Let the complex variable $s=\sigma+it$, then from the following identity valid for $\sigma=\Re s>1$ $$\zeta(s)=s\int_1^\infty \frac{[x]}{x^{s+1}}dx$$ where $\zeta(s)$ is the Riemann Zeta function, I've computed by differentition under the integral sign $$ \frac{\zeta(s)}{s}- \frac{d\zeta(s)}{d\sigma} =s(\sigma+1)\int_1^\infty\frac{[x]\log x}{x^{\sigma+2}}e^{-it\log x} dx.$$
I don't know if my deduction was right, and I believe that previous makes sense for $\sigma>2$, and on the strip $ \left\{ x+iy:a\leq x\leq b \right\} $ thus with $a>2$, since $\frac{\zeta(s)}{s}- \frac{d\zeta(s)}{d\sigma}$ is a bounded function of $s$ defined on the strip, holomorphic in the interior of the strip and continuous on the whole strip, defining $$M(x)=\sup_y \left| \frac{\zeta(x+iy)}{x+iy}- \frac{d\zeta(x+iy)}{dx} \right| ,$$ the Hadamard three-lines theorem proves that $\log M(x)$ is a convex function on $\left[ a,b \right]$ (here as I've said $a>2$).
Question. Were rights, do make sense all my claims looking a simple application of the Hadamard three-lines theorem for a function on a strip to the right of $\sigma>2$? Thanks in advance.