TITLE: some series about :On some strange summation formulas by R. William Gosper
QUESTION [5 upvotes]: I read the paper On some strange summation formulas by R. William Gosper which includes
$$\sum_{k = 1}^{\infty}\frac{(-1)^k\cos\big(\sqrt{k^{2}\pi^{2} -a^2})}{k^{2}}=\frac{\pi^2}{12}\big({-\cosh(a)}+\frac{3}{a}\sinh(a)\big)$$
I was looking at the following series maple could sum. Any idea how to prove it, thanks. Define
$$S_1(a)=\sum _{k=1}^{\infty } \frac{(-1)^k \cos \left(\sqrt{k^2\pi ^2-a}\right)}{\left(k^2+1\right)}$$
$$S_2(a)=\sum _{k=1}^{\infty } \frac{(-1)^k \cos \left(\sqrt{k^2\pi ^2-a}\right)}{\left(k^2+1\right)^2}$$
Let $\beta=\sqrt{-a-\pi^2}.\,$ Are the following true?:
$$S_1(a)=\frac{1}{2} \big(\pi\,  \text{csch}(\pi ) \cos ( \beta)-\cosh \left(\sqrt{a}\right)\big)$$
$$S_2(a)=\frac{1}{4} \Big(\pi\, \text{csch}(\pi ) \cos ( \beta) \big[1+\pi  \coth (\pi )\big] -2 \cosh \left(\sqrt{a}\right)-\frac{\pi ^3 \text{csch}(\pi ) \sin ( \beta)}{ \beta}\Big)$$

REPLY [4 votes]: Note that $$S_{1}\left(a\right)=\sum_{k\geq1}\frac{\left(-1\right)^{k}\cos\left(\sqrt{k^{2}\pi^{2}-a}\right)}{k^{2}+1}=\frac{1}{2}\sum_{k\in\mathbb{Z}}\frac{\left(-1\right)^{k}\cos\left(\sqrt{k^{2}\pi^{2}-a}\right)}{k^{2}+1}-\frac{\cosh\left(\sqrt{a}\right)}{2}
 $$ and now we can apply this well known summation formula $$\sum_{n\in\mathbb{Z}}\left(-1\right)^{n}f\left(n\right)=-\sum\left\{ \textrm{residues of }\pi\csc\left(\pi z\right)f(z)\textrm{ at }f\left(z\right)\textrm{'s poles}\right\} 
 $$ and so since we have poles at $z=\pm i
 $ we have that $$S_{1}\left(a\right)=\color{red}{\frac{\pi\cosh\left(\sqrt{a+\pi^{2}}\right)\textrm{csch}\left(\pi\right)}{2}-\frac{\cosh\left(\sqrt{a}\right)}{2}}.
 $$ In the same spirit we can evaluate $S_{2}\left(a\right)$ and get $$S_{2}\left(a\right)=\color{blue}{-\frac{\pi\textrm{csch}\left(\pi\right)}{4}\left(\frac{\pi^{2}\sinh\left(\gamma\right)}{\gamma}-\cosh\left(\gamma\right)-\pi\cosh\left(\gamma\right)\coth\left(\pi\right)\right)-\frac{\cosh\left(\sqrt{a}\right)}{2}}.$$
where $\gamma=\sqrt{a+\pi^{2}}$.