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On page 30, he writes that $\xi(0)=-\zeta(0)=1/2$, but on page 16 he writes that:

$\xi(s)=1/2 s(s-1)\pi^{-1/2s}\Gamma(1/2s)\zeta(s)$ in eq.(2.1.12); so if I plug into this equation $s=0$ then I get that it should vanish, shouldn't it?

What's wrong here?

https://books.google.co.il/books?id=1CyfApMt8JYC&printsec=frontcover#v=onepage&q&f=false

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  • $\begingroup$ What do you know about $\Gamma(0)$? $\endgroup$ – Gerry Myerson Jul 16 '15 at 8:58
  • $\begingroup$ $\Gamma(s)$ has a simple pole at s=0, so it's get cancelled with $s$. $\endgroup$ – MathematicalPhysicist Jul 16 '15 at 9:18
  • $\begingroup$ Right, though I'd rather say it's the zero of $s$ that gets cancelled by the pole of $\Gamma$. So, what makes you think $\zeta$ should vanish at zero? $\endgroup$ – Gerry Myerson Jul 16 '15 at 11:52
  • $\begingroup$ It doesn't vanish at zero, I thought that $\Gamma(s)$ vanishes at zero. $\endgroup$ – MathematicalPhysicist Jul 16 '15 at 12:22
  • $\begingroup$ But you just told me that $\Gamma(s)$ has a pole at $s=0$. Are you working under the impression that having a pole is the same thing as vanishing? $\endgroup$ – Gerry Myerson Jul 16 '15 at 23:18
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For the functional equation $$ \xi(s) = \xi(1 - s) $$ we have the definition $$ \xi(s) = \frac{1}{2}\pi^{-s/2}s(s-1)\Gamma\left(\frac{s}{2}\right)\zeta(s).\! $$ The functional equation just gives $\xi(0)=\xi(1)$. The function $Z(s)=\frac{1}{2}\pi^{-s/2}\Gamma(\frac{s}{2})\zeta(s)$ has a meromorphic continuation to the whole $s$-plane, with simple poles at $s=0$ and $s=1$. So the question is, what the value of $\xi(0)$ is. Following your argument, we should have $\xi(0)=0$, and not $\xi(0)=1/2$. However, we have to take into account the simple poles, too.

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