What is the close form of: $\sum\limits_{k=1}^{\infty}\log\left(\frac{1}{k^2}+1\right)$ Is there a close form for of this series 
$$\sum_{k=1}^{\infty}\log\left(\frac{1}{k^2}+1\right) =\log \prod_{k=1}^{\infty}\left(\frac{1}{k^2}+1\right)$$
I know it converges in fact since $ \log(x+1)\le x$ for $x>0$ we have,
$$\sum_{k=1}^{\infty}\log\left(\frac{1}{k^2}+1\right)\le\sum_{k=1}^{\infty}\frac{1}{k^2} =  \frac{\pi^2}{6}$$
converges and its sum is less that $\frac{\pi^2}{6}$
 A: $$\log\left(\frac{\sinh \pi}{\pi}\right) $$
by the Weierstrass product for the (hyperbolic) sine function.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\bbox[#ffc,5px]{\ds{\forall\ n \in \mathbb{N}_{\geq 1}}} &
\\[1mm]
\prod_{k = 1}^{n}\pars{{1 \over k^{2}} + 1} & =
\prod_{k = 1}^{n}{1 \over k}\,{1 \over k}\pars{k - \ic}\pars{k + \ic} =
{\verts{\pars{1 - \ic}^{\large\overline{n}}}^{2} \over \pars{n!}^{2}} =
{1 \over \pars{n!}^{2}}\,
\verts{\Gamma\pars{n + 1 - \ic} \over \Gamma\pars{1 - \ic}}^{2}
\\[5mm] & =
{1 \over \pars{n!}^{2}}\,
{\verts{\Gamma\pars{n + 1 - \ic}}^{\,2} \over \bracks{\ic\,\Gamma\pars{\ic}}\Gamma\pars{1 - \ic}} =
-\,{\ic \over \pars{n!}^{2}}\,{\verts{\Gamma\pars{n + 1 - \ic}}^{2} \over \pi/\sin\pars{\pi\ic}}
\\[5mm] & = {\sinh\pars{\pi} \over \pi}\,
\verts{\pars{n - \ic}! \over n!}^{2}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}  \,\,\,
{\sinh\pars{\pi} \over \pi}\,
\verts{\root{2\pi}\pars{n - \ic}^{n + 1/2 - \ic}\expo{-\pars{n  - \ic}} \over \root{2\pi}n^{n + 1/2}\expo{-n}}^{2}
\\[5mm] & \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}  \,\,\,
{\sinh\pars{\pi} \over \pi}\,
\verts{n^{n + 1/2 - \ic}\,\pars{1 - \ic/n}^{\,n}\expo{\ic} \over
n^{n + 1/2}}^{2} \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}
{\sinh\pars{\pi} \over \pi}\,
\underbrace{\verts{\expo{-\ic\ln\pars{n}}\expo{-\ic}\expo{\ic}}^{2}}_{\ds{=\ 1}}
\\[5mm] & \implies
\bbox[15px,#ffe,border:1px dotted navy]{\ds{%
\prod_{k = 1}^{\infty}\pars{{1 \over k^{2}} + 1} = {\sinh\pars{\pi} \over \pi}}}
\\[5mm] & \implies
\bbox[15px,#ffe,border:1px dotted navy]{\ds{%
\sum_{k = 1}^{\infty}\ln\pars{{1 \over k^{2}} + 1} =
\ln\pars{\sinh\pars{\pi} \over \pi}}}
\end{align}
