Prove that $\rho=1/2+it$

Question

Given, $$\dfrac{\int_1^\infty\dfrac{1-x+[x]}{x^2}x^{1-\rho}dx}{\int_1^\infty\dfrac{1-x+[x]}{x^2}x^\bar{\rho}dx} = \dfrac{\int_1^\infty\dfrac{1-x+[x]}{x^2}x^{\rho}dx}{\int_1^\infty\dfrac{1-x+[x]}{x^2}x^{1-\bar{\rho}}dx}$$ where $[x]$ is the greatest integer function and $\rho\in \mathbb{C}$ is a constant , $0<\Re(\rho)<1$.

Prove that $\Re(\rho)=1/2$

My try:

Let $$ I = \dfrac{\int_1^\infty\dfrac{1-x+[x]}{x^2}x^{1-\rho}dx}{\int_1^\infty\dfrac{1-x+[x]}{x^2}x^\bar{\rho}dx}$$

and $$J= \dfrac{\int_1^\infty\dfrac{1-x+[x]}{x^2}x^{\rho}dx}{\int_1^\infty\dfrac{1-x+[x]}{x^2}x^{1-\bar{\rho}}dx}$$

$$J= \frac{1}{\overline{I}}$$

$$I=J$$ gives $$ I= \frac{1}{\overline{I}}$$

$$\mid I \mid^2 =1$$

$$\mid I\mid=1$$

$$\left|\dfrac{\int_1^\infty\dfrac{1-x+[x]}{x^2}x^{1-\rho}dx}{\int_1^\infty\dfrac{1-x+[x]}{x^2}x^\bar{\rho}dx}\right|=1$$ Note that $\rho=1/2+it_0$ satisfies the above integral.

$$\mid \frac{\int_1^\infty\frac{1-x+[x]}{x^2}x^{1-\rho}dx}{\int_1^\infty\frac{1-x+[x]}{x^2}x^\bar{\rho}dx}\mid=1$$ $$\mid \frac{\int_1^\infty\frac{1-x+[x]}{x^2}x^{1-2{\Re({\rho})}}x^{\overline{\rho}}dx}{\int_1^\infty\frac{1-x+[x]}{x^2}x^\bar{\rho}dx}\mid=1$$ Can we apply mean value theorem for improper integrals to conclude that there exists some $c\in(1,\infty)$ such that $$c^{1-2\Re(\rho)}=1$$

$$c>1$$ gives $$\Re(\rho)=1/2$$ How to prove $\Re(\rho)=1/2$.?

Answer 1

One more time $$F(s)=\int_1^\infty (\lfloor x\rfloor-x+1)x^{-s-1}dx= \frac{\zeta(s)}{s}-\frac{1}{s-1}+\frac1{s}$$

And $$|F(s)|^2=|F(1-s)|^2\ne 0$$ has plenty of solutions away from $\Re(s)=1/2$.

  F(s)= zeta(s)/s+1/s-1/(s-1); t=var('t'); plot(abs(F(0.48+i*t))^2-abs(F(1-0.48-i*t))^2,(13,14.5))

https://cocalc.com/projects/d7fa91cb-9f60-4060-8df0-eeb0459e729d/files/2020-12-21-122308.sagews?session=default