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
    $\begingroup$ 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

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.

  • $\begingroup$ 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

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?

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.