📂Functions

Riemann Hypothesis and Trivial Roots of the Riemann Zeta Function

Formula

The following is called the Riemann functional equation. $$\zeta (s) = 2^{s} \pi^{s - 1} \sin \left( {{ \pi s } \over { 2 }} \right) \Gamma (1-s) \zeta (1-s)$$

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• $\Gamma$ is the Gamma function.
• $\zeta$ is the Riemann zeta function.

Description

In the Riemann functional equation, if $s \in 2 \mathbb{Z}$ then $\displaystyle \sin \left( {{ \pi s } \over { 2 }} \right) = 0$, so naturally one might think $\zeta (s) = 0$. However, when $s = 0$, the right-hand side has $\zeta (1 - 0)$, thus it doesn’t even get defined, let alone being a root, and when $s > 0$,

Simple pole of the gamma function: The domain of the Gamma function as a complex function, $\Gamma$, is as follows. $$\mathbb{C} \setminus \left( \mathbb{Z} \setminus \mathbb{N} \right) = \mathbb{C} \setminus \left\{ 0 , -1, -2, \cdots \right\}$$ Moreover, the set of singular points of $\Gamma$, $\left( \mathbb{Z} \setminus \mathbb{N} \right)$, is a set of simple poles. $$\Gamma (1-z) = {\frac{ \pi }{ \Gamma (z) \sin \pi z }}$$

According to this, the simple pole $\Gamma (1-s)$ of $\displaystyle {\frac{ 1 }{ \sin \pi s }}$ cancels out the sine term $\displaystyle \sin \left( {{ \pi s } \over { 2 }} \right)$, preventing it from being a root. Hence, all the remaining negative even numbers $s \in - 2\mathbb{N}$ become the roots of $\zeta$, which are called trivial roots of the Riemann zeta function. The famous Riemann Hypothesis is a hypothesis about the non-trivial roots, excluding these trivial ones.

Derivation¹

Definition and symmetry of the Riemann Xi function: The function defined as follows, $\xi$, is called the Riemann Xi function. $$\xi (s) := {{ 1 } \over { 2 }} s ( s-1) \pi^{-s/2} \zeta (s) \Gamma \left( {{ s } \over { 2 }} \right)$$ $$\xi ( 1 - s) = \xi (s)$$

By symmetry, $$\pi^{-s/2} \Gamma \left( {{ s } \over { 2 }} \right) \zeta (s) = \pi^{-1/2 + s/2} \Gamma \left( {{ 1 } \over { 2 }} - {{ s } \over { 2 }} \right) \zeta (1-s)$$ Multiplying both sides by $\pi^{s/2}$, $$\Gamma \left( {{ s } \over { 2 }} \right) \zeta (s) = \pi^{-1/2 + s} \Gamma \left( {{ 1 } \over { 2 }} - {{ s } \over { 2 }} \right) \zeta (1-s)$$

Euler’s reflection formula: $$\Gamma (p) \Gamma (1-p) = {{ \pi } \over { \sin (\pi p) }}$$

When $\displaystyle p = s/2$ in Euler’s reflection formula, $$\Gamma (s/2) \Gamma (1-s/2) = {{ \pi } \over { \sin (\pi s/2) }}$$ Thus, $$\begin{align*} \zeta (s) =& \pi^{-1/2 + s} \Gamma \left( {{ 1 } \over { 2 }} - {{ s } \over { 2 }} \right) \zeta (1-s) {{ 1 } \over { \pi }} \Gamma (1-s/2) \sin (\pi s / 2) & \\ =& \pi^{-3/2 + s} \Gamma \left( {{ 1 - s } \over { 2 }} \right) \Gamma \left( 1 - {{ s } \over { 2 }} \right) \sin \left( {{ \pi s } \over { 2 }} \right) \zeta (1-s) & \cdots (\ast) \end{align*}$$

Legendre duplication formula: $$\Gamma (2r) = {{2^{2r-1} } \over { \sqrt{ \pi } } } \Gamma \left( r \right) \Gamma \left( {{1} \over {2}} + r \right)$$

When $\displaystyle r= {{ 1-s } \over { 2 }}$ in the Legendre duplication formula, $$\Gamma (1-s) = 2^{-s} \pi^{-1/2} \Gamma \left( {{ 1-s } \over { 2 }} \right) \Gamma \left( 1- {{ s } \over { 2 }} \right)$$ Thus, substituting for $(\ast)$, $$\begin{align*} \zeta (s) =& \pi^{-3/2 + s} \Gamma \left( {{ 1 - s } \over { 2 }} \right) \Gamma \left( 1 - {{ s } \over { 2 }} \right) \sin \left( {{ \pi s } \over { 2 }} \right) \zeta (1-s) \\ =& \pi^{-3/2 + s} 2^{s} \pi^{+1/2} \Gamma ( 1-s) \sin \left( {{ \pi s } \over { 2 }} \right) \zeta (1-s) \\ =& 2^{s} \pi^{s-1} \sin \left( {{ \pi s } \over { 2 }} \right) \Gamma ( 1-s) \zeta (1-s) \end{align*}$$

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