Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Edit Major rewrite since Johan Andersson observed the original question is trivial because of vanishing of coefficients.

From this question

$$ \zeta(1-x) = \frac{2(\zeta'(x)\Gamma(x/2)+\Gamma((1-x)/2) \zeta'(1-x)\pi^{x-1/2}) )}{\Gamma((1-x)/2) \pi^{-1/2+x}(2\log\pi -\psi((1-x)/2)-\psi(x/2))} \qquad (1)$$

On the critical line $\zeta'(1/2 - i t) = |\zeta'(1/2+ i t)|^2 / \zeta'(1/2+ i t)$ so $\zeta'(1/2 - i t)$ can be eliminated, leaving on $\zeta(1/2+i t),\zeta'(1/2+i t)$ and modulus of the derivative,leading to

$$ \zeta \left( 1/2+it \right) = 2\,{\frac {|\zeta'\left(1/2+it \right)|^2/\zeta'\left(1/2+it \right) \Gamma \left( 1/4-1/2\,it \right) +\Gamma \left( 1/4+1/2\,i t \right) \zeta' \left(1/2+it \right) {\pi }^{-it}}{\Gamma \left( 1/4+1/2\,it \right) {\pi }^{-it} \left( 2\,\ln \left( \pi \right) - \psi \left( 1/4+1/2\,it \right) -\psi \left( 1/4-1/2\,it \right) \right) }} \qquad (D) $$

So on the critical line $\zeta(1/2+it)=f(\zeta'(1/2+it))$ though the complex modulus complicates things.

(Fishing expedition) Can (D) be solved for zeta, unlikely giving alternate expression for zeta on the critical line?

sage/mpmath code in case of typos

def difide1(t):
    differential equation via |zeta'(s)| on the critical line
    should return zero for real $t$
    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 ) )) - zeta(1/2+J*t)
share|improve this question
I did not check this too carefully, but I suspect that (1) and (14) are equivalent and thus this does not give a non trivial differential equation (since the coefficients in front of $\zeta'$ and $\zeta$ in that case will vanish). Please look at those coefficients. I would also like to point out that it is well known that the Riemann zeta-function is hypertranscendental, en.wikipedia.org/wiki/Hypertranscendental_function and results such as Voronin universality gives restrictions on how such differential equations will look like. –  Johan Andersson Jun 27 '13 at 11:30
@JohanAndersson you appear to be right, the coefficients indeed vanish numerically, so this is true but useless. –  joro Jun 27 '13 at 11:44
@JohanAndersson is modulus allowed in differential equations? On the critical line $\zeta'(1/2 - i t) = |\zeta'(1/2+ i t)|^2 / \zeta'(1/2+ i t)$ and put in (1) or (14) –  joro Jun 27 '13 at 14:42

Your Answer


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

Browse other questions tagged or ask your own question.