This question is closely related to this question I posted before, but I will state it concisely. In fact, I want to decompose that $$\sum_{n_1\geq 1}\sum_{n_2\geq 1}\sum_{n_3\geq 1} \frac{\cos 2\pi n_1x_1\cos 2\pi n_2x_2\cos 2\pi n_3x_3}{n_1^2+n_2^2+n_3^2} =F(x_1,x_2,x_3)+R(x_1,x_2,x_3)$$ where $F(x_1,x_2,x_3)$ is a combination of some elementary functions, and $R(x_1,x_2,x_3)$ is a smooth function, i.e. $C^{\infty}(\mathbb{R}^3)$ function.
By invoking \begin{equation} \sum_{n_2\geq 1}\frac{\cos 2\pi n_2 x_2}{n_1^2+n_2^2}=\left\{ \begin{aligned} & -\frac{1}{2n_1^2}+\frac{\pi}{2n_1}\frac{\cosh \pi(2x_2-1)n_1}{\sinh \pi n_1} &n_1\neq 0\\ &\frac{\pi^2}{6}-\pi^2x_2+\pi^2x_2^2 & n_1=0 \end{aligned}\right. \end{equation} $$\sum_{n_1\geq 1}\frac{\cos 2\pi n_1x_1}{n_1}e^{-2\pi n_1x_2}=\pi x_2-\log 2-\frac{1}{2}\log (\sinh^2 \pi x_2 +\sin^2\pi x_1)$$ the first identity is proved by contour integral, i.e. $$\sum_{n_2\geq 1}\frac{\cos 2\pi n_2 x_2}{n_1^2+n_2^2}=\frac{1}{2}\sum_{n_2 \in \mathbb{Z}}\frac{\cos 2\pi n_2 x_2}{n_1^2+n_2^2}-\frac{1}{n_1^2}$$ and by residue theorem $$\sum_{n_2 \in \mathbb{Z}}\frac{\cos 2\pi n_2 x_2}{2\pi i(n_1^2+n_2^2)}+\mathcal{Re}\left[\mbox{residue of }\frac{e^{i2\pi x_2z}}{(n_1^2+z^2)(e^{i2\pi z}-1)} \mbox{ at zeros of }n_1^2+z^2 \right]=\int_{C_n} \frac{e^{i2\pi x_2z}}{(n_1^2+z^2)(e^{i2\pi z}-1)}dz \rightarrow 0$$ the second identity is simply the Taylor expansion of $\log(1+z)$.
Now we calculate \begin{equation} \begin{aligned} \sum_{n_1\geq 1}\cos 2\pi n_1x_1&\sum_{n_2\geq 1}\cos 2\pi n_2x_2 \sum_{n_3\geq 1}\frac{\cos 2\pi n_3x_3}{n_1^2+n_2^2+n_3^2} \\ =\ &-\frac{1}{2}\sum_{n_1\geq 1}\cos 2\pi n_1x_1\sum_{n_2\geq 1}\frac{\cos 2\pi n_2x_2}{n_1^2+n_2^2} \\ &\ +\frac{\pi}{2} \sum_{n_1\geq 1}\cos 2\pi n_1x_1\sum_{n_2\geq 1}\frac{\cos2\pi n_2x_2}{\sqrt{n_1^2+n_2^2}}\frac{\cosh \pi(2x_3-1)\sqrt{n_1^2+n_2^2}}{\sinh \pi \sqrt{n_1^2+n_2^2}} \\ =\ & G_1+\frac{\pi}{2}G_2 \end{aligned} \end{equation} $G_1$ is easily handled. Now calculate $G_2$: \begin{equation} \begin{aligned} \sum_{n_1\geq 1}&\cos 2\pi n_1x_1\sum_{n_2\geq 1}\frac{\cos2\pi n_2x_2}{\sqrt{n_1^2+n_2^2}}\frac{\cosh \pi(2x_3-1)\sqrt{n_1^2+n_2^2}}{\sinh \pi \sqrt{n_1^2+n_2^2}} \\ =\ &\sum_{n_1\geq 1}\cos 2\pi n_1x_1\sum_{n_2\geq 1}\frac{\cos2\pi n_2x_2}{\sqrt{n_1^2+n_2^2}}\left(\frac{\cosh \pi(2x_3-1)\sqrt{n_1^2+n_2^2}}{\sinh \pi \sqrt{n_1^2+n_2^2}}-e^{-2\pi \sqrt{n_1^2+n_2^2}x_3} \right) \\ &+\sum_{n_1\geq 1}\cos 2\pi n_1x_1\sum_{n_2\geq 1}\frac{\cos2\pi n_2x_2}{\sqrt{n_1^2+n_2^2}}e^{-2\pi \sqrt{n_1^2+n_2^2}x_3} \\ =\ &G_3+G_4 \end{aligned} \end{equation} the terms in $G_3$ exponential decays, so $G_3$ is smooth. Now my question is how to calculate $G_4$? $$G_4=\sum_{n_1\geq 1}\cos 2\pi n_1x_1\sum_{n_2\geq 1}\frac{\cos2\pi n_2x_2}{\sqrt{n_1^2+n_2^2}}e^{-2\pi \sqrt{n_1^2+n_2^2}x_3}$$