# 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.

 $\displaystyle\zeta(s)$ $\displaystyle=$ $\displaystyle\frac{1}{1-2^{1-s}}\sum_{n=1}^{\infty}(-1)^{n+1}n^{-s}\qquad% \sigma>0$ (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 $[x]$ denotes the largest integer $\leq x$, 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 $0:

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

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

 $\displaystyle\int_{0}^{1}\frac{tdt}{(u+t)^{s+1}}$ $\displaystyle=$ $\displaystyle\int_{0}^{1}(u+t)^{-s}dt-\int_{0}^{1}u(u+t)^{-s-1}dt$ $\displaystyle=$ $\displaystyle\frac{(u+1)^{1-s}-u^{1-s}}{1-s}+\frac{u\left[(u+1)^{-s}-u^{-s}% \right]}{s}$

so the right side of (4) is

 $\frac{-s}{1-s}\left[(u+1)^{1-s}-u^{1-s}\right]-u\left[(u+1)^{-s}-u^{-s}\right]% -\frac{u^{1-s}}{1-s}+\frac{(u+1)^{1-s}}{1-s}$
 $=(u+1)^{-s}\left[\frac{-s(u+1)}{1-s}-u+\frac{u+1}{1-s}\right]+u^{-s}\left[% \frac{us}{1-s}+u-\frac{u}{1-s}\right]$
 $=(u+1)^{-s}\cdot 1+u^{-s}\cdot 0$

and the lemma is proved.

Now take $u=1$ and let $v\to\infty$ 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 $\sigma\geq\epsilon$ for any $\epsilon>0$, and the claim (2) follows.

We have

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

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

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

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

 $f(y)=\int_{1}^{y}((x))dx$

and integrate by parts:

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

The first two terms on the right are zero, and the integral converges  for $\sigma>-1$ because $f$ 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 $-1<\sigma<2$. 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 $((x))$ 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  FormulaeForZetaInTheCriticalStrip 2013-03-22 13:28:14 2013-03-22 13:28:14 mathcam (2727) mathcam (2727) 11 mathcam (2727) Theorem msc 11M99 CriticalStrip ValueOfTheRiemannZetaFunctionAtS0 AnalyticContinuationOfRiemannZeta