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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}. $

  • 5
    Well by the above $\zeta(0)$ is undefined. The value of $\zeta(0)$ comes from Riemann's functional equation2012-12-24

2 Answers 2

1

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.

  • 0
    @DaoyiPeng: Welcome! Glad to hear that.2012-12-25
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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?

  • 0
    Could you explain more? I take logarithmic derivative but when letting $z \to 1$, it seems to diverge.2018-05-09