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From analytic continuation of zeta function,

\begin{align} \zeta(s) = \frac{\pi^{\frac{s}{2}}}{\Gamma\left(\frac{s}{2}\right)} \left[ \frac{1}{s(s-1)} + \int_1^{\infty} \left( x^{\frac{s}{2}-1} + x^{-\frac{s}{2}-1} \right) \left(\frac{\theta(x) - 1}{2}\right) dx \right] \end{align}

From this I obtain a functional equation \begin{align} \pi^{-\frac{s}{2}} \Gamma\left(\frac{s}{2}\right) \zeta(s) = \pi^{-\frac{1-s}{2}} \Gamma\left(\frac{1-s}{2}\right) \zeta(1-s) \end{align} My next step is to prove reflection functional equation

\begin{align} \zeta(1-s) = 2 (2\pi)^{-s} \cos\left(\frac{s\pi}{2}\right) \Gamma(s)\zeta(s) \end{align}

Reordering a functional equation for $\zeta(s)$, I have \begin{align} \zeta(1-s) = \pi^{\frac{1}{2}} \frac{\Gamma\left(\frac{s}{2}\right)}{\Gamma\left(\frac{1-s}{2}\right)} \zeta(s) \end{align} I have problem with showing above two equations are indeed same...

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Let $$F(s) = \Gamma\left(1-\frac{s}{2}\right) \left( \Gamma\left(\frac{1-s}{2}\right) 2 (2\pi)^{-s}\Gamma(s) -\frac{\pi^{\frac{1}{2}}\Gamma\left(\frac{s}{2}\right)}{ \cos\left(\frac{s\pi}{2}\right)}\right)$$ From $\Gamma(s) = \int_0^\infty t^{s-1}e^{-t}dt$,$ \Gamma(s+1) = s \Gamma(s), \Gamma(1) = 1$ we obtain that $F$ is $4$-periodic.

From $\Gamma(1/2) = \sqrt{\pi}$ we find the poles cancel each other so that $F$ is entire.

$F$ is bounded because $\Gamma(s+1)$ is the Fourier transform of $e^u e^{-e^{u}}$ whose all derivatives are $L^1$, thus $\Gamma(s+1)$ is rapidly decreasing on vertical lines $\Re(s ) > 0$ and $ \Gamma(s+1) = s \Gamma(s)$ implies it is rapidly decreasing on all vertical lines.

That is $F$ is a bounded entire function, thus it is constant, $F(s) = F(i\infty) = 0$.

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Use \begin{align} &\frac{\pi^{\frac{1}{2}}}{2^{s-1}} \Gamma (s) = \Gamma \left(\frac{s}{2}\right) \Gamma \left(\frac{s+1}{2}\right) \\ & \Gamma\left(\frac{1+s}{2}\right) \Gamma\left(\frac{1-s}{2}\right) = \frac{\pi}{\cos\left(\frac{\pi s}{2}\right)} \end{align} which comes from Legndre duplication formula and Euler's Reflection formula of gamma function.

Then I can prove the relation well!

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