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We know that $$\xi(s)=\frac{s(s-1)\pi^{\frac{{-s}}2}}{2}\Gamma\left(\frac{s}{2}\right)\zeta(s)$$

I want to prove the functional equation $$\xi(s)=\xi(1-s)$$ I could not find it anywhere. My approach is to show that this equality holds: $$\frac{s(s-1)\pi^{\frac{{-s}}2}}{2}\Gamma\left(\frac{s}{2}\right)\zeta(s)=\frac{(1-s)(-s)\pi^{\frac{{s-1}}2}}{2}\Gamma\left(\frac{1-s}{2}\right)\zeta(1-s)$$ We can get rid of the polynomials straight away because they are clearly the same: $$\pi^{\frac{{-s}}2}\Gamma\left(\frac{s}{2}\right)\zeta(s)=\pi^{\frac{{s-1}}2}\Gamma\left(\frac{1-s}{2}\right)\zeta(1-s)$$

Now, since $\zeta(s)=2^s\pi^{s-1}\text{sin}\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s)$ we have

$$\pi^{\frac{{-s}}2}\Gamma\left(\frac{s}{2}\right)2^s\pi^{s-1}\text{sin}\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s)=\pi^{\frac{{s-1}}2}\Gamma\left(\frac{1-s}{2}\right)\zeta(1-s)$$ Dividing through $\zeta(1-s)$ $$\pi^{\frac{{s-2}}2}\Gamma\left(\frac{s}{2}\right)2^s\text{sin}\left(\frac{\pi s}{2}\right)\Gamma(1-s)=\pi^{\frac{{s-1}}2}\Gamma\left(\frac{1-s}{2}\right)$$ Using Euler's reflection formula $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\texttt{sin}(\pi z)}$ we have$$\pi^{\frac{{s}}2}2^s\frac{\Gamma(1-s)}{\Gamma\left(1-\frac{s}{2}\right)}=\pi^{\frac{{s-1}}2}\Gamma\left(\frac{1-s}{2}\right)$$

At this point I am not sure what to do next. Wolfram seems to agree with me enter image description here

hence perhaps maybe this is the way forward. Could anyone help me out how to finish this? or maybe give me completely new proof, I do not mind either way.

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Never mind guys, You use the gamma duplication formula

$$\Gamma(z)\Gamma(z+\frac{1}{2})=2^{1-2z}\pi^{\frac{1}{2}}\Gamma(2z)$$

and the result follows.

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  • $\begingroup$ ...and never trust WA too much... $\endgroup$ – DonAntonio Mar 10 '18 at 16:50

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