# 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 $\frac{\pi^2}{6}$

• You may like this link: math.stackexchange.com/questions/790314/… – Sangchul Lee Oct 30 '17 at 17:27
• Taylor expanding the log gives you a sum of zeta functions at even integer arguments. Further simplification probably takes some more sophisticated machinery along the lines of Weierstrass factorization. – Ian Oct 30 '17 at 17:27

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}
$$\log\left(\frac{\sinh \pi}{\pi}\right)$$ by the Weierstrass product for the (hyperbolic) sine function.