Show that the first derivative of the Riemann Zeta function $\zeta'(s) < 0$ if $s \in (1-\epsilon,1)$ and $\epsilon > 0$ is sufficiently small.

Question

Show that $\zeta'(s) < 0$ if $s \in (1-\epsilon,1)$ and $\epsilon > 0$ is sufficiently small.

Using the fact that

\begin{align} \zeta(s) = \frac{s}{s-1}-s\int_1^\infty\frac{\{t\}}{t^{s+1}}dt \end{align}

I get to the point where

\begin{align} \zeta'(s)&=\frac{(s-1)-s}{(s-1)^2}-\int_1^\infty\frac{\{t\}}{t^{s+1}}dt-s\int_1^\infty-\ln t\frac{\{t\}}{t^{s+1}}dt\\ &=\frac{-1}{(s-1)^2} -\int_1^\infty\frac{\{t\}}{t^{s+1}}dt+s\int_1^\infty\ln t\frac{\{t\}}{t^{s+1}}dt \end{align}

It is clear that $\frac{-1}{(s-1)^2}<0$ but how do I show that $-\int_1^\infty\frac{\{t\}}{t^{s+1}}dt+s\int_1^\infty\ln t\frac{\{t\}}{t^{s+1}}dt \leq 0$? Do I make use of the fact that $s \in(1-\epsilon,1)$ where $\epsilon$ is sufficiently small?

Any hints and suggestions are appreciated.

Answer 1

You do not need to prove that $$\int_{1}^{\infty}(s\log t-1)\frac{\{t\}}{t^{s+1}}\,dt $$ is negative, you only need to prove that its value is bounded when $s$ approaches $1$.

For instance, by assuming $1-\varepsilon\leq s<1$ we have: $$\int_{1}^{\infty}(s\log t-1)\frac{\{t\}}{t^{s+1}}\,dt \leq \int_{1}^{+\infty}\frac{\log t-1}{t^{s+1}}\,dt=\frac{1-s}{s^2}\leq\frac{\varepsilon}{s^2},$$ hence $\varepsilon\leq\frac{1}{2}$ grants: $$-\frac{1}{(1-s)^2}+\int_{1}^{\infty}(s\log t-1)\frac{\{t\}}{t^{s+1}}\,dt \leq -2 <0.$$