Let $~\rho(x)~=~\dfrac{\Gamma(x)\cdot\zeta(2x)}{\pi^x}.~$ Then $~\rho\Big(\tfrac14+x\Big)~=~\rho\Big(\tfrac14-x\Big)~$ constitutes the functional

equation for the Riemann $\zeta$ function. The presence of the product $~\Gamma(x)\cdot\zeta(2x)~$ 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 $\tfrac14,$ since this value stands midway between $\rho$'s two poles.

What poses serious difficulties from an intuitive perspective, however, is the presence of $~\dfrac{\zeta(2x)}{\pi^x}~$

instead of the expected $~\dfrac{\zeta(2x)}{\pi^{\color{red}2x}},~$ given the fact that $\zeta(2k)$ always possesses a known closed

form in terms of $\pi^{\color{red}2k}$ rather than merely $\pi^k$ for integer values of the argument k. One can, of course,

always write $~\rho(x/2)~=~\dfrac{\Gamma(x/2)~}{~\Gamma(1/2)^x}\cdot\zeta(x),~$ 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.