Is it possible to go from one variation of the Riemann functional equation to another?

Question

I saw online that the functional equation can be written as $$\zeta(s)=\pi^{s-1/2}\frac{\Gamma(\frac{1-s}{2})}{\Gamma(s/2)}\zeta(1-s)$$ However, the more well known version is $$\zeta(s)=2^s\pi^{s-1}\sin\frac{\pi s}{2}\Gamma(1-s)\zeta(1-s)$$ My question is whether it is possible to go from the first to second. I tried using the Euler reflection formula but I’m not sure where to go from there, or if that’s even the way to go.

Answer 1

As pointed out by Thomas Andrews, we can appeal to Legendre's Duplication Formula as given by

$$\Gamma(z)\Gamma(z+1/2)=2^{1-2z}\sqrt {\pi}\Gamma(2z)\tag1$$

Setting $z=(1-s)/2$ in $(1)$ reveals

$$\Gamma\left(\frac {1-s}2\right)\Gamma\left(1-\frac{s}{2}\right)=2^{s}\sqrt {\pi}\Gamma(1-s)\tag2$$

Next, we use Euler's Reflection Formula $\Gamma(s/2)\Gamma(1-s/2)=\frac{\pi}{\sin(\pi s/2)}$ in $(2)$ to obtain

$$\Gamma\left(\frac {1-s}2\right)\frac{\pi}{\sin(\pi s/2)\Gamma(s/2)}=2^{s}\sqrt {\pi}\Gamma(1-s)\tag3$$

Rearranging $(3)$ we see that

$$\frac{\Gamma\left(\frac {1-s}2\right)}{\Gamma(s/2)}=2^{s}\sin(\pi s/2)\pi^{-1/2}\Gamma(1-s)\tag4$$

Finally, substituting $(4)$ in the functional equation

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

yields the modified, "more well known," functional equation

$$\bbox[5px,border:2px solid #C0A000]{\zeta(s)=2^{s}\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s)}$$