How to calculate $\sum_{x \in \mathbb{Z}^2} \frac{2\pi(1 - 2x_1)}{|2\pi x + \hat{\pi}|^4}$? Consider the series
$$\sum_{x=-\infty}^\infty \frac{2\pi(1 - 2x)}{(2\pi x + \pi)^4}.$$
This series converges to $\frac{\pi}{12}$ as can be seen in WolframAlpha . Now instead of a scalar $x$ I would like to consider instead a vector $x$ in $\mathbb{Z}^2$ and calculate the value of
$$\sum_{x \in \mathbb{Z}^2} \frac{2\pi(1 - 2x_1)}{|2\pi x + \hat{\pi}|^4},$$
where $\hat{\pi} = [\pi, \pi]^T$. So this could be viewed as a generalization of the previous summation to two dimensions.
How can the value that this series converges to be calculated?

Pulling out the $\pi$ factor,
$$ -\frac{2}{\pi^3}\sum_{x\in\mathbb{Z}}\frac{(2x-1)}{(2x+1)^4}=-\frac{2}{\pi^3}\sum_{n\geq 0}\left(\frac{(2n-1)}{(2n+1)^4}+\frac{-3-2n}{(2n+1)^4} \right)$$
equals
$$ \frac{8}{\pi^3}\sum_{n\geq 0}\frac{1}{(2n+1)^4} =\frac{8}{\pi^3}\left(\zeta(4)-\frac{1}{16}\zeta(4)\right)=\frac{8}{\pi^3}\cdot\frac{15}{16}\cdot\frac{\pi^4}{90}=\frac{\pi}{12}.$$
In a similar way, the second series equals
$$ \frac{16}{\pi^3}\sum_{x\geq 0}\sum_{y\geq 0}\frac{1}{\left((2x+1)^2+(2y+1)^2\right)^2}$$
that is deeply related with a Dirichlet series. It is:
$$ \frac{16}{\pi^3}\cdot\!\!\!\!\!\!\!\!\sum_{\substack{n\geq 1 \\ n\equiv 2\!\pmod{\!\!4}}}\!\!\!\!\!\! \frac{r_2(n)}{n^2}$$
where $r_2(n)$ stands for the number of ways for writing $n$ as a sum of two squares. Its explicit value is given by:
$$ \frac{7\sinh(\pi)\,\zeta(3)-\pi^3}{4 \pi ^2 (1+\cosh\pi)}. $$
That can be achieved through the Poisson summation formula.
