Proving $\left\vert\int_n^{n+1}s\int_n^xt^{-s-1}dt\,dx\right\vert<\vert sn^{-s-1}\vert$, for $n \in \mathbb{N}$, $x \in [n;n+1]$, $s\in \mathbb{C}$

Question

Let's get out of silence? I try to prove the Riemann hypothesis because I have the humble intuition that it is true, whatever you may think of this step, with my little skills. I have time, so...

I try several ways, one is to use the Dirichlet eta function: for $s\in \mathbb{C}, \Re(s)>0, \displaystyle \eta(s) = \sum \limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s}$.

We have, for $s \in \mathbb{C}, 0<\Re(s)<1$, $\zeta(s) = 0 \Rightarrow \eta(s) = 0 \Rightarrow \Im(\eta(s))=0$. And:

$$\displaystyle 0=\Im(\eta(s))= \lim \limits_{N \rightarrow \infty} \sum \limits_{n=1}^{N} (-1)^{n} ~n^{-\Re(s)} \sin(\Im(s)\ln(n)) = \lim \limits_{N \rightarrow \infty} \sum \limits_{n=1}^{N} \cos(n\pi) ~n^{-a} \sin(b\ln(n)),$$

where $a = \Re(s), b = \Im(s)$.

I've read the following, and it may help me but I don't know how to prove it, for $n \in \mathbb{N}$, $ s\in \mathbb{C}$, $ 0<\Re(s)<1$ : $$\displaystyle \left\vert n^{-s} - \int_{n}^{n+1} x^{-s} \right\vert < \left\vert\int_n^{n+1} s \int_n^xt^{-s-1}dt\,dx\right\vert<| sn^{-s-1}|, $$ which implies that: $$ \displaystyle \left\vert \sum \limits_{n=1}^{N} n^{-s} - \int_{n}^{n+1} x^{-s} \right\vert < \sum \limits_{n=1}^{N} | sn^{-s-1}| $$

I thought I could use the idea of this inequality to study the sum $\displaystyle \sum \limits_{n=1}^{N} \cos(n\pi) ~n^{-a} \sin(b\ln(n))$, beginning by something like this: $$\displaystyle \left\vert \cos(n\pi) ~n^{-a} \sin(b\ln(n)) - \int_{n}^{n+1} \cos(\pi x) ~x^{-a} \sin(b\ln(x)) \right\vert < ~...$$

So, how to prove this simple double integral inequality?

For $n \in \mathbb{N}$, $x \in [n;n+1]$, $s\in \mathbb{C}$, $$\left\vert\int_n^{n+1}s\int_n^xt^{-s-1}dt\,dx\right\vert< | sn^{-s-1} | $$

I've tried to say : $n \le t \le x \le n+1$, and to use the triangle inequality.

Answer 1

This is a partial answer for a few cases in case it is helpful.

Case: $s = 0$. It's not true for $s=0$ because you have equality.

Case: $\text{Re}(s) > -1$ and $s\ne 0$. Using the triangle inequality for integrals, we have $$\left\vert\int_n^{n+1}s\int_n^xt^{-s-1}dt\,dx\right\vert \le \int_n^{n+1} |s| \int_n^x \left\vert t^{-s-1}\right\vert dt\,dx < \int_n^{n+1} |s| \int_n^{n+1} \left\vert t^{-s-1}\right\vert dt\,dx $$ When $\text{Re}(s) > -1$, $f(t)=| t^{-s-1}|$ is a decreasing function, so its maximum in $[n,n+1]$ is $f(n) = |n^{-s-1}|$, which gives you the result.