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The answer to an integral involved $\frac{\zeta'(2)}{\zeta(2)}$, but I am stuck trying to find this number - either to a couple decimal places or exact value.

In general the logarithmic deriative of the zeta function is the dirichlet series of the van Mangolt function:

$$\frac{\zeta'(s)}{\zeta(s)} = \sum_{n \geq 0} \Lambda(n) n^{-s} $$

Let's cheat: Wolfram Alpha evaluates this formula as:

$$ \frac{\zeta'(2)}{\zeta(2)} = - 12 \log A + \gamma + \log 2 + \log \pi \tag{$\ast$}$$

This formula features some interesting constants:

Wikipedia even says that $A$ and $\pi$ are defined in similar ways... which is an interesting philosophical point.


Do we have a chance of deriving $(\ast)$?

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By differentiating both sides of the functional equation$$ \zeta(s) = \frac{1}{\pi}(2 \pi)^{s} \sin \left( \frac{\pi s}{s} \right) \Gamma(1-s) \zeta(1-s),$$ we can evaluate $\zeta'(2)$ in terms of $\zeta'(-1)$ and then use the fact that a common way to define the Glaisher-Kinkelin constant is $\log A = \frac{1}{12} - \zeta'(-1)$.

Differentiating both sides of the functional equation, we get

$$\begin{align} \zeta'(s) &= \frac{1}{\pi} \log(2 \pi)(2 \pi)^{s} \sin \left( \frac{\pi s}{2} \right) \Gamma(1-s) \zeta(1-s) + \frac{1}{2} (2 \pi)^{s} \cos \left(\frac{\pi s}{2} \right) \Gamma(1-s) \zeta(1-s)\\ &- \frac{1}{\pi}(2 \pi)^{s} \sin \left(\frac{\pi s}{2} \right)\Gamma^{'}(1-s) \zeta(1-s) - \frac{1}{\pi}(2 \pi)^{s} \sin \left(\frac{\pi s}{2} \right)\Gamma(1-s) \zeta'(1-s). \end{align}$$

Then letting $s =-1$, we get $$\zeta'(-1) = -\frac{1}{2\pi^{2}}\log(2 \pi)\zeta(2) + 0 + \frac{1}{2 \pi^{2}}(1- \gamma)\ \zeta(2) + \frac{1}{2 \pi^{2}}\zeta'(2)$$ since $\Gamma'(2) = \Gamma(2) \psi(2) = \psi(2) = \psi(1) + 1 = -\gamma +1. \tag{1}$

Solving for $\zeta'(2)$,

$$ \begin{align} \zeta'(2) &= 2 \pi^{2} \zeta'(-1) + \zeta(2)\left(\log(2 \pi)+ \gamma -1\right) \\ &= 2 \pi^{2} \left(\frac{1}{12} - \log (A) \right) + \zeta(2)\left(\log(2 \pi)+ \gamma -1\right) \\ &= \zeta(2) - 12 \zeta(2) \log(A)+ \zeta(2) \left(\log(2 \pi)+ \gamma -1\right) \tag{2} \\ &= \zeta(2) \left(-12 \log(A) + \gamma + \log(2 \pi) \right). \end{align}$$

$(1)$ https://en.wikipedia.org/wiki/Digamma_function

$(2)$ Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$

EDIT:

If you want to show that indeed $$\zeta'(-1)= \frac{1}{12}- \lim_{m \to \infty} \left( \sum_{k=1}^{m} k \log k - \left(\frac{m^{2}}{2}+\frac{m}{2} + \frac{1}{12} \right) \log m + \frac{m^{2}}{4} \right) = \frac{1}{12}- \log(A),$$ you could differentiate the representation $$\zeta(s) = \lim_{m \to \infty} \left( \sum_{k=1}^{m} k^{-s} - \frac{m^{1-s}}{1-s} - \frac{m^{-s}}{2} + \frac{sm^{-s-1}}{12} \right) \ , \ \text{Re}(s) >-3. $$

This representation can be derived by applying the Euler-Maclaurin formula to $\sum_{k=n}^{\infty} {k^{-s}}$.

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  • $\begingroup$ Why do we consider $\zeta'(-1)$ when we have $ \zeta'(2)$. I mean $\zeta'(2)$ is better to consider, because of the series and integral representations. $\endgroup$ – Zaid Alyafeai Aug 6 '16 at 22:49
  • $\begingroup$ @ZaidAlyafeai The Glaisher–Kinkelin constant is frequently defined in terms of $\zeta'(-1)$. And the functional equation of the Riemann zeta function provides a way to relate $\zeta'(2)$ to $\zeta'(-1)$, thus showing how $\zeta'(2)$ is related to $A$. $\endgroup$ – Random Variable Aug 7 '16 at 1:39

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