Series involving Glaisher–Kinkelin constant

[alyafey22]

Prove the following

\(\displaystyle \sum_{k\geq 2}\frac{\log(k)}{k^2}=\zeta(2)\left(\log A^{12}-\gamma-\log(2\pi) \right)\)

 

[polygamma]

Spoiler

$\displaystyle \sum_{k=1}^{\infty} \frac{\ln k}{k^{2}} = \zeta'(2) $I'm going to use the closed-form expression $$\log A = \frac{1}{12} - \zeta'(-1)$$

and the functional equation $$\zeta(s) = 2^{s} \pi^{s-1} \sin \left( \frac{\pi s}{s} \right) \Gamma(1-s) \zeta(1-s)$$Then

$$ \zeta'(s) = 2^{s} \log (2) \ \pi^{s-1} \sin \left(\frac{\pi s}{2} \right) \Gamma(1-s) \zeta(1-s) + 2^{s} \pi^{s-1} \log (\pi) \sin \left(\frac{\pi s}{2} \right) \Gamma(1-s) \zeta(1-s)$$

$$ + \ 2^{s} \pi^{s-1} \frac{\pi}{2} \cos \left( \frac{\pi s}{2} \right) \Gamma(1-s) \zeta(1-s) - 2^{s} \pi^{s-1} \sin \left(\frac{\pi s}{2} \right) \Gamma'(1-s) \zeta(1-s) $$

$$ - 2^{s} \pi^{s-1} \sin \left(\frac{\pi s}{2} \right) \Gamma(1-s) \zeta'(1-s)$$So

$$ \zeta'(-1) =\frac{\log 2}{2 \pi^{2}} (-1)(1) \zeta(2) + \frac{\log \pi }{2 \pi^{2}} (-1)(1) \zeta(2) + 0 + \frac{1}{2 \pi^{2}}(-1) (1 - \gamma) \zeta(2) + \frac{1}{2 \pi^{2}}(-1)(1) \zeta'(2)$$$$ \implies \zeta'(2) = \zeta(2) \big( - \log(2 \pi) + 1- \gamma \big) - 2 \pi^{2} \zeta'(-1) $$

$$ =\zeta(2) \big( - \log(2 \pi) +1 -\gamma \big) - 2 \pi^{2} \Big( \frac{1}{12} - \log A \Big) $$

$$ = \zeta(2) \big( - \log(2 \pi) +1 - \gamma \big) - \zeta(2) + 12 \zeta(2) \log A $$

$$ = \zeta(2) \Big( 12 \log A - \gamma - \log(2 \pi) \Big)$$