Integral sum thing 500

$\begin{aligned} S & = \sum_{n=2}^\infty \frac {\zeta(n) \color{#3D99F6}\cos \left(\frac {\pi n} 2 \right)}n & \small \color{#3D99F6} \text{Note that }\cos \left(\frac {\pi n} 2 \right) = \begin{cases} 1 & \text{if }n \bmod 4 = 0 \\ 0 & \text{if }n \bmod 4 = 1 \\ -1 & \text{if }n \bmod 4 = 2 \\ 0 & \text{if }n \bmod 4 = 3 \end{cases} \\ & = \sum_{n=1}^\infty \frac {(-1)^n \zeta(2n)}{2n} \\ & = \frac 12 \sum_{n=1}^\infty \frac {(-1)^n \sum_{k=1}^\infty \frac 1{k^{2n}}}n \\ & = \frac 12 \sum_{k=1}^\infty \sum_{n=1}^\infty \frac {(-1)^n}n \left(\frac 1{k^2}\right)^n \\ & = - \frac 12 \sum_{k=1}^\infty \log \left(1 + \frac 1{k^2}\right) \\ & = - \frac 12 \log \left({\color{#3D99F6} \prod_{k=1}^\infty \left(1 + \frac 1{k^2}\right)}\right) & \small \color{#3D99F6} \text{Using }\frac {\sin (\pi s)}{\pi s} = \prod_{n=1}^\infty \left(1-\frac {s^2}{n^2}\right) \\ & = - \frac 12 \log \left({\color{#3D99F6}\frac {\sin {\pi i}}{\pi i}}\right) & \small \color{#3D99F6} \text{Using }\sin x = \frac i2 \left(e^{-xi} - e^{xi} \right) \\ & = - \frac 12 \log \left({\color{#3D99F6}\frac {e^\pi - e^{-\pi}}{2\pi}}\right) \\ & = - \frac 12 \log \left(\frac {\sinh \pi }\pi \right) \\ & = \frac {\log \left(\pi \text{ csch}(\pi) \right)}2 \end{aligned}$

Therefore, $a = \boxed{2}$ .