Series Thing

Simple solution thanks to @Jasper Braun

Considering the following identity:

$\begin{aligned} \prod_{n=1}^\infty \left(1-\frac {x^2}{n^2\pi^2}\right) & = \frac {\sin x}x & \small \color{#3D99F6} \text{Putting }x = \frac \pi 5 \\ \prod_{n=1}^\infty \left(1- \left(\frac 1{5n}\right)^2 \right) & = \frac {\sin \frac \pi 5}{\frac \pi 5} \\ \sum_{n=1}^\infty \ln \left(1- \left(\frac 1{5n}\right)^2 \right) & = \ln \left(\frac 5 \pi {\color{#3D99F6} \sin \frac \pi 5} \right) \\ & = \ln \left(\frac 5 \pi {\color{#3D99F6} \sqrt{\frac {5-\sqrt 5}8}} \right) & \small \color{#3D99F6} \text{See note.} \end{aligned}$

$\implies a+b = 5+8 = \boxed{13}$

---

Note:

Using the identity ( see proof ):

$\begin{aligned} \cos \frac \pi 5 + {\color{#3D99F6} \cos \frac {3\pi}5} & = \frac 12 & \small \color{#3D99F6} \text{As }\cos (\pi - x) = - \cos x \\ \cos \frac \pi 5 - {\color{#3D99F6} \cos \frac {2\pi}5} & = \frac 12 \\ \cos \frac \pi 5 - 2 \cos^2 \frac \pi 5 + 1 & = \frac 12 \\ 4 \cos^2 \frac \pi 5 - 2\cos \frac \pi 5 - 1 & = 0 \end{aligned}$

Solving the quadratic equation:

$\begin{aligned} \cos \frac \pi 5 & = \frac {2 + \sqrt{20}}{8} = \frac {1+\sqrt 5}4 \\ \implies \sin \frac \pi 5 & = \sqrt {1-\cos^2 \frac \pi 5} = \sqrt{\frac {5-\sqrt 5}8} \end{aligned}$

$\displaystyle\sum _{n=1}^{\infty } \left(\log \left(n^2\right)-\log \left(n^2-y^2\right)\right)=\displaystyle\sum _{n=1}^{\infty } \int_0^y \frac{2 x}{n^2-x^2} \, dx=\displaystyle\int_0^y \left(\sum _{n=1}^{\infty } \frac{2 x}{n^2-x^2}\right) \, dx=\displaystyle\int_0^y \frac{1-\pi x \cot (\pi x)}{x} \, dx=\log (y \csc (\pi y))+\log (\pi )$