# Is the imaginary part of $t\mapsto\zeta(1/2+it)$ close to the derivative of its real part?

Plotting $$t\mapsto\zeta(1/2+it)$$ on Wolfram alpha, it seems that the maxima of its real part are close to the zeros of its imaginary part, while the maxima of the latter seem close to the inflection points of the former. Can this be made precise? For example, is there a canonical notion of distance between those two functions that attains only small values?

• This can be rephrased as follows: the Hilbert transform of the real part is close to the derivative. Or, in yet another form: the Fourier transform of $t \mapsto \Re \zeta(1/2+i t)$ is concentrated around $\{-1, 0, 1\}$. This indeed is the case, and can be quantified in some sense; see, for example, this answer. – Mateusz Kwaśnicki Feb 11 at 13:17
• Well, I should have rather written "this indeed may be the case, and possibly could be quantified". I do not claim I can quantify this at that moment. – Mateusz Kwaśnicki Feb 11 at 13:23

## 1 Answer

(An extended comment.) The derivative of the real part does not really seem to be close to the imaginary part, as seen in the following picture generated by Mathematica:

Code: Plot[{Re[I Zeta'[1/2 + I t]], Im[Zeta[1/2 + I t]]}, {t, 0, 80}]

The corresponding zeroes of the two functions indeed seem to be reasonably close to each other. This is no surprise, however: $$\zeta(\tfrac{1}{2} + i t)$$ essentially circles around (mostly in the right half-plane). In each "circle" the distance between a maximum or minimum of the real part and the corresponding zero of an imaginary part is roughly as large as the distance froth the "center" to the real axis. And most "circles" seem to be centered near the real axis.

• Thank you for your answer. Denoting respectively the derivative of the real part and the imaginary part by $Zr'$ and $Zi$, could one expect to get a result along the lines of $\int_{x_{0}}^{x}(Zr'(t)-Zi(t))dt\ll_{\varepsilon}x^{\varepsilon}$ for some $x_{0}>1$ and all $\varepsilon>0$? – Sylvain JULIEN Feb 11 at 19:56
• I doubt this is any more regular than the integral of $\Im z(t)$, but honestly I know next to nothing about the Riemann zeta function. – Mateusz Kwaśnicki Feb 11 at 20:16
• Do you think my question in my previous comment should be asked separately? – Sylvain JULIEN Feb 11 at 20:29
• @SylvainJULIEN: Again, I do not know, you may ask someone more knowledgeable. – Mateusz Kwaśnicki Feb 11 at 20:42
• There certainly does seem to be some sort of correlation: consider ListPlot@Table[{Re[I Zeta'[1./2 + I t]], Im[Zeta[1./2 + I t]]}, {t, 0, 800}] and ListPlot@Table[{Re[I Zeta'[1./2 + I t]], Im[Zeta[1./2 + I t]]}, {t, 700, 800, 0.1}]. – Greg Martin Feb 12 at 0:21