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$$\begin {aligned}\pi^{-s/2}\Gamma(s/2)\zeta(s)=&\pi^{-(1-s)/2}\Gamma((1-s)/2)\zeta(1-s) \\ \zeta(0) =&\frac{\pi^{-1/2}\Gamma(1/2)\zeta(1)}{\pi^{0}\Gamma(0)}=\frac{\zeta(1)}{\Gamma(0)}\end {aligned}$$

so that $\zeta(0)$ is the ratio of the residues of $\zeta(s)$ at $s=1$ end $\Gamma(s)$ at $s=0$. The latter is just $1$ since

$$\lim_{s\rightarrow 0} s\Gamma(s) = \Gamma(1)=1.$$

To calculate the residue of $\zeta(s)$ at $s=1$, I simply used the expression, proved in Stein & Shakarchi Complex Analysis p. 171

$$\pi^{-s/2}\Gamma(s/2)\zeta(s) = \frac 1{s-1} -\frac1s +\frac12\int_1^\infty(u^{(-s/2)-1/2}+u^{s/2-1})[\theta(u)-1]du$$

I argued that since the integral on the RHS is holomorphic at $s=1$, the residue of the LHS at $s=1$ must be $1$ (the coefficient of the first fraction). Since $\pi^{-1/2}\Gamma(1/2)$ cancel each other, it should be the case that $\operatorname {res}_{s=1} \zeta(s)=1$, but then my argument says that $\zeta(0) = 1$. Where did I miss the $-\frac12$?

Rodrigo
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  • You seem to be using a weird, to say the least, form of the functional equation, and your idea that $;\zeta(0);$ is the ratio of that residues is odd for me. Look at my answer here:http://math.stackexchange.com/questions/493216/how-to-compute-zeta-0 – DonAntonio Jan 19 '14 at 09:41

1 Answers1

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Where did I miss the $−\frac{1}{2}$?

In two parts. The factor $\frac{1}{2}$ comes from the fact that you divide by $\Gamma\left(\frac{s}{2}\right)$, and not $\Gamma(s)$. So you'd divide by the residue of $F(s) = \Gamma\left(\frac{s}{2}\right)$ in $0$, which is $2\cdot \operatorname{Res}(\Gamma; 0) = 2$. The factor $-1$ comes from the fact that you have $\zeta(1-s)$ on the right hand side, so you'd have to use the residue of $G(s) = \zeta(1-s)$ in $0$, and from

$$\zeta(s) = \frac{1}{s-1} + h(s) \Rightarrow G(s) = \frac{1}{(1-s)-1} + h(1-s) = -\frac{1}{s} + h(1-s)$$

we see that $\operatorname{Res}(G; 0) = -1$.

Daniel Fischer
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