Hyperbolic Cotangents And Sums Of Squares

Let $C_N$ be the positively oriented square contour with vertices at $\xi(N+\tfrac12) + i\eta(N + \tfrac12)$ for $\xi,\eta = 1,-1$ .

For any $b \in \mathbb{N}$ the function $f_b(z) \,=\, \frac{\pi \cot \pi z}{(z^2 + b^2)^2}$ is meromorphic with simple poles at every integer, and with double poles at $\pm ib$ . Thus $\begin{array}{rcl} \displaystyle \frac{1}{2\pi i}\int_{C_N} f_b(z)\,dz & = & \displaystyle \mathrm{Res}_{z=ib}f_b(z) + \mathrm{Res}_{z=-ib}f_b(z) + \sum_{a=-N}^N \mathrm{Res}_{z=a}f_b(z) \\ & = & \displaystyle \frac{\pi^2}{2b^2} - \frac{\pi \coth\pi b}{2b^3} - \frac{\pi^2 \coth^2\pi b}{2b^2} + \frac{1}{b^4} + 2\sum_{a=1}^N \frac{1}{(a^2+b^2)^2} \end{array}$ Since $|f_b(z)| \,=\, O(N^{-4})$ uniformly on $C_N$ , it follows, letting $N \to \infty$ , that $\sum_{a=1}^\infty \frac{1}{(a^2+b^2)^2} \; = \; \frac{\pi \coth\pi b}{4b^3} + \frac{\pi^2 \coth^2\pi b}{4b^2} - \frac{1}{2b^4} - \frac{\pi^2}{4b^2}$ for any $b \in \mathbf{N}$ , and hence it follows that $\sum_{a,b=1}^\infty \frac{1}{(a^2+b^2)^2} \; = \; \tfrac14\pi\sum_{b=1}^\infty \frac{\coth\pi b}{b^3} + \tfrac14\pi^2\sum_{b=1}^\infty \frac{\coth^2\pi b}{b^2} - \tfrac12\zeta(4) - \tfrac14\pi^2\zeta(2)$

Next, the function $g(z) = \frac{\pi \cot \pi z \coth\pi z}{z^3}$ is meromorphic with simple poles at $n$ and $in$ for any nonzero integer $n$ , and a quintuple pole at $0$ . Thus $\begin{array}{rcl} \displaystyle \frac{1}{2\pi i} \int_{C_N} g(z)\,dz & = & \displaystyle \mathrm{Res}_{z=0}g(z) + \sum_{b=1}^N \big(\mathrm{Res}_{z=b} + \mathrm{Res}_{z=-b} + \mathrm{Res}_{z=ib} + \mathrm{Res}_{z=-ib}\big) g(z) \\ & = & \displaystyle 4\sum_{b=1}^N \frac{\coth\pi b}{b^3} - \tfrac{7}{45}\pi^3 \end{array}$ Since $|g(z)| = O(N^{-3})$ uniformly on $C_N$ , letting $N \to \infty$ gives us that $\sum_{b=1}^\infty \frac{\coth \pi b}{b^3} \; = \; \tfrac{7}{180}\pi^3$ and hence $\sum_{a,b=1}^\infty \frac{1}{(a^2+b^2)^2} \; = \; \tfrac14\pi^2 \sum_{b=1}^\infty \frac{\coth^2\pi b}{b^2} - \tfrac{3}{80}\pi^4$

Finally, using results from Number Theory we know that $\sum_{a,b=1}^\infty \frac{1}{(a^2+b^2)^2} \; = \; \zeta(2)G - \zeta(4) \; = \; \tfrac16\pi^2 G - \tfrac{1}{90}\pi^4$ where $G$ is Catalan's constant. Putting this all together gives $\sum_{b=1}^\infty \frac{\coth^2\pi b}{b^2} \; =\; \tfrac23G + \tfrac{19}{180}\pi^2$ making the answer $2 + 3 + 19 + 180 + 2 = 206$ .

Consider

$\displaystyle \int_C f(z)dz = \int_C \frac{\pi \cot(\pi z)\coth^2(\pi z)}{z^2} dz$

Poles

At $z=0$ order $5$

At $z=n , n \in Z-\{0\}$ order $1$

At $z=ki , k \in Z-\{0\}$ order $2$

Evaluation

At $z=0$ , Laurent series of $f(z)=\frac{1}{\pi^2 z^5}+\frac{1}{3z^3}-\frac{8\pi^2}{45z}+\cdots$

Residue is $-\frac{8\pi^2}{45}$

At $z=n$

$\displaystyle \lim_{z \to n} \frac{(z-n)\pi\cot(\pi z)\coth^2(\pi z)}{z^2}=\frac{\coth^2(\pi z)}{n^2}$

At $z=ki$

$\displaystyle \lim_{z \to ki}\frac{d}{dz}\left( \frac{(z-ki)^2\pi\cot(\pi z)\coth^2(\pi z)}{z^2} \right)=-\frac{(k\pi+\sinh(2k\pi))\cosech^2(k\pi)}{\pi k^3}$

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$\displaystyle \int_C \frac{\pi \cot(\pi z)\coth^2(\pi z)}{z^2} dz=0$ $\displaystyle \int_C \frac{\pi \cot(\pi z)\coth^2(\pi z)}{z^2} dz=-\frac{8\pi^2}{45} + 2\sum_{n\geq 1}\frac{\coth^2(\pi n)}{n^2}-2\sum_{k\geq 1}\frac{(k\pi+\sinh(2k\pi))\cosech^2(k\pi)}{\pi k^3}$

But $\displaystyle \sum_{k\geq 1} \frac{(k\pi+\sinh(2k\pi))\cosech^2(k\pi)}{\pi k^3}=\frac23G+\frac{1}{10}\zeta(2)=\frac23G+\frac{\pi^2}{60}$

Thus, $\displaystyle \sum_{n\geq 1}\frac{\coth^2(\pi n)}{n^2} = \frac{4\pi^2}{45}+\frac23G+\frac{\pi^2}{60}$

$\displaystyle \sum_{n\geq 1}\frac{\coth^2(\pi n)}{n^2}=\frac23G+\frac{19\pi^2}{180}$