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](https://mathoverflow.net/questions/89518/are-the-semi-trivial-zeros-of-zetas-pm-zeta1-s-all-on-the-critical-li), 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](https://math.stackexchange.com/questions/468334/an-integer-counting-function-nx)):


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


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.



Are the zeros of the sum/difference of these integrals all on the critical line?
**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

```

