# 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

$$\log\left(\frac{\sinh \pi}{\pi}\right)$$ by the Weierstrass product for the (hyperbolic) sine function.