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

Proof. Recall that one of the formulas for the Riemann zeta function in the critical strip is given by

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

where denotes the integer part of .

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

where $\Gamma$ denotes the gamma function.

The only pole of $\zeta$ occurs at . Therefore, $\zeta$ is analytic, and thus continuous, at .

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

$\zeta(0)$	$=\displaystyle \lim_{s \to 0^+} \zeta(s)$

	$=\displaystyle \lim_{s \to 0^+} 2^s \pi^{s-1} \sin \frac{\pi s}{2} \Gamma(1-s) \zeta(1-s)$

	$=\displaystyle \lim_{s \to 0^+} 2^s \pi^{s-1} \left( \sum_{n=0}^{\infty} \frac... ...rac{1}{(1-s)-1}+1-(1-s) \int_1^{\infty} \frac{x-[x]}{x^{(1-s)+1}} \, dx \right)$

	$=\displaystyle \lim_{s \to 0^+} 2^s \pi^{s-1} \left( \frac{\pi s}{2} \right) \... ...\left( \frac{1}{-s}+1-(1-s) \int_1^{\infty} \frac{x-[x]}{x^{2-s}} \, dx \right)$

	$=\displaystyle \lim_{s \to 0^+} 2^s \pi^{s-1} \left( \frac{\pi}{2} \right) \le... ...\left( \frac{-1}{s}+1-(1-s) \int_1^{\infty} \frac{x-[x]}{x^{2-s}} \, dx \right)$

	$=\displaystyle \lim_{s \to 0^+} 2^{s-1} \pi^s \left( 1+\sum_{n=1}^{\infty} \fr... ...mma(1-s) \left( -1+s-s(1-s) \int_1^{\infty} \frac{x-[x]}{x^{2-s}} \, dx \right)$

	$=\displaystyle \left( \lim_{s \to 0^+} 2^{s-1} \pi^s \Gamma(1-s) \left( -1+s-s... ...=1}^{\infty} \frac{(-1)^n}{(2n+1)!} \left( \frac{\pi s}{2} \right)^{2n} \right)$

	$=\displaystyle \left( 2^{0-1} \pi^0 \Gamma(1-0) \left( -1+0-0(1-0) \int_1^{\in... ...infty} \frac{(-1)^n}{(2n+1)!} \left( \frac{\pi \cdot 0}{2} \right)^{2n} \right)$

	$=\displaystyle \left( \frac{1}{2} \cdot 1 \cdot \Gamma(1) \cdot (-1+0-0) \right) \left( 1+\sum_{n=1}^{\infty} 0 \right)$
	$=\displaystyle \frac{-1}{2}$ .

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