Computing in closed form $\sum_{n=1}^{\infty}\frac{\operatorname{Ci}\left(\frac{3}{4}\zeta(2) \space n\right)}{n^2}$

$\def\cosi{{\rm Ci}}$ It is well-known, (see for instance[ 1,$\S 5.2$]), that for positive real numbers $x$ we have $$\cosi(x)=\gamma+\ln x -\int_0^x\frac{1-\cos t}{t}\, d t.$$ It follows that, for $0<a\leq 2\pi$ and $n\geq 1$ we have $$\cosi(a n)=\gamma+\ln a +\ln n -\int_0^a\frac{1-\cos(n t)}{t}\, d t.$$ and consequently, $$\sum_{n=1}^\infty\frac{\cosi(a n)}{n^2}=(\gamma +\ln a)\sum_{n=1}^\infty\frac{1}{n^2}+ \sum_{n=1}^\infty\frac{\ln n}{n^2}-\int_0^a\frac{1}{t}\left(\sum_{n=1}^\infty\frac{1-\cos nt}{n^2}\right) d t,$$ where, in the last term we used the positivity of the integrand to justify interchanging summation and integration. We conclude that $$\sum_{n=1}^\infty\frac{\cosi(a n)}{n^2}=\frac{\pi^2}{6}(\gamma +\ln a)- \zeta^\prime(2)-\int_0^a\frac{1}{t}\left(\sum_{n=1}^\infty\frac{1-\cos nt}{n^2}\right) d t.\tag{1}$$ On the other hand, it is well-known, that for $0\leq t\leq 2\pi$ that $$\sum_{n=1}^\infty\frac{\cos nt}{n^2}=\frac{\pi^2}{6}-\frac{\pi t}{2}+\frac{t^2}{4},$$ and consequently, $$\sum_{n=1}^\infty\frac{1-\cos nt}{n^2}=\frac{\pi t}{2}-\frac{t^2}{4}.$$ Therefore, for $0\leq a\leq 2\pi$, we obtain $$\int_0^a\frac{1}{t}\left(\sum_{n=1}^\infty\frac{1-\cos nt}{n^2}\right)\, d t =\int_0^a\left(\frac{\pi }{2}-\frac{t}{4}\right)\, d t=\frac{\pi a }{2}-\frac{a^2}{8}.$$ Replacing this in $(1)$ we obtain the desired formula : $$\sum_{n=1}^\infty\frac{\cosi(a n)}{n^2}=\frac{\pi^2}{6}(\gamma +\ln a)- \zeta^\prime(2)-\frac{\pi a }{2}+\frac{a^2}{8}, \quad\hbox{for $0<a\leq 2\pi$}$$

Remark. This problem is enlisted as an open problem in the old Siam website here.

Remark. Of course, I forgot to say that the case we are considering corresponds to $a=\dfrac{3}{4}\zeta(2)=\dfrac{\pi^2}{8}$.