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 thisuseful 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, ifthe integral in (2) is analytic for $\sigma >0$,then (2) holds for $\sigma >0$.But $x-[x]$ is bounded, so the integral convergesuniformly 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 integralconverges 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’sfunctional equation in the strip .By analytic continuation, it follows that the functionalequation holds everywhere.One way to prove it in the strip is to decompose thesawtooth function $((x))$ into a Fourier series, anddo a termwise integration.But the proof gets rather technical, because thatseries does not converge uniformly.