+2\pi i\frac{e^{\pi i/a}}{e^{\pi i/a}-1}\Big)$$
$$-Res_{z=-1/2a}f(z)=-\frac{8\pi ia^4}{e^{-\pi i/a}-1}\Big(-6a-2\pi ia+2\pi i\frac{e^{-\pi i/a}}{e^{-\pi i/a}-1}\Big)$$
Taking all together,
$$S(a)=16a^4(\pi^2/3-\pi^2a+2\pi^2a^2-8a^2)-\frac{8\pi^2a^4}{\sin^2\frac{\pi}{2a}}-48\pi a^5\cot\frac{\pi}{2a}$$
Finally,
$$S_0=\frac{1}{4\pi^2}\frac{1}{16a^4}\frac{1}{2}S(a)=\frac{1}{24}-\frac{a}{8}+\frac{a^2}{4\pi^2}(\pi^2-4)-\frac{1}{16\sin^2\frac{\pi}{2a}}-\frac{3}{8\pi}\,a\cot\frac{\pi}{2a}$$
