Given $$\frac{\mid \int_{x=1}^\infty(\frac{1-x+[x]}{x^2})x^{1-\bar{\rho}} dx\mid}{ \mid \int_{x=1}^\infty(\frac{1-x+[x]}{x^2})x^\rho dx\mid }=1$$ where [x] denotes the greatest integer function and $\rho$ and $\bar{\rho}$ are complex constants such that $0<\Re(\rho)<1$. Prove that $\Re(\rho)=1/2$.
My try: Define $$\phi(x)= \frac{1-x+[x]}{x^2},\quad f(x)= x^{1-\bar{\rho}}\quad\text{and}\quad g(x)= x^{{\rho}},$$ then $\phi(x)\geq 0$.
And f(x) and g(x) are continuous in $[1,\infty)$. So by mean value theorem for improper integrals there exist some c$\in(1,\infty)$ such that $$\frac{\mid \int_{x=1}^\infty(\frac{1-x+[x]}{x^2})x^{1-\bar{\rho}} dx\mid}{ \mid \int_{x=1}^\infty(\frac{1-x+[x]}{x^2})x^\rho dx\mid }=\mid\frac{c^{1-\bar{\rho}}}{c^{\rho}}\mid=1$$ $$\implies c^{1-2\Re(\rho)}=1$$ $$\implies(1-2\Re(\rho))\log c=0$$ $c>1$, gives $$\Re(\rho)=1/2.$$
It is not true. For $\Re(s) > 1$ and by analytic continuation for $\Re(s) >0$ $$F(s)=\int_1^\infty (1-x+\lfloor x\rfloor)x^{-s-1}dx=\frac{\zeta(s)}{s}-\frac1{s-1}+\frac1s$$
Then $|F(1/4)/F(1-1/4)| > 1$ and $|F(1/4+2i)/F(1-1/4-2i)| < 1$ thus any curve $1/4\to 1/4+2i$ passes through a point where $|F(s)/F(1-s)|=1$.