Fourier expansion of the complexified Gram series

Question

Consider the Riemann's R function, also known as the Gram series: $$\text{R}(x)=1+\sum_{k=1}^{\infty}\frac{\left(\log x\right)^{k}}{kk!\zeta(k+1)}$$ Now consider the form: $$\text{R}\left(e^{2\pi ix} \right )+\text{R}\left(e^{-2\pi ix} \right )$$ By the definition of $\text{R}(x)$ an easy calculation shows that: $$\text{R}\left(e^{2\pi ix} \right )+\text{R}\left(e^{-2\pi ix} \right )=2\sum_{m=1}^{\infty}\frac{\mu(m)}{m}\text{Ci}\left(\frac{2\pi x}{m}\right)$$ Where $\mu(m)$ is the $\text{M}\ddot{\text{o}}\text{bius}$ function. and $\text{Ci}(\cdot)$ is the cosine integral function. This function is clearly an even periodic function with period 1, and thus it has a Fourier series expansion of the form: $$\text{R}\left(e^{2\pi ix} \right )+\text{R}\left(e^{-2\pi ix} \right )=a_{0}+\sum_{n=1}^{\infty}a_{n}\cos(2\pi n x)$$ My question: i am having a trouble trying to find the constants $a_{n}$. Any idea on how to do that?