# value of the Riemann zeta function at $s=0$

## Primary tabs

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\newtheorem*{thm*}{Theorem}
\begin{document}
\begin{thm*}
Let $\zeta$ denote the meromorphic extension of the Riemann zeta function to the \PMlinkescapetext{entire} complex plane.  Then $\displaystyle \zeta(0)=\frac{-1}{2}$.
\end{thm*}

\begin{proof}
Recall that one of the \PMlinkescapetext{formulas} for the Riemann zeta function in the critical strip is given by

$$\zeta(s)=\frac{1}{s-1}+1-s\int_1^{\infty} \frac{x-[x]}{x^{s+1}} \,dx,$$

where $[x]$ denotes the integer part of $x$.

Also recall the functional equation

$$\zeta(s) = 2^s \pi^{s-1} \sin \frac{\pi s}{2} \Gamma(1-s) \zeta(1-s),$$

where $\Gamma$ denotes the gamma function.

The only \PMlinkname{pole}{Pole} of $\zeta$ occurs at $s=1$.  Therefore, $\zeta$ is analytic, and thus continuous, at $s=0$.

Let $\displaystyle \lim_{s \to 0^+}$ denote the limit as $s$ approaches $0$ along any path contained in the region $\operatorname{Re}(s)>0$.  Thus:

\begin{center}
\begin{tabular}{ll}
$\zeta(0)$ & $=\displaystyle \lim_{s \to 0^+} \zeta(s)$ \\
\\
& $=\displaystyle \lim_{s \to 0^+} 2^s \pi^{s-1} \sin \frac{\pi s}{2} \Gamma(1-s) \zeta(1-s)$ \\
\\
& $=\displaystyle \lim_{s \to 0^+} 2^s \pi^{s-1} \left( \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \left( \frac{\pi s}{2} \right)^{2n+1} \right) \Gamma(1-s) \left( \frac{1}{(1-s)-1}+1-(1-s) \int_1^{\infty} \frac{x-[x]}{x^{(1-s)+1}} \, dx \right)$ \\
\\
& $=\displaystyle \lim_{s \to 0^+} 2^s \pi^{s-1} \left( \frac{\pi s}{2} \right) \left( \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \left( \frac{\pi s}{2} \right)^{2n} \right) \Gamma(1-s) \left( \frac{1}{-s}+1-(1-s) \int_1^{\infty} \frac{x-[x]}{x^{2-s}} \, dx \right)$ \\
\\
& $=\displaystyle \lim_{s \to 0^+} 2^s \pi^{s-1} \left( \frac{\pi}{2} \right) \left( 1+\sum_{n=1}^{\infty} \frac{(-1)^n}{(2n+1)!} \left( \frac{\pi s}{2} \right)^{2n} \right) \Gamma(1-s) s \left( \frac{-1}{s}+1-(1-s) \int_1^{\infty} \frac{x-[x]}{x^{2-s}} \, dx \right)$ \\
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& $=\displaystyle \lim_{s \to 0^+} 2^{s-1} \pi^s \left( 1+\sum_{n=1}^{\infty} \frac{(-1)^n}{(2n+1)!} \left( \frac{\pi s}{2} \right)^{2n} \right) \Gamma(1-s) \left( -1+s-s(1-s) \int_1^{\infty} \frac{x-[x]}{x^{2-s}} \, dx \right)$ \\
\\
& $=\displaystyle \left( \lim_{s \to 0^+} 2^{s-1} \pi^s \Gamma(1-s) \left( -1+s-s(1-s) \int_1^{\infty} \frac{x-[x]}{x^{2-s}} \, dx \right) \right) \left( \lim_{s \to 0^+} 1+\sum_{n=1}^{\infty} \frac{(-1)^n}{(2n+1)!} \left( \frac{\pi s}{2} \right)^{2n} \right)$ \\
\\
& $=\displaystyle \left( 2^{0-1} \pi^0 \Gamma(1-0) \left( -1+0-0(1-0) \int_1^{\infty} \frac{x-[x]}{x^{2-0}} \, dx \right) \right) \left(1+\sum_{n=1}^{\infty} \frac{(-1)^n}{(2n+1)!} \left( \frac{\pi \cdot 0}{2} \right)^{2n} \right)$ \\
\\
& $=\displaystyle \left( \frac{1}{2} \cdot 1 \cdot \Gamma(1) \cdot (-1+0-0) \right) \left( 1+\sum_{n=1}^{\infty} 0 \right)$ \\
& $=\displaystyle \frac{-1}{2}$. \end{tabular}
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\end{proof}
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