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

\begin{eqnarray} \zeta(s)&=& \frac{1}{1-2^{1-s}}\sum_{n=1}^\infty (-1)^{n+1}n^{-s} \qquad\sigma>0 \label{eq:one} \\ \zeta(s)&=& \frac{1}{s-1}+1-s\int_1^\infty \frac{x-[x]}{x^{s+1}}dx \qquad\sigma>0 \label{eq:two} \\ \zeta(s)&=& \frac{1}{s-1}+\frac{1}{2}-s\int_1^\infty \frac{((x))}{x^{s+1}}dx \quad\sigma>-1 \label{eq:three} \end{eqnarray}where $[x]$ denotes the largest integer $\le x$ , and $((x))$ denotes $x-[x]-\frac{1}{2}$ .

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

Lemma: For integers $u$ and $v$ such that $0<u<v$ : \begin{equation*} \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} \end{equation*} Proof: If we can prove the special case $v=u+1$ , namely \begin{equation} (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} \label{eq:four} \end{equation}then the lemma will follow by summing a finite sequence of cases of (). The integral in () is \begin{eqnarray*} \int_0^1 \frac{tdt}{(u+t)^{s+1}} &=& \int_0^1 (u+t)^{-s}dt - \int_0^1 u(u+t)^{-s-1}dt \\ &=& \frac{(u+1)^{1-s}-u^{1-s}}{1-s} +\frac{u\left[(u+1)^{-s}-u^{-s}\right]}{s} \end{eqnarray*}so the right side of () 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 () holds for $\sigma>1$ . By the principle of analytic continuation, if the integral in () is analytic for $\sigma>0$ , then () holds for $\sigma>0$ . But $x-[x]$ is bounded, so the integral converges uniformly on $\sigma\ge\epsilon$ for any $\epsilon>0$ , and the claim () follows.

We have \begin{equation*} \frac{1}{2}s\int_1^\infty x^{-1-s}dx=\frac{1}{2} \end{equation*}Adding and subtracting this quantity from (), we get () 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 () in a later version of this entry.

Using formula (), 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.