MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For $\Re s = 1/2$ numerical evidence suggest: $$ \Re \zeta'(s)/\zeta(s) = 1/2 \log(\pi) - 1/2 \Re \psi(s/2) \qquad (1) $$

How this was found. Consider the symmetrized zeta function $\zeta^*(x)= \pi^{-x/2}\Gamma(x/2)\zeta(x)$.

Experimentally $\Re{ \zeta^*{'}(1/2 + it)} $ vanishes. Taking derivative and solving for $\psi(s/2)$ gives (1).

Is (1) true?

Any bounds for $\Re \zeta'(s)/\zeta(s)$ on the critical line?

Experimentally it is $- 1/2 \log{t}$ I suppose larger than that (minus a small constant) will contradict RH.

share|cite|improve this question
Fixed a critical typo \log => \Re – joro Jun 18 '13 at 11:48
Write $\zeta(z) = r(z) e^{i \theta(z)}$, where $r(z)$ and $\theta(z)$ are real valued. So for the logarithmic derivative, we obtain $\Re (\log \zeta \; )' (z) = r(z)'/r(z)$. Since $\zeta$ does vanish on $\Re s= 1/2$ and the $\Gamma$-function doesn't, the derivative of $\Gamma$ has to vanish. So you claim that the zeros of $\Gamma'$ on $\Re s =1/2$ coincide with the zeros of $\zeta$ on $\Re s=1/2$? I would guess that cannot be the case by counting the zeros asymptotically and assuming RH. – Marc Palm Jun 18 '13 at 11:57
@Marc thanks for the deleted answer. btw, fixed a critical typo with \Re, hope it is correct now. I don't claim anything about the zeros, claim that get numeric evidence for the equality. Do you have a counterexample? – joro Jun 18 '13 at 12:17
Ah okay, I found the nitpick in my argument. $\zeta'/\zeta$ can have a pole without its real part having one:( – Marc Palm Jun 18 '13 at 15:06
up vote 5 down vote accepted

Yes, your formula is true (and no RH is needed),

Your symmetrized zeta-function is symmetrized such that the functional equation

$$ \zeta^* (s)=\zeta^* (1-s) $$ holds. Thus we have that $\zeta^* (s)-\zeta^* (1-s)= 0 $ and on the critical line for $ s = 1/2+it $ this gives us that

$$ 2 \Im \left( \zeta^* ( \frac 1 2 + it ) \right) = \zeta^* ( \frac 1 2 + it ) - \zeta^* ( \frac 1 2 - it ) = 0, $$ i.e. $\zeta^*(1/2+it)$ is real valued (in fact it equals the Hardy $Z$-function).

Taking the derivative of both sides gives us $ { \zeta^* }' (s)= - { \zeta^* }' (1-s) $ and thus $ { \zeta^* }' (s) + { \zeta^* } '(1-s)=0 $. On the critical line this gives us

$$ 2 \Re \left( { \zeta^* } '( \frac 1 2+it) \right) = { \zeta^* } '( \frac 1 2 + it ) + { \zeta^* } '( \frac 1 2 -it )=0, $$ and thus ${\zeta^*}' (1/2+it)$ is imaginary. That answers your question (1).

Now taking the logarithmic derivative of the identity

$$ \zeta^*(s)=\pi^{-s/2} \Gamma(s/2) \zeta(s) $$ gives us that

$$ \frac{ { \zeta^ * } '(s) } { \zeta ^ * (s) } = -(\log \pi)/2 + \frac {\Gamma'(s/2)}{2\Gamma(s/2)} + \frac{ \zeta '(s) } { \zeta (s) } $$

By the fact that the left hand side is imaginary (imaginary divided by real) on the critical line, taking the real part of the last identity yields your numerically verified result.

Since the di-gamma function $ \psi(s)=\Gamma'(s)/\Gamma(s) $ is a nice function with asymptotic expansion and by the fact that the first order term $s^{-1}$ vanishes when taking the real part we see that $ \Re( \psi(1/4+it/2)) =\log (t/2) +O(t^{-2}) $, but if we want more terms we can easily get an error term of order $t^{-N}$ for any $N>0$. Thus your formula gives for example:

$$ \Re \left( \frac{ \zeta' (\frac 1 2 +it ) } { \zeta (\frac 1 2 + it ) } \right) = - \frac 1 2 \log \left(\frac t {2 \pi} \right) +O(t^{-2}). $$

share|cite|improve this answer
Thank you. The Micah paper gives this unconditionally too because of cancellation on the critical line. – joro Jun 18 '13 at 14:26
You don't make exception for simple zeta zeros. Does it hold for simple zeta zeros too? – joro Jun 19 '13 at 10:03
You are right, I guess that you need to assume that $\zeta(1/2+it) \neq 0$ in order for your formula and my argument to work (since otherwise the expression is not even defined - division by zero), but for all $t$ such that $\zeta(1/2+it) \neq 0$ your formula holds. The proof method does not really use the zeros though, just the functional equation. – Johan Andersson Jun 19 '13 at 11:37

Your formula follows from Hadamard's product formula for $\zeta(s)$ and the corresponding partial fraction decomposition of $\zeta'(s)/\zeta(s)$. See, for instance, section 2 of Soundararajan's paper:

share|cite|improve this answer
Thank you. I think the result on p.4 is unconditionally equivalent to the question. – joro Jun 18 '13 at 14:46
Edited accordingly... – Micah Milinovich Jun 18 '13 at 14:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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