The periodic Green's function $G^p$ for Laplacian $\Delta$ in $\mathbb{R}^3$ is defined by lattice sum: $$G^p(x)=-\sum_{n\in \mathbb{Z}^3-\{0\}}\frac{e^{2\pi i n\cdot x}}{4\pi^2|n|^2}$$ and it is well known that the fundamental solution $G$ for $\Delta$ in $\mathbb{R}^3$ is $$G(x)=-\frac{1}{4\pi |x|}$$
Problem The function $$R=G^p-G$$ belongs in $C^{\infty}(\mathbb{R}^3)$, and has its Taylor expansion: $$R(x)=R(0)-\frac{1}{6}|x|^2+O(|x|^4)$$
My attempt By writing \begin{equation} \begin{aligned} G^p(x)=\ & -\sum_{n\in \mathbb{Z}^3-\{0\}}\frac{e^{2\pi i n\cdot x}}{4\pi^2|n|^2}= -\frac{1}{4\pi^2} \sum_{n\in \mathbb{Z}^3-\{0\}}\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} \\ =\ &-\frac{2}{\pi^2}\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}{\pi^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{1}{2\pi^2} \sum_{n_3\geq 1} \frac{\cos 2\pi n_3x_3}{n_3^2} \\ &\ -\frac{1}{\pi^2}\sum_{n_1\geq 1}\cos 2\pi n_1x_1\sum_{n_3\geq 1} \frac{\cos 2\pi n_3x_3}{n_1^2+n_3^2}-\frac{1}{2\pi^2} \sum_{n_2\geq 1} \frac{\cos 2\pi n_2x_2}{n_2^2} \\ &\ -\frac{1}{\pi^2}\sum_{n_2\geq 1}\cos 2\pi n_2x_2\sum_{n_3\geq 1} \frac{\cos 2\pi n_3x_3}{n_2^2+n_3^2}-\frac{1}{2\pi^2} \sum_{n_1\geq 1} \frac{\cos 2\pi n_1x_1}{n_1^2} \\ :=\ &\Lambda(x)+\Theta_1(x)+\Theta_2(x)+\Theta_3(x) \end{aligned} \end{equation}
By invoking the summation identities: \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} $$\sum_{n_1\geq 1}\frac{\cos 2\pi n_1x_1}{n_1}e^{-2\pi n_1x_2}=\pi x_2-\log 2-\frac{1}{2}\log (\sinh^2 \pi x_2 +\sin^2\pi x_1)$$ I have calculated \begin{equation} \begin{aligned} \Theta_1(x)&:=-\frac{1}{\pi^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{1}{2\pi^2} \sum_{n_3\geq 1} \frac{\cos 2\pi n_3x_3}{n_3^2} \\ &=\frac{1}{12}+\frac{\log 2}{2\pi}-\frac{x_1+x_2}{2}+\frac{x_1^2}{4} +\frac{\log (\sinh^2 \pi x_2 +\sin^2\pi x_1)}{8\pi}+r_1(x) \end{aligned} \end{equation} where $r_1(x)\in C^{\infty}(\mathbb{R}^3)$ and $r_1(x)=O(|x|^4)$. therefore, we have \begin{equation} \begin{aligned} \Theta_1(x)&+\Theta_2(x)+\Theta_3(x)= \frac{1}{4}+\frac{3\log 2}{2\pi}-(x_1+x_2+x_3)+\frac{x_1^2+x_2^2+x_3^2}{4}+r_1(x)+r_2(x)+r_3(x) \\ &+\frac{\log (\sinh^2 \pi x_2 +\sin^2\pi x_1)+\log (\sinh^2 \pi x_3 +\sin^2\pi x_1)+\log (\sinh^2 \pi x_3 +\sin^2\pi x_2)}{8\pi} \end{aligned} \end{equation} where $r_1(x)+r_2(x)+r_3(x)=O(|x|^4) $ and belongs to $C^{\infty}(\mathbb{R}^3)$.
My question
- How to estimate $$\frac{\log (\sinh^2 \pi x_2 +\sin^2\pi x_1)+\log (\sinh^2 \pi x_3 +\sin^2\pi x_1)+\log (\sinh^2 \pi x_3 +\sin^2\pi x_2)}{8\pi}$$
- How to deal with $\Lambda(x)$ term.
Any help will be appreciated.