TITLE: Evaluating $\sum_{k=1}^{\infty} \frac{\cos^2(k)}{4k^2-1}$
QUESTION [8 upvotes]: What options do I have for this series? No idea how to do it.
$$\sum_{k=1}^{\infty} \frac{\cos^2(k)}{4k^2-1}$$

REPLY [4 votes]: Avoiding contour integration:
$$
\begin{align}
\sum_{k=1}^\infty\frac{\cos^2(kx)}{4k^2-1}
&=\sum_{k=1}^\infty\frac{\cos(2kx)+1}{2(4k^2-1)}\\
&=\frac12\mathrm{Re}\left(\sum_{k=1}^\infty\frac{e^{2ikx}+1}{4k^2-1}\right)\\
&=\frac14\mathrm{Re}\left(\sum_{k=1}^\infty\frac{e^{2ikx}+1}{2k-1}-\frac{e^{2ikx}+1}{2k+1}\right)\\
&=\frac14\mathrm{Re}\left(e^{2ix}+1+\sum_{k=1}^\infty\frac{e^{2i(k+1)x}-e^{2ikx}}{2k+1}\right)\\
&=\frac14\mathrm{Re}\left(2+(e^{2ix}-1)+(e^{2ix}-1)\sum_{k=1}^\infty\frac{e^{2ikx}}{2k+1}\right)\\
&=\frac14\mathrm{Re}\left(2+(e^{2ix}-1)\sum_{k=0}^\infty\frac{e^{2ikx}}{2k+1}\right)\\
&=\frac14\mathrm{Re}\left(2+(e^{ix}-e^{-ix})\sum_{k=0}^\infty\frac{e^{i(2k+1)x}}{2k+1}\right)\\
&=\frac14\mathrm{Re}\left(2+i\sin(x)\log\left(\frac{1+e^{ix}}{1-e^{ix}}\right)\right)\\
&=\frac14\mathrm{Re}\left(2+i\sin(x)\log\left(i\tan(x/2)\right)\right)\\
&=\frac14\mathrm{Re}\left(2-\frac\pi2\sin(x)+i\sin(x)\log(\tan(x/2))\right)\\
&=\frac12-\frac\pi8\sin(x)
\end{align}
$$
Plugging in $x=1$ yields
$$
\sum_{k=1}^\infty\frac{\cos^2(k)}{4k^2-1}=\frac12-\frac\pi8\sin(1)
$$