In Titchmarch's book the functional equation is given as $$\zeta(s)=2^s\pi^{s-1}\sin \left( \frac{s\pi}2\right) \Gamma(1-s)\zeta(1-s).$$ However, in the third proof, he derives a following equation $$\pi^{-\frac{s}2}\Gamma\left( \frac{s}2\right) \zeta(s)=\pi^{-\frac12+\frac{s}2}\Gamma\left( \frac12-\frac{s}2\right) \zeta(1-s)$$ and simply states that our initial equation follows from our derived equation. Can somebody explain to me how exactly?

  • 2
    $\begingroup$ Hint: This boils down to a Gamma function relation. Apply the Gamma function duplication formula dlmf.nist.gov/5.5.E5 for $z=(1-s)/2$. This gives you two terms $\Gamma(1/2-s/2)$ and $\Gamma(1-s/2),\;$ for the last use the reflection formula. $\endgroup$ – gammatester Feb 18 '16 at 13:49

We obviously have $$ \zeta(s) = \pi^{- \frac{1}{2} + s} \frac{\Gamma(\frac{1}{2} - \frac{s}{2})}{\Gamma(\frac{s}{2})} \zeta(1 - s).$$ We now use the duplication formula in the form $\Gamma(z + \frac{1}{2}) = 2^{1 - 2z} \sqrt{\pi} \frac{\Gamma(2z)}{\Gamma(z)}$, where $z = -\frac{s}{2}$. This yields $$\pi^{- \frac{1}{2} + s} \frac{\Gamma(\frac{1}{2} - \frac{s}{2})}{\Gamma(\frac{s}{2})} = 2^{1 + s} \pi^s \frac{\Gamma(-s)}{\Gamma(-\frac{s}{2})\Gamma(\frac{s}{2})}.$$ Now using Euler's reflection formula $\Gamma(z) = \frac{\pi}{\sin(\pi z) \Gamma(1 - z)}$ for $z = -\frac{s}{2}$ gives us \begin{align*}2^{1 + s} \pi^s \frac{\Gamma(-s)}{\Gamma(-\frac{s}{2})\Gamma(\frac{s}{2})} &= 2^{1 + s} \pi^{s-1} \frac{\Gamma(-s)\Gamma(1 + \frac{s}{2})}{\Gamma(\frac{s}{2})} \sin(-\frac{\pi s}{2}) \\ &= 2^{1 + s} \pi^{s-1} \frac{\frac{s}{2} \Gamma(-s)\Gamma(\frac{s}{2})}{\Gamma(\frac{s}{2})} \sin(-\frac{\pi s}{2})\\ &= -2^{s} \pi^{s-1} \Gamma(1-s)\sin(-\frac{\pi s}{2}) \\ &= 2^{s} \pi^{s-1} \Gamma(1-s)\sin(\frac{\pi s}{2}).\end{align*}

  • $\begingroup$ Please, how you obtain the second identity after that you use Euler's reflection formula? $\endgroup$ – user243301 Feb 18 '16 at 14:13
  • $\begingroup$ $\Gamma(z + 1) = z\Gamma(z)$, or what do you mean? $\endgroup$ – Paul K Feb 18 '16 at 14:14
  • $\begingroup$ Yes, now I understand! $\endgroup$ – user243301 Feb 18 '16 at 14:17

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