Does $\sum\limits_{k \geq 0} \frac{1}{(4k+1)(4k+2)} = \frac{\log(2)}{4} + \frac{\pi}{8}$ hold

Question

Context: I'm interested in trying to find as many explicit formulae as possible for:
$$ S_n(s) = \sum_{k \geq 0} \frac{1}{\prod_{i \in s} (nk+i)} $$
for all $ n \geq 2 $ and for all $ s $ subset of $ \{ 1, \dots, n \} $ such that $ \mbox{Card}(s) = 2 $.
So far, I can establish
$$ S_2(\{ 1, 2 \}) = \log(2) $$
$$ S_3(\{ 1, 2 \}) = \frac{\pi \sqrt{3}}{9},
S_3(\{ 2, 3 \}) = \frac{\log\left(3\right)}{2}-\frac{\pi\sqrt{3}}{18},
S_3(\{ 1, 3 \}) = \frac{\log\left(3\right)}{4}+\frac{\pi\sqrt{3}}{36} $$
$$ S_4(\{ 1, 3 \}) = \frac{\pi}{8}, S_4(\{ 2, 4 \}) = \frac{\log(2)}{4} $$
But I just guess that
$$ S_4(\{ 1, 2 \}) = \frac{\log(2)}{4} + \frac{\pi}{8} $$
Is there a generic method to establish all formulae?

Answer 1

$$S=\sum_{0}^{\infty} \frac{1}{(4k+1)(4k+2)}=\sum_{k=0}^{\infty}\left(\frac{1}{4k+1}-\frac{1}{4k+2}\right)$$ $$=\sum_{k=0}^{\infty}\int_{0}^{\infty} [e^{-(4k+1)x}-e^{-(4k+2)x}]=\int_{0}^{\infty} \left( \frac{e^{-x}}{1-e^{=4x}}-\frac{e^{-2x}}{1-e^{-4x}}\right) dx $$ $$=\int_{0}^{\infty} \frac{e^{-x}(1-e^{-x})}{1-e^{-4x}}=\int_{0}^{1}\frac{(1-t)}{1-t^4}dt$$ $$=\frac{1}{2}\int_{0}^{1} \left(\frac{1}{1+t}+\frac{1}{1+t^2}-\frac{2t}{2(1+t^2)}\right) dt=\frac{1}{4}\ln 2+\frac{\pi}{8}.$$