Value of the double sum of product of cosines For $m \neq n$, $a \in (0,1/2)$
\begin{equation}
\sum_{n\geq 1}^{\infty} \sum_{m\geq 1}^{\infty}\int_{a \pi}^{\pi/2}  \frac{\cos(2mx)\cos(2nx)}{mn} \, dx
\end{equation}
I know that the value of the integral is
\begin{equation}
\sum_{n\geq 1}^{\infty} \sum_{m\geq 1}^{\infty} \left(-\frac{\left(n-m\right)\sin\left(2{\pi} a n+2{\pi}a m\right)+\left(n+m\right)\sin\left(2{\pi}a n -2{\pi}a m\right)}{4mn\left(n^2-m^2\right)} \right)
\end{equation}
Does this double sum have a close form? If so, what is it?

Since $\sum_{m\geq 1}\frac{\cos(2mx)}{m}$ is pointwise convergent to $-\log(2\sin x)$ on $\left(0,\frac{\pi}{2}\right)$ we have:
$$\int_{a\pi}^{\pi/2}\sum_{m,n\geq 1}\frac{\cos(2mx)\cos(2nx)}{mn}\,dx= \int_{a\pi}^{\pi/2}\log^2(2\sin x)\,dx \tag{1}$$
and
$$ \int_{a\pi}^{\pi/2}\sum_{m=1}^{+\infty}\frac{\cos^2(2mx)}{m^2}\,dx = \int_{a\pi}^{\pi/2}\left[\frac{\pi^2}{6}-x(\pi-2x)\right]\,dx \tag{2}$$
is an elementary integral. $(1)$, however, depends on $\text{Li}_2$ (the dilogarithm function).
In the special case $a=0$ we have
$$ \int_{0}^{\pi/2}\log^2(2\sin x)\,dx = \int_{0}^{1}\frac{\log^2(2)+\log^2(x)+2\log(2)\log(x)}{\sqrt{1-x^2}}\,dx \tag{3}$$
and the RHS of $(3)$ can be computed by substituting $x=\sqrt{z}$ and by applying Feynman's trick to Euler's Beta function, leading to:
$$ \int_{0}^{\pi/2}\log^2(2\sin x)\,dx =\frac{\pi^3}{24}.\tag{4}$$
