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The answers given to the question whether all zeros in the critical strip of $\zeta(s)\pm\zeta(1-s)$ lie on the critical line, suggest that this can indeed be proven, however only for those zeros where $s \ne \rho$ (to be more precise; those zeros occur when $\chi(s)=2^s \pi^{s-1} \sin(\pi s/2) \phantom. \Gamma(1-s) = \pm 1$).

Now assume $s \in \mathbb{C}$, $\Re(s) \ge 0$ and take the known expression:

$$\zeta(s) = \dfrac{s}{s-1} - \frac12+s \int_1^\infty \frac{1/2-\{x\}}{x^{s+1}}\,\mathrm{d}x$$

and substitute the fractional part of $\{x\}$ by a closed form (derived here):

$$\displaystyle \{x\} = x - \lfloor x \rfloor = \frac12 + \frac{i}{2 \pi} \ln \left(-\mathrm{e}^{-2 \pi i x} \right)$$

which gives:

$$\displaystyle \zeta(s) = \dfrac{s}{s-1} - \frac12-\frac{s i}{2 \pi} \int_1^\infty \frac{\ln \left(-\mathrm{e}^{-2\pi i x}\right)\,}{x^{s+1}}\mathrm{d}x $$

From the discussion in the comments section below, it has become clear that the various CAS-packages give different outcomes when evaluating this integral. When the integral is finite from 1 to $N$, both Maple and Mathematica give the correct outcome, but Sage seems to struggle. The difference probably can be explained from CAS picking the correct (principal?) branch of the multi-valued $\ln(-e)$ element in the integral. However, despite 2 CAS results continuously improving in accuracy with increasing $N$, in all CAS the integral is yielding a very wrong outcome at $\infty$. This is not only the case for the $\ln(-e)$ integral, but also for the integral with $\{x\}$ (which is a proven formula for $\zeta(s)$). I now wonder if this has something to do with how the various CAS evaluate the fractional part at infinity. In any case, CAS are not going to give us the answer and some real pen and paper math is required to assess what exactly happens at $\infty$. Any thoughts are welcome.

Since I used a finite integral in Maple to test my conjecture below (EDIT: some zeros in the strip have been found and the conjecture has been proven wrong), I decided to be more precise in the OP and replaced $\infty$ by an as large as you like $N$ in the integral below.

Isolate the integral part,

$$I(s) =\frac{s i}{2 \pi} \int_1^N \frac{\ln \left(-\mathrm{e}^{-2\pi ix}\right)\,}{x^{s+1}}\mathrm{d}x $$

and I like to conjecture, that in the critical strip, all zeros of:

$$I(s) \pm I(1-s)$$

are on the critical line $\Re(s)=\frac12$, however now with the certainty that when $s= \rho$ then $I(s) \ne 0$.

Via the reflective relation $\zeta(s) = \chi(s)\phantom . \zeta(1-s)$, this can be simplified into the following relation:

$$I(s) \pm I(1-s) =0 \text{ when } \displaystyle \chi(s)= \frac{\frac{s}{1-s} + \frac12 \pm -I(1-s)}{\frac{1-s}{s} + \frac12 + \hspace{3 mm} I(1-s)}$$

Appreciate any thoughts on possible approaches to proof this conjecture, e.g. based on the symmetry between the two integrals $I(s)$ and $I(1-s)$ or the symmetry around $\chi(s)$.

Thanks.

