Looks like on the critical line one can compute $\zeta(1/2+it)$ from $\zeta^{'}(1/2+it)$ and simpler functions.


$$ \begin{aligned} f(t)= & 2\, \left( {\frac { \left( \left| \zeta^{'} \left( 1/2+it \right) \right| \right) ^{2}\Gamma \left( 1/4-1/2\,it \right) }{\zeta^{'} \left( 1/2+it \right) }}+\Gamma \left( 1/4+1/2\,it \right) \zeta^{'} \left( 1/2+it \right) {\pi }^{-it} \right) \\ & \left( \Gamma \left( 1 /4+1/2\,it \right) \right) ^{-1} \left( {\pi }^{-it} \right) ^{-1} \left( 2\,\ln \left( \pi \right) -\psi \left( 1/4+1/2\,it \right) - \psi \left( 1/4-1/2\,it \right) \right) ^{-1} \end{aligned} $$

For real $t$ we have $\zeta(1/2+it)=f(t)$, except possibly at zeros of $\zeta^{'}(1/2+it)$.

So we can compute $\zeta(1/2+it)$ given $\zeta^{'}(1/2+it)$ and computing simpler functions like the digamma $\psi$.

Q1 Is this really a differential equation for zeta on the critical line?

Q2 In case of positive answer to Q1, what is its classification?

Plot: enter image description here Sage code in machine readable form:

  def f(t):
    import mpmath
    from mpmath import gamma,zeta,log,psi
    return 2*( ( mpmath.fabs(zeta(1/2+J*t,derivative=1))**2/zeta(1/2+J*t,derivative=1) )*gamma(1/4-J*t/2)+gamma(1/4+J*t/2)*zeta(1/2+J*t,derivative=1)*Pi**(-J*t))/(gamma(1/4+J*t/2)*Pi**(-J*t)*(2*log(Pi)-psi(0,1/4+J*t/2)-psi(0,1/4-J*t/2 ) )) # should equal zeta(1/2+J*t) for real $t$

The zeta function can be written $\zeta(1/2+it)=Z(t)e^{-i\vartheta(t)}$ where $Z(t)$ are real analytic and $\vartheta(t)$ is a simple function related to $\Gamma(s)$. The relation between $\zeta(1/2+it)$ and $\zeta'(1/2+it)$ can be written in a simpler form $$-2\vartheta'(t)\zeta(1/2+it)=\zeta'(1/2+it)+e^{-2i\vartheta(t)}\zeta'(1/2-it)$$ in which you may change $\zeta'(1/2-it)=|\zeta'(1/2+it)|^2/\zeta'(1/2+it)$. This is proved in page 222 of the paper ``On the exact location of the non-trivial zeros of Riemann's zeta function'', Acta Arithm. 163 (2004) 215--245.

The proof do not depend on the nature of $Z(t)$. So this ``differential equation'' is satisfied by $f(1/2+it) =e^{-i\vartheta(t)} u(t)$ with $u$ any even real analytic function. Therefore it has no information about the zeros of zeta.

  • $\begingroup$ Thank you the given $f(t)$ depends on the right $Z(t)$, so it computes zeta zeros. $\endgroup$ – joro Jun 14 '15 at 4:30
  • 1
    $\begingroup$ More generally, zeta function universality basically ensures that the only differential equations obeyed by zeta are those coming from meromorphicity and the functional equation: en.wikipedia.org/wiki/Zeta_function_universality $\endgroup$ – Terry Tao Aug 20 '17 at 15:35

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