PlanetMath: formulae for zeta in the critical strip

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formulae for zeta in the critical strip

(Theorem)

Let us use the traditional notation $s=\sigma+it$ for the complex variable, where $\sigma$ and are real numbers.

$\displaystyle \zeta(s)$	$\displaystyle =$	$\displaystyle \frac{1}{1-2^{1-s}}\sum_{n=1}^\infty (-1)^{n+1}n^{-s} \qquad\sigma>0$	(1)
$\displaystyle \zeta(s)$	$\displaystyle =$	$\displaystyle \frac{1}{s-1}+1-s\int_1^\infty \frac{x-[x]}{x^{s+1}}dx \qquad\sigma>0$	(2)
$\displaystyle \zeta(s)$	$\displaystyle =$	$\displaystyle \frac{1}{s-1}+\frac{1}{2}-s\int_1^\infty \frac{((x))}{x^{s+1}}dx \quad\sigma>-1$	(3)

where

denotes the largest integer $\le x$ , and

denotes $x-[x]-\frac{1}{2}$ .

We will prove (2) and (3) with the help of this useful lemma:

Lemma: For integers and such that :

$\displaystyle \sum_{n=u+1}^v n^{-s} = -s\int_u^v \frac{x-[x]}{x^{s+1}}dx +\frac{v^{1-s}-u^{1-s}}{1-s}$

Proof: If we can prove the special case , namely

$\displaystyle (u+1)^{-s} = -s\int_u^{u+1} \frac{x-[x]}{x^{s+1}}dx +\frac{(u+1)^{1-s}-u^{1-s}}{1-s}$

(4)

then the lemma will follow by summing a finite sequence of cases of (4). The integral in (4) is

$\displaystyle \int_0^1 \frac{tdt}{(u+t)^{s+1}}$	$\displaystyle =$	$\displaystyle \int_0^1 (u+t)^{-s}dt - \int_0^1 u(u+t)^{-s-1}dt$
	$\displaystyle =$	$\displaystyle \frac{(u+1)^{1-s}-u^{1-s}}{1-s} +\frac{u\left[(u+1)^{-s}-u^{-s}\right]}{s}$

so the right side of (4) is

$\displaystyle \frac{-s}{1-s}\left[(u+1)^{1-s}-u^{1-s}\right] -u\left[(u+1)^{-s}-u^{-s}\right] -\frac{u^{1-s}}{1-s} +\frac{(u+1)^{1-s}}{1-s}$

$\displaystyle =(u+1)^{-s}\left[\frac{-s(u+1)}{1-s}-u+\frac{u+1}{1-s}\right] +u^{-s}\left[\frac{us}{1-s}+u-\frac{u}{1-s}\right]$

$\displaystyle =(u+1)^{-s}\cdot 1+u^{-s}\cdot 0$

and the lemma is proved.

Now take and let $v\to \infty$ in the lemma, showing that (2) holds for $\sigma>1$ . By the principle of analytic continuation, if the integral in (2) is analytic for $\sigma>0$ , then (2) holds for $\sigma>0$ . But is bounded, so the integral converges uniformly on $\sigma\ge\epsilon$ for any $\epsilon>0$ , and the claim (2) follows.

We have

$\displaystyle \frac{1}{2}s\int_1^\infty x^{-1-s}dx=\frac{1}{2}$

Adding and subtracting this quantity from (2), we get (3) for $\sigma>0$ . We need to show that

$\displaystyle \int_1^\infty \frac{((x))}{x^{s+1}}dx$

is analytic on $\sigma>-1$ . Write

$\displaystyle f(y)=\int_1^y ((x))dx$

and integrate by parts:

$\displaystyle \int_1^\infty \frac{((x))}{x^{s+1}}dx =\lim_{x\to\infty}f(x)x^{-1-s} - f(1)x^{-1-1}+(s+1) \int_1^\infty\frac{f(x)}{x^{s+2}}dx$

The first two terms on the right are zero, and the integral converges for $\sigma>-1$ because

is bounded.

Remarks: We will prove (1) in a later version of this entry.

Using formula (3), one can verify Riemann's functional equation in the strip $-1<\sigma<2$ . By analytic continuation, it follows that the functional equation holds everywhere. One way to prove it in the strip is to decompose the sawtooth function into a Fourier series, and do a termwise integration. But the proof gets rather technical, because that series does not converge uniformly.

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Cross-references: series, Fourier series, function, functional equation, converges, converges uniformly, bounded, analytic, analytic continuation, side, right, integral, finite sequence, summing, proof, integer, real numbers, variable, complex
There are 2 references to this entry.

This is version 8 of formulae for zeta in the critical strip, born on 2003-02-15, modified 2003-10-09.
Object id is 4040, canonical name is FormulaeForZetaInTheCriticalStrip.
Accessed 2717 times total.

Classification:

AMS MSC:

11M99 (Number theory :: Zeta and $L$-functions: analytic theory :: Miscellaneous)

Pending Errata and Addenda

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