# Riemann Zeta Has No Zeros With Real Part One

## Theorem

Let $\zeta$ be the Riemann zeta function.

Then for all $t \in \R$, $\zeta(1 + it) \neq 0$

## Proof

Throughout, the complex variable $s$ is $s = \sigma + it$.

We have, for $\sigma > 1$,

 $\displaystyle -\frac{\zeta'(s)}{\zeta(s)}$ $=$ $\displaystyle \sum_{n\geq 1} \Lambda(n) n^{-s}$ By Logarithmic Derivative of Riemann Zeta Function $\displaystyle$ $=$ $\displaystyle \sum_{n\geq 1} \Lambda(n) n^{-\sigma} \exp\left( -it\log n\right)$ Where $s = \sigma + it$ $\displaystyle$ $=$ $\displaystyle \sum_{n\geq 1} \Lambda(n) n^{-\sigma} \left( \cos(t\log n) - i \sin(t\log n) \right)$ By Euler's Formula

Therefore,

$\displaystyle -\Re\left( \frac{\zeta'(s)}{\zeta(s)} \right) = \sum_{n\geq 1} \Lambda(n) n^{-\sigma} \cos(t\log n) \qquad (1)$

Now observe that

$3 + 4\cos\theta + \cos(2\theta) = 2(1+ \cos\theta)^2 \geq 0$

Because for all $n \geq 1$ we have $\Lambda(n)n^{-\sigma} \geq 0$, we have

 $\displaystyle 0$ $\leq$ $\displaystyle \sum_{n\geq 1} \Lambda(n) n^{-\sigma} \left\{ 3 + 4\cos(t\log n) + \cos(2t\log n) \right\}$ $\displaystyle$ $=$ $\displaystyle -\Re\left( 3\frac{\zeta'(\sigma)}{\zeta(\sigma)} + 4\frac{\zeta'(\sigma + it)}{\zeta(\sigma + it)} + \frac{\zeta'(\sigma + 2it)}{\zeta(\sigma + 2it)} \right)$ By $(1)$

Now let

$\eta(s) = \zeta(s)^3\cdot \zeta(s+it)^4\cdot \zeta(s+2it)$

Then the above computation has shown that

$\displaystyle\Re\left( \frac{\eta'(s)}{\eta(s)} \right) \leq 0$

By Poles of Riemann Zeta Function we know that $\zeta$ has a simple pole at $s=1$ with residue $1$.

Suppose that $1+it$ is a zero of $\zeta$ of order $d \geq 1$.

Therefore, at $s = 1$, $\eta$ has a zero of order $4d - 3 \geq 0$, that is,

$\displaystyle \eta(s) \sim (s-1)^{4d -3}$

as $s \to 1^+$, where $\sim$ indicates asymptotic equality, and superscript $+$ denotes a limit from the right along the real line. Therefore

$\displaystyle \frac{\eta'(s)}{\eta(s)} \sim \frac{4d -3}{s-1}$

as $s \to 1^+$. Since $\displaystyle \Re\left( \frac{4d -3}{s-1}\right) \to + \infty$ as $s \to 1^+$, it follows that

$\displaystyle \Re\left(\frac{\eta'(s)}{\eta(s)}\right) \to \infty$

as $s \to 1^+$. But we have already shown that

$\displaystyle\Re\left( \frac{\eta'(s)}{\eta(s)} \right) \leq 0$

$\blacksquare$