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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?

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    What is the absolute value of a vector? – 2017-02-14
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    And I suppose WolframAlpha computed the first limit by expanding the quartic, and solving the series for each individual term. You might try the same approach here. – 2017-02-14
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    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 $$ \frac{8}{\pi^3}\sum_{n\geq 0}\frac{1}{(2n+1)^4} =\frac{8}{\pi^3}\left(\zeta(4)-\frac{1}{16}\zeta(4)\right)=\frac{8}{\pi^3}\cdot\frac{15}{16}\cdot\frac{\pi^4}{90}=\frac{\pi}{12}.$$ – 2017-02-14
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    @TMM In the second series we are taking the magnitude of the vector. – 2017-02-14
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    @JackD'Aurizio Thanks, very nice to see how you evaluated that! Have you any ideas for the second one? – 2017-02-14
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    @sonicboom: that is less trivial but still has a nice closed form (just see below). Are you confident with Dirichlet series? – 2017-02-14

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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.

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    Thanks again, I am not experienced with Dirichlet series, I will have to look them up. I haven't needed to work with series so much but now that I'm dealing with biperiodic lattices they are appearing everywhere. What initial step are you taking to split the series from $$ -\frac{2}{\pi^3}\sum_{x\in\mathbb{Z}}\frac{(2x-1)}{(2x+1)^4}$$ into $$-\frac{2}{\pi^3}\sum_{n\geq 0}\left(\frac{(2n-1)}{(2n+1)^4}+\frac{-3-2n}{(2n+1)^4} \right)$$ – 2017-02-14
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    Exactly, the preliminary steps involve a reduction to the first quadrant. – 2017-02-14
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    sorry but I'm not sure what mapping is being used to change from $x$ to $n$ when you split the summation into two separate summation. I imagined it was simply $x \to n$ for $x >0$ and $x \to -n$ for $x < 0$ but that would give $$-\frac{2}{\pi^3}\sum_{n \ge 0} \bigg( \frac{(2n - 1)}{(2n + 1)^4} + \frac{(-2n - 1)}{(1 - 2n)^4}\bigg).$$ So how did you get your expression? – 2017-02-17