Unsymmetric Functional Equation for Riemann Zeta Function

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\zeta$ be the Riemann zeta function.

Let $\Gamma$ be the gamma function.

Then for all $s \in \C$,

$\displaystyle \map \zeta {1 - s} = 2^{1 - s} \pi^{-s} \, \map \cos {\frac {\pi s} 2} \map \Gamma s \, \map \zeta s$


Proof

We have for $s \notin \Z$ Euler's Reflection Formula:

$\map \Gamma s \, \map \Gamma {1 - s} = \dfrac \pi {\map \sin {\pi s} }$

Replacing $s \mapsto \dfrac {1 + s} 2$ we deduce:

\(\displaystyle \map \Gamma {\frac {1 + s} 2} \, \map \Gamma {\frac {1 - s} 2}\) \(=\) \(\displaystyle \frac \pi {\map \sin {\pi \paren {1 + s} / 2} }\) substituting $s \mapsto \dfrac {1 + s} 2$
\(\displaystyle \) \(=\) \(\displaystyle \frac \pi {\map \cos {\pi s / 2} }\) Sine and Cosine are Periodic on Reals

Also, we have Legendre's Duplication Formula for $z \notin -\dfrac 1 2 \N_0$:

$\map \Gamma s \, \map \Gamma {s + \dfrac 1 2} = 2^{1 - 2 s} \sqrt \pi \map \Gamma {2 s}$

Replacing $s \mapsto s / 2$ this yields:

$\map \Gamma {\dfrac s 2} \, \map \Gamma {\dfrac {1 + s } 2} = 2^{1 - s} \sqrt \pi \, \map \Gamma s$

Together these give:

$(1): \quad \dfrac {\map \Gamma {s / 2} } {\map \Gamma {\paren {1 - s} / 2} } = 2^{1 - s} \pi^{-1/2} \, \map \Gamma s \, \map \cos {\pi s / 2}$


Now we take the Functional Equation for Riemann Zeta Function:

$\pi^{-s/2} \, \map \zeta s \, \map \Gamma {s / 2} \, \map \Gamma {\dfrac {1 - s} 2}^{-1} = \pi^{\paren {s - 1} / 2} \, \map \zeta {1 - s}$

and substitute $(1)$ to give:

$\pi^{\paren {s - 1} / 2} \, \map \zeta {1 - s} = \pi^{-\paren {s + 1} / 2} \, \map \zeta s 2^{1 - s} \, \map \Gamma s \, \map \cos {\pi s / 2}$

Multiplying by $\pi^{\paren {s - 1} / 2}$ this becomes:

$\map \zeta {1 - s} = \pi^{-s} 2^{1 - s} \, \map \cos {\pi s / 2} \, \map \Gamma s \, \map \zeta s$

as desired.

$\blacksquare$