Consider the set $$U=\{s\in \mathbb{C}:\ \Re(s)>0\}\setminus \left\{1+(\ln(2))^{-1}2\pi ik:\ k\in \mathbb{Z}\right\}$$ Clearly $U$ is open and connected, and so $\zeta$ has a unique analytic continuation from $$V=\{s\in \mathbb{C}:\ \Re(s)>1\}$$ to the set $U$.
Let $\{x\}=x-[x]$ be the fractional part of $x>0$. It is well-known that $$\zeta(s)=\underbrace{\frac{s}{s-1}-s\int_1^\infty \frac{\{x\}}{x^{s+1}}dx}_{f(s)}\qquad \forall s\in V$$ $$\zeta(s)=\frac{1}{1-2^{1-s}}\underbrace{\sum_{n\in \mathbb{N}}\frac{(-1)^{n-1}}{n^s}}_{\eta(s)}\qquad \forall s\in V$$ It's easy to see that both $f$ and $\frac{1}{1-2^{1-s}}\eta$ are analytic on $U,V$. Hence, by uniqueness of analytic continuation to connected open sets, $f(s)=\frac{1}{1-2^{1-s}}\eta(s)$ for all $s\in U$. My question is, how do we prove this by direct computation?
My attempt:
Since the integral in $f$ is convergent, we have that $$\begin{aligned} &\ s\int_0^\infty\frac{x-[x]}{x^{s+1}}dx \\ =&\ s\sum_{n\in \mathbb{N}}\int_n^{n+1}\frac{x-[x]}{x^{s+1}}dx \\ =&\ \sum_{n\in \mathbb{N}}s\int_n^{n+1}\frac{x-n}{x^{s+1}}dx \\ =&\ \sum_{n\in \mathbb{N}}\left[s\int_n^{n+1}\frac{1}{x^s}dx+\int_n^{n+1}n\left(\frac{-s}{x^{s+1}}\right)dx\right] \\ =&\ \sum_{n\in \mathbb{N}}\left[\left.\frac{s}{1-s}\frac{1}{x^{s-1}}\right|_{x=n}^{x=n+1}+\left.\frac{n}{x^s}\right|_{x=n}^{x=n+1}\right] \\ =&\ \sum_{n\in \mathbb{N}}\left[\frac{s}{1-s}\left(\frac{1}{(n+1)^{s-1}}-\frac{1}{n^{s-1}}\right)+\left(\frac{n}{(n+1)^s}-\frac{n}{n^s}\right)\right] \\ =&\ \sum_{n\in \mathbb{N}}\left[\frac{s}{1-s}\left(\frac{1}{(n+1)^{s-1}}-\frac{1}{n^{s-1}}\right)+\left(\frac{1}{(n+1)^{s-1}}-\frac{1}{(n+1)^s}-\frac{1}{n^{s-1}}\right)\right] \\ =&\ \sum_{n\in \mathbb{N}}\left[\left(\frac{s}{1-s}+1\right)\left(\frac{1}{(n+1)^{s-1}}-\frac{1}{n^{s-1}}\right)-\frac{1}{(n+1)^s}\right] \\ =&\ \sum_{n\in \mathbb{N}}\left[\frac{1}{s-1}\left(\frac{1}{n^{s-1}}-\frac{1}{(n+1)^{s-1}}\right)-\frac{1}{(n+1)^s}\right] \\ =&\ 1+\sum_{n\in \mathbb{N}}\left[\frac{1}{s-1}\left(\frac{1}{n^{s-1}}-\frac{1}{(n+1)^{s-1}}\right)-\frac{1}{n^s}\right] \\ =&\ \frac{s}{s-1}-\frac{1}{s-1}+\sum_{n\in \mathbb{N}}\left[\frac{1}{s-1}\left(\frac{1}{n^{s-1}}-\frac{1}{(n+1)^{s-1}}\right)-\frac{1}{n^s}\right] \end{aligned}$$ Therefore, $$f(s)=\frac{1}{s-1}+\sum_{n\in \mathbb{N}}\left[\frac{1}{n^s}-\frac{1}{s-1}\left(\frac{1}{n^{s-1}}-\frac{1}{(n+1)^{s-1}}\right)\right]$$ Now it suffices to show that for all $s\in U$, $$\frac{1}{s-1}+\sum_{n\in \mathbb{N}}\left[\frac{1}{n^s}-\frac{1}{s-1}\left(\frac{1}{n^{s-1}}-\frac{1}{(n+1)^{s-1}}\right)\right]=\frac{1}{1-2^{1-s}}\sum_{n\in \mathbb{N}}\frac{(-1)^{n-1}}{n^s}$$
But I haven't found a way to do it so far. Can anyone offer any hint or help? Thanks!
The proof I can think to, for $\Re(s)\in (0,1)$: $$\eta(s)= \sum_{n\ge 1} (-1)^{n+1}s\int_n^\infty x^{-s-1}dx=s\int_1^\infty (\sum_{n\le x}(-1)^{n+1})x^{-s-1}dx$$
$$\frac{s}{s-1}-s\int_1^\infty \{x\}x^{-s-1}dx=-s\int_0^\infty \{x\}x^{-s-1}dx$$ $$2^{1-s} s\int_0^\infty \{x\}x^{-s-1}dx= s\int_0^\infty 2\{x/2\}x^{-s-1}dx$$
$$\sum_{n\le x}(-1)^{n+1}=2\{x/2\}-\{x\}$$