I want to show $\zeta(s)=\frac{1}{s-1}+\frac{1}{2}+ (s+1)s \int_{1}^{\infty} \frac{ G(x)}{x^{s+2}} dx.$ With $G(x)= \int_{0}^{x} (\{u\} -1/2) du$ for Re(s)>-1
By the Mac Laurin sommation formula, I have $\zeta(s)=\frac{1}{s-1}+\frac{1}{2}- s \int_{1}^{\infty} \frac{ (\{x\}-1/2)}{x^{s+1}} dx.$
Here is what I have tried so far: $- s \int_{1}^{\infty} \frac{ (\{x\}-1/2)}{x^{s+1}} dx=- s \int_{1}^{\infty} \frac{G^{(1)}(x)}{x^{s+1}} dx+s \int_{1}^{\infty}\frac{dx}{2 x^{s+1}} $= $ - s \int_{1}^{\infty} \frac{G^{(1)}(x)}{x^{s+1}} dx+ \frac{1}{2}$. I then integrated by part.
$ - s \int_{1}^{\infty} \frac{G^{(1)}(x)}{x^{s+1}} dx+ \frac{1}{2}=-s \left( \frac{G(x)}{x^{s+1}} |_1^{\infty}+ (s+1)\int_{1}^{\infty} \frac{G(x) dx}{x^{s+2}} \right)$
I am stuck after this. Can anyone help?
You want to show the Abel summation formula, for $Re(s) > 1$ : $$\zeta(s) = \sum_{n=1}^\infty n^{-s} = \int_{1-\epsilon}^\infty (\sum_{n=1}^\infty \delta(x-n)) x^{-s} dx = \int_1^\infty \lfloor x \rfloor s x^{-s-1}dx$$ That is just an integration by parts with a distribution $\sum_{n=1}^\infty \delta(x-n) = \frac{d}{dx} \lfloor x \rfloor$. If you don't like distributions, no problem, write that $\int_n^{n+1} \lfloor x \rfloor s x^{-s-1} dx = n (n^{-s}-(n+1)^{-s}) = n^{1-s}-(n+1)^{1-s}+ (n+1)^{-s}$ so that $\int_1^N \lfloor x \rfloor s x^{-s-1} dx = \sum_{n=1}^{N-1} (n^{1-s}-(n+1)^{1-s}+ (n+1)^{-s}) = -N^{1-s}+\sum_{n=1}^N n^{-s}$ and the result follows.
Once this is done, you get $\zeta(s) = \frac{s}{s-1}-s\int_1^\infty \{x \} x^{-s-1}dx = 1+\frac{1}{s-1}-s\int_1^\infty \{x \} x^{-s-1}dx$ $= 1+\frac{1}{s-1}-1/2-s\int_1^\infty (\{x \}-1/2) x^{-s-1}dx$ and integrating by parts you get your formula (up to a minus sign)
$$\zeta(s) = \frac{1}{s-1}+1/2-s(s+1)\int_1^\infty (\int_0^x(\{t \}-1/2)dt) x^{-s-2}dx$$
Finally, since $\int_0^x(\{t \}-1/2)dt = \mathcal{O}(1)$ it converges and it is analytic for $Re(s)> -1$ and hence it is the analytic continuation of $\zeta(s)$.