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][1], 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][2]):

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

which gives:

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

Isolate the integral part,

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

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.

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