formulae for zeta in the critical strip

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

$\zeta (s)$ $=$ ${\displaystyle \frac{1}{1-2^{1-s}}}{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{(-1)}^{n+1}{n}^{-s}\mathit{\hspace{1em}\hspace{1em}}\sigma >0$ (1)

$\zeta (s)$ $=$ ${\displaystyle \frac{1}{s-1}}+1-s{\displaystyle {\int }_1^{\mathrm{\infty }}}{\displaystyle \frac{x-[x]}{x^{s+1}}}𝑑x\mathit{\hspace{1em}\hspace{1em}}\sigma >0$ (2)

$\zeta (s)$ $=$ ${\displaystyle \frac{1}{s-1}}+{\displaystyle \frac{1}{2}}-s{\displaystyle {\int }_1^{\mathrm{\infty }}}{\displaystyle \frac{((x))}{x^{s+1}}}𝑑x\mathit{\hspace{1em}}\sigma >-1$ (3)

where $[x]$ denotes the largest integer $\le x$, and $((x))$ denotes $x-[x]-\frac{1}{2}$.

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

Lemma: For integers $u$ and $v$ such that :

$$\sum _{n=u+1}^v{n}^{-s}=-s{\int }_u^v\frac{x-[x]}{x^{s+1}}𝑑x+\frac{v^{1-s}-u^{1-s}}{1-s}$$

Proof: If we can prove the special case $v=u+1$, namely

$${(u+1)}^{-s}=-s{\int }_u^{u+1}\frac{x-[x]}{x^{s+1}}𝑑x+\frac{{(u+1)}^{1-s}-u^{1-s}}{1-s}$$ (4)

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

${\displaystyle {\int }_0^1}{\displaystyle \frac{tdt}{{(u+t)}^{s+1}}}$ $=$ ${\displaystyle {\int }_0^1}{(u+t)}^{-s}𝑑t-{\displaystyle {\int }_0^1}u{(u+t)}^{-s-1}𝑑t$

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

so the right side of (4) is

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

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

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

and the lemma is proved.

Now take $u=1$ and let $v\to \mathrm{\infty }$ 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 $x-[x]$ is bounded, so the integral converges uniformly on $\sigma \ge ϵ$ for any $ϵ>0$, and the claim (2) follows.

We have

$$\frac{1}{2}s{\int }_1^{\mathrm{\infty }}{x}^{-1-s}𝑑x=\frac{1}{2}$$

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

$${\int }_1^{\mathrm{\infty }}\frac{((x))}{x^{s+1}}𝑑x$$

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

$$f(y)={\int }_1^y((x))𝑑x$$

and integrate by parts:

$${\int }_1^{\mathrm{\infty }}\frac{((x))}{x^{s+1}}𝑑x=\underset{x\to \mathrm{\infty }}{lim}f(x)x^{-1-s}-f(1)x^{-1-1}+(s+1){\int }_1^{\mathrm{\infty }}\frac{f(x)}{x^{s+2}}𝑑x$$

The first two terms on the right are zero, and the integral converges for $\sigma >-1$ because $f$ 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 . 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 $((x))$ into a Fourier series, and do a termwise integration. But the proof gets rather technical, because that series does not converge uniformly.

Title	formulae for zeta in the critical strip
Canonical name	FormulaeForZetaInTheCriticalStrip
Date of creation	2013-03-22 13:28:14
Last modified on	2013-03-22 13:28:14
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	11
Author	mathcam (2727)
Entry type	Theorem
Classification	msc 11M99
Related topic	CriticalStrip
Related topic	ValueOfTheRiemannZetaFunctionAtS0
Related topic	AnalyticContinuationOfRiemannZeta