1
$\begingroup$

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?

$\endgroup$
  • 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
3
$\begingroup$

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*}

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.