Hello wonderful people
It seems that the Riemann zeta function in the critical strip may be given by: \begin{gather} \zeta(s)=\frac{1}{s-1}+1-s \int\limits_{1}^{+\infty} \frac{x-[x]}{x^{s+1}} \mathrm{d}x \end{gather} Does someone have a nice and pedagogical proof for a public of engineers?
Thx in advance
If a "public of engineers" wants a much more detailed proof, see below.
We assume $Re(s) > 1$ until indicated otherwise. \begin{align*} \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} &= \sum_{n=1}^{1} \frac{n}{n^s} + \sum_{n=2}^{\infty} \frac{1}{n^s} = \sum_{n=1}^{1} \frac{n}{n^s} + \sum_{n=2}^{\infty} \frac{n - (n-1)}{n^s} \\ &= \sum_{n=1}^{1} \frac{n}{n^s} + \sum_{n=2}^{\infty} \frac{n}{n^s} - \sum_{n=2}^{\infty} \frac{n-1}{n^s} = \sum_{n=1}^{\infty} \frac{n}{n^s} - \sum_{n=2}^{\infty} \frac{n-1}{n^s} \\ &= \sum_{n=1}^{\infty} \frac{n}{n^s} - \sum_{n=1}^{\infty} \frac{n}{(n+1)^s} = \sum_{n=1}^{\infty} n\left[ \frac{1}{n^s} - \frac{1}{(n+1)^s}\right] \\ & = s \sum_{n=1}^{\infty} n \int_n^{n+1} x^{-s-1}\,dx. \end{align*} Since $[x] = n$ for any $x$ in the interval $[n,n+1)$, we have \begin{align*} &= s \sum_{n=1}^{\infty} \int_n^{n+1} [x]x^{-s-1}\,dx = s \int_1^{\infty} [x]x^{-s-1}\,dx \\ &= s \left[ \int_1^{\infty} x^{-s}\,dx \right] - s \int_1^{\infty} \{x\} x^{-s-1}\,dx \quad\text{(because $[x] = x − \{x\}$)}.\\ &= s \left[ \frac{x^{-s+1}}{-s+1} \Bigg|_1^{\infty} \right] - s \int_1^{\infty} \{x\} x^{-s-1}\,dx, \end{align*} allowing the following simplification \begin{align*} \zeta(s) &= \frac{s}{s-1} - s\int_{1}^{\infty} \frac{\{x\} }{x^{s+1}}\,dx \quad Re(s) > 1. \end{align*} Because $0 \leq \{x\} \leq 1$, the last integral converges and is holomorphic on $Re(s) > 0$. But that means the full equation is meromorphic on $Re(s) > 0$, and thus provides an analytic continuation of $\zeta(s)$ on the half plane $Re(s) > 0$. The $s/(s-1)$ term gives a simple pole at $s = 1$ with residue $1$.
$\Re(s) >1 $ and $a_n=1$ $$\sum_n a_n n^{-s}= \sum_n a_n s \int_n^\infty x^{-s-1}dx=s \int_1^\infty (\sum_{n \le x} a_n)x^{-s-1}dx$$ so that $$\zeta(s)=s\int_1^\infty [x]x^{-s-1}dx=s\int_1^\infty ([x]-x)x^{-s-1}dx+ \frac{s}{s-1}$$ The latter expression is the analytic continuation to $\Re(s) > 0,s\ne 1$.
Another derivation makes use of the derivative property of the Hurwitz zeta function.
We may rewrite $$\zeta(s)-1-\frac{1}{s-1}=[x\zeta(s,x+1)]_0^1-\int_0^1\zeta(s,x+1)dx,$$ then by IBP this equals $$\int_0^1x\frac{\partial}{\partial x}\zeta(s,x+1)dx=-s\int_0^1x\zeta(s+1,x+1)dx=-s\int_0^1x\sum_{k=1}^\infty\frac{1}{(k+x)^{s+1}}dx.$$ Since $k\geq 1$ and $x\in(0,1)$ this is equivalent to $$-s\int_1^\infty\frac{x-[x]}{x^{s+1}}dx.$$
You can take Abel's partial summation formula:
$$ \sum_{1 \leq n \leq t} a_n f(n) = C(t) f(t) - \int_1^s C(x) f'(x) dx $$
where $C(t) = \sum_{1 \leq n \leq t} a_n$ and $f(t)$ is a complex valued function defined for $t \in \mathbb{R}^+$. Which is easily derived here. Then you choose $a_n=1$ and $f(t) = t^{-s}$:
$$ \sum_{1 \leq n \leq t} \frac{1}{n^s} = \frac{\lfloor t \rfloor}{t^s} + s \int_1^t \frac{\lfloor x \rfloor}{x^{s+1}} dx $$
For $\Re (s) > 1$ we can take the limit $t \rightarrow \infty$ resulting in:
$$ \zeta (s) = \sum_{n= 1}^\infty \frac{1}{n^s} = s \int_1^\infty \frac{\lfloor x \rfloor}{x^{s+1}} dx $$