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  • $\begingroup$ Agno, I can't compute $I(s)$. Would you please give the values of $I(3), I(1/2 + 13.35 i), I(1/2 + 20 i)$, the method to compute them and to what precision they are correct? Thank you. $\endgroup$
    – joro
    Commented Aug 19, 2013 at 14:45
  • $\begingroup$ Joro. I found the problem. The $\frac12$ should not be subtracted form the $x$. I tried it in Wolfram Alpha and this works. Here is the code for $\zeta(3)$ till $x=9$ : 3/(3-1)-1/2-(3*i)/(2*π)*(integral from 1..9 of ln(-e^(-2*π*i*(x)))/(x^(3+1))) $\endgroup$
    – Agno
    Commented Aug 19, 2013 at 18:37
  • $\begingroup$ Agno, computing I(s) via zeta got some zeros off the critical line and edited the answer. $\endgroup$
    – joro
    Commented Aug 21, 2013 at 12:29
  • $\begingroup$ Many thanks, Joro. Always a pleasure to see the Master of Root finding in action. Looking at some of my scars from earlier discussions with you, I decided this time to explicitly limit my conjecture to the Critical Strip only ;-) Delighted to see that you did not discover (yet) any other zeros in the strip than those lying on the critical line! $\endgroup$
    – Agno
    Commented Aug 21, 2013 at 21:43
  • $\begingroup$ Agno sorry, my mistake, didn't notice the critical strip. Edited the answer with zeros in the critical strip. $\endgroup$
    – joro
    Commented Aug 22, 2013 at 6:04

1 Answer 1

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EDIT

Since $I(s)$ is hard to compute using the integral, searched for zeros in the critical strip expressing $I(s)$ with zeta.

Zeros of $I(s)-I(1-s)$:

 (0.542373937181871507937660440099538246914 -/+ 169.9110890356259176158839120274631129439j)

Zeros of $I(s)+I(1-s)$:

 (0.7898290418725801070374041640775529742698 + 111.370199942379561610316560047101576977j)
 (0.2101709581274198929625958359224470257302 + 111.370199942379561610316560047101576977j)
 (0.2752069724061228509354681961440878521307 + 150.4900933771384115043090398407167418796j)
 (0.7247930275938771490645318038559121478693 + 150.4900933771384115043090398407167418796j)
 (0.7409784227009039316446697155153005654462 + 169.6169782129636093132302613533815909668j)
 (0.8060767456413526455641207946358500080546 + 185.1541122304347183385135208635099380142j)

Their location agree with the formula involving $\chi$.

sage/maxima found closed form for the indefinite integral (edit: in the initial revision of the question).

$$\int \frac{\log\left(e^{\left(-i \, {\left(2 \, x - 1\right)} \pi\right)}\right)}{x^{s + 1}} dx = -\frac{x^{-s} \log\left(-e^{\left(-2 i \, \pi x\right)}\right)}{s} + \frac{2 i \, \pi x^{-s + 1}}{{\left(s - 1\right)} s} + C$$

I doubt zeta can be expressed via limits of such simple expression.

Session:

sage: var('x,s');assume(s != 0);assume(s-1 != 0);ex=log(exp(-2*pi*I*(x-1/2)))/x^(1+s)
(x, s)
sage: inte=integrate(ex,x);inte
-x^(-s)*log(-e^(-2*I*pi*x))/s + 2*I*pi*x^(-s + 1)/((s - 1)*s)
sage: (diff(inte,x)-ex).full_simplify()
0
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  • $\begingroup$ There is indeed seems to be a strange ambivalence in the integral occuring at infinity, however try using an ever increasing finite integral and you'll see that it perfectly approaches $\zeta(s)$. Don't know why the behavior "tilts" at infinity. $\endgroup$
    – Agno
    Commented Aug 19, 2013 at 8:52
  • $\begingroup$ @Agno are you sure you don't have typos and the formula for zeta is correct? For s=2 the limit of the integral at x=0 is infinity... Numerically and according to maple the integral doesn't converge, might be wrong. $\endgroup$
    – joro
    Commented Aug 19, 2013 at 9:02
  • $\begingroup$ Pretty sure there are no typos. Try using a finite integral up to say x=100 (I used Maple). Please note that it only works for $\re(s) \ge 0$ (should have stated that!), but it certainly should work in the critical strip. $\endgroup$
    – Agno
    Commented Aug 19, 2013 at 9:10
  • $\begingroup$ @Agno do you agree that the indefinite integral is correct? $\endgroup$
    – joro
    Commented Aug 19, 2013 at 9:11
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    $\begingroup$ @Agno since log(exp(x))==x, why don't you simplify the integral? Do you work with the principal branch of log() as all CAS do? (there is closed form too after the simplification). $\endgroup$
    – joro
    Commented Aug 19, 2013 at 10:34

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