Integral representation of $\zeta(s)$ in terms of the fractional part of $x$

Question

I derived an integral representation of the Riemann zeta function, which is: $$\zeta(s)=\frac{s}{s-1}-s\int_{1}^{\infty}\{x\}x^{-s-1}dx$$ where $\{x\}$ is the fractional part of $x$. Please verify my proof. Here it is:

$$\zeta(s)=\sum_{n=1}^{\infty}n\left(\frac{1}{n^s}-\frac{1}{(n+1)^s}\right)=s\sum_{n\geq1}n\int_{n}^{n+1}x^{-s-1}dx$$ We know that $x=[x]+\{x\}$ where $[x]$ is the integral part of $x$. Since $[x]=n$ for all $x\in[n,n+1)$, we have $$\zeta(s)=s\sum_{n\geq1}n\int_{n}^{n+1}x^{-s-1}dx=s\int_{1}^{\infty}\{x\}x^{-s-1}dx$$ By writing $[x]=x-\{x\}$ and simplifying, we get: $$\zeta(s)=\frac{s}{s-1}-s\int_{1}^{\infty}\{x\}x^{-s-1}dx,\quad\mathrm{Re}(s)>1$$

Did I make any mistake in my proof? If yes, please tell it in the answers.

Answer 1

More directly:

$$I=\int_{1}^{\infty} (x-[x]) x^{-1-s} dx $$

$$ = \int_{1}^{\infty} x^{-s} dx- \int_{1}^{\infty} [x] x^{-1-s} dx $$

$$ =-\frac{1}{1-s}-\sum_{k=1}^{\infty} k \int_{k}^{k+1} x^{-1-s} dx$$ $$\implies I=-\frac{1}{1-s}-\frac{1}{s}\sum_{k=1}^{\infty} k [(k+1)^{-s}-k^{-s}]$$ $$I=-\frac{1}{1-s}-\frac{1}{s}\sum_{k=1}^{\infty} [(k+1)^{1-s}- (k+1)^{-s}-k^{1-s}].$$ $$I=-\frac{1}{1-s}-\frac{1}{s}[-1-\sum_{k=1}^{\infty} (k+1)^{-s}.$$ $$\implies I=-\frac{1}{1-s}-\frac{1}{s}\zeta (s), s>1.$$ From here the result follows.