Unsymmetric Functional Equation for Riemann Zeta Function

Theorem
Let $\zeta$ be the Riemann zeta function.

Let $\Gamma$ be the gamma function.

Then for all $s \in \C$,


 * $\displaystyle \zeta(1-s) = 2^{1-s}\pi^{-s} \cos\left(\frac12 s\pi \right)\Gamma(s)\zeta(s)$

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


 * $\displaystyle \Gamma \left({s}\right) \Gamma \left({1 - s}\right) = \frac \pi {\sin \left({\pi s}\right)}$

Replacing $s \mapsto (1+s)/2$ we deduce:

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


 * $\displaystyle \Gamma \left({s}\right) \Gamma \left (s + \frac 1 2 \right) = 2^{1-2s} \sqrt \pi \Gamma \left({2 s}\right)$

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


 * $\displaystyle \Gamma \left({\frac s2}\right) \Gamma \left (\frac{1+s}2 \right) = 2^{1-s} \sqrt \pi \Gamma \left({s}\right)$

Together these give:


 * $\displaystyle \frac{\Gamma(s/2)}{\Gamma((1-s)/2)} = 2^{1-s}\pi^{-1/2} \Gamma \left({s}\right) \cos\left(\pi s/2\right) \qquad (1)$

Now we take the Functional Equation for Riemann Zeta Function:


 * $\displaystyle \pi^{-s/2}\zeta(s)\Gamma(s/2)\Gamma\left(\frac{1-s}2\right)^{-1} = \pi^{(s-1)/2}\zeta(1-s)$

and substitute $(1)$ to give:


 * $\displaystyle \pi^{(s-1)/2}\zeta(1-s) = \pi^{-(s+1)/2}\zeta(s) 2^{1-s}\Gamma \left({s}\right) \cos\left(\pi s/2\right)$

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


 * $\displaystyle \zeta(1-s) = \pi^{-s} 2^{1-s}\cos\left(\pi s/2\right)\Gamma \left({s}\right)\zeta(s)$

as desired.