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How do I prove that $\zeta'(0)/\zeta(0)=\log(2\pi)$ ?

I can get $\zeta(0)=-\frac{1}{2}$, but I don't know how to calculate $\zeta'(0)=-\frac{1}{2}\log(2\pi)$ ? Can you help me ?

Here $\zeta(s)$ is Riemann zeta function: $$\zeta(s):=\sum_{n=1}^{\infty}\frac{1}{n^s}. $$

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    $\begingroup$ Well by the above $\zeta(0)$ is undefined. The value of $\zeta(0)$ comes from Riemann's functional equation $\endgroup$ – Nameless Dec 24 '12 at 14:57
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Maybe you're interested to check $(38)$ here.

The Wallis formula may also be written as $$\left(4^{\zeta{(0)}} \cdot e^{-\zeta'{(0)}}\right)^2=\frac{\pi}{2}$$ Chris.

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  • $\begingroup$ Thank you very much ! It's helpful. $\endgroup$ – Dao yi Peng Dec 25 '12 at 3:59
  • $\begingroup$ @DaoyiPeng: Welcome! Glad to hear that. $\endgroup$ – user 1357113 Dec 25 '12 at 8:57
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Begin with $$ \zeta(1-z) = 2 (2\pi)^{-z}\cos\frac{\pi z}{2} \Gamma(z)\;\zeta(z) $$ then take logarithmic derivative. Can you finish?

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  • $\begingroup$ Could you explain more? I take logarithmic derivative but when letting $z \to 1$, it seems to diverge. $\endgroup$ – Edward Wang May 9 '18 at 17:38

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