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

Theorem.

Let $\zeta$ denote the meromorphic extension of the Riemann zeta function to the complex plane. Then $\displaystyle\zeta(0)=\frac{-1}{2}$.

Proof.

Recall that one of the 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 pole (http://planetmath.org/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:

 $\zeta(0)$ $=\displaystyle\lim_{s\to 0^{+}}\zeta(s)$ $=\displaystyle\lim_{s\to 0^{+}}2^{s}\pi^{s-1}\sin\frac{\pi s}{2}\Gamma(1-s)% \zeta(1-s)$ $=\displaystyle\lim_{s\to 0^{+}}2^{s}\pi^{s-1}\left(\sum_{n=0}^{\infty}\frac{(-% 1)^{n}}{(2n+1)!}\left(\frac{\pi s}{2}\right)^{2n+1}\right)\Gamma(1-s)\left(% \frac{1}{(1-s)-1}+1-(1-s)\int_{1}^{\infty}\frac{x-[x]}{x^{(1-s)+1}}\,dx\right)$ $=\displaystyle\lim_{s\to 0^{+}}2^{s}\pi^{s-1}\left(\frac{\pi s}{2}\right)\left% (\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}\left(\frac{\pi s}{2}\right)^{2n}% \right)\Gamma(1-s)\left(\frac{1}{-s}+1-(1-s)\int_{1}^{\infty}\frac{x-[x]}{x^{2% -s}}\,dx\right)$ $=\displaystyle\lim_{s\to 0^{+}}2^{s}\pi^{s-1}\left(\frac{\pi}{2}\right)\left(1% +\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(2n+1)!}\left(\frac{\pi s}{2}\right)^{2n}% \right)\Gamma(1-s)s\left(\frac{-1}{s}+1-(1-s)\int_{1}^{\infty}\frac{x-[x]}{x^{% 2-s}}\,dx\right)$ $=\displaystyle\lim_{s\to 0^{+}}2^{s-1}\pi^{s}\left(1+\sum_{n=1}^{\infty}\frac{% (-1)^{n}}{(2n+1)!}\left(\frac{\pi s}{2}\right)^{2n}\right)\Gamma(1-s)\left(-1+% s-s(1-s)\int_{1}^{\infty}\frac{x-[x]}{x^{2-s}}\,dx\right)$ $=\displaystyle\left(\lim_{s\to 0^{+}}2^{s-1}\pi^{s}\Gamma(1-s)\left(-1+s-s(1-s% )\int_{1}^{\infty}\frac{x-[x]}{x^{2-s}}\,dx\right)\right)\left(\lim_{s\to 0^{+% }}1+\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(2n+1)!}\left(\frac{\pi s}{2}\right)^{2% n}\right)$ $=\displaystyle\left(2^{0-1}\pi^{0}\Gamma(1-0)\left(-1+0-0(1-0)\int_{1}^{\infty% }\frac{x-[x]}{x^{2-0}}\,dx\right)\right)\left(1+\sum_{n=1}^{\infty}\frac{(-1)^% {n}}{(2n+1)!}\left(\frac{\pi\cdot 0}{2}\right)^{2n}\right)$ $=\displaystyle\left(\frac{1}{2}\cdot 1\cdot\Gamma(1)\cdot(-1+0-0)\right)\left(% 1+\sum_{n=1}^{\infty}0\right)$ $=\displaystyle\frac{-1}{2}$.

Title value of the Riemann zeta function at $s=0$ ValueOfTheRiemannZetaFunctionAtS0 2013-03-22 16:07:17 2013-03-22 16:07:17 Wkbj79 (1863) Wkbj79 (1863) 20 Wkbj79 (1863) Theorem msc 11M06 CriticalStrip FormulaeForZetaInTheCriticalStrip