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Let $\Lambda(s)=\pi^{-s/2} \Gamma\left(\frac{s}{2}\right)\zeta(s).$ Then $\Lambda\left(\frac{1}{2} + s\right) = \Lambda\left(\frac{1}{2} - s\right)$ constitutes the functional equation for the Riemann $\zeta$ function. The presence of the product $\Gamma\left(\frac{s}{2}\right) \zeta(s)$ is perfectly understandable, inasmuch as the poles of the former coincide with the (trivial) zeroes of the latter; as is also the symmetry with regard to $\frac{1}{2}$ since this value stands midway between the two poles of $\Lambda$.

What poses serious difficulties from an intuitive perspective, however, is the presence of $\pi^{-s/2} \zeta(s)$ instead of the expected $\pi^{-s} \zeta(s)$ given the fact that $\zeta(2k)$ always possesses a known closed form in terms of $\pi^{2k}$ rather than merely $\pi^k$ for integer values of the argument $k$. One can, of course, always write $\Lambda(s) = \Gamma\left(\frac{1}{2}\right)^{-s} \Gamma\left(\frac{s}{2}\right) \zeta(s)$ but, for all its niceness, the latter appears somewhat contrived, inasmuch as the power of $\pi$ is clearly a contribution of Riemann's $\zeta$ rather than Euler's $\Gamma$ function.

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    $\begingroup$ See also this question: mathoverflow.net/questions/7656/…. $\endgroup$ – Peter Humphries Apr 20 '17 at 17:15
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    $\begingroup$ Many people have suggested that the ancient Greeks made a mistake when they identified $\pi$ as a fundamental constant; they should have chosen $2\pi$ instead. Had they only known about the zeta-function, perhaps they would have chosen $\sqrt\pi$. $\endgroup$ – Gerry Myerson Apr 20 '17 at 23:17
  • $\begingroup$ @GerryMyerson: Actually, they were torn between $2\pi$ and $\dfrac\pi2,$ so they eventually ended up choosing their geometric mean, since circles are geometric shapes, after all... $\endgroup$ – Lucian Apr 27 '17 at 9:03
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Your intuition breaks down because $\zeta(1-2k)$ has a closed form in terms of Bernoulli numbers, but no powers of $\pi$ at all. This was known to Euler (via Abel summation as the series is of course divergent). To get to something that might look symmmetric in powers of $\pi$, you need to move "half" of the $\pi$ to the other side of the equation.

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To be fair, the functional equation of $\zeta(s)$ is not really $\Lambda\left(1 - s\right)=\Lambda\left( s\right)$, but

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

And in this one the inutition in terms of special values at the even integers is clear.

The proper expression for $\Lambda\left( s\right)$ at $s=2k$ is

$$\Lambda\left( s\right)=\pi^{-s/2}\Gamma(s/2)\zeta(s)=\alpha\pi^{s/2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \alpha\in\mathbb{Q}$$

Using this closed form (instead of the one for $\zeta(s)$), the intuition works again.

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  • $\begingroup$ $\dfrac{\zeta(1-s)}{2^{1-s}}=\Gamma(s)\cos\left(\dfrac\pi2s\right)\dfrac{\zeta(s)}{\pi^s}$ has a certain niceness to it, but leaves the exponential factor on the left unexplained. Perhaps an influence of the Dirichlet $\eta$ function, which helps expand Riemann's $\zeta$ on $(0,1)$ ? $\endgroup$ – Lucian Apr 20 '17 at 18:31
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    $\begingroup$ @Lucian See this proof (Démonstration) $\zeta(s) = \chi(s) \zeta(1-s)$ where $\chi(s) = \frac{s}{\pi} \int_0^\infty x^{-s-1} \sin(\pi x)dx$ $\endgroup$ – reuns Apr 21 '17 at 3:07

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