In an analytic number theory textbook we derived the formula $$\zeta(s)=\frac{s}{s-1}-s\int_1^\infty \{x\}x^{-s-1}\mathrm{dx}$$
and the textbook says, that the integral on the right side converges uniformly on any compact subset of $\{s\mid \mathrm{Re}(s)>0\}$ and is a holomorphic function. But there is no reasoning why any of this should be true, so it can't be that hard..
That the Integral converges uniformyl means what $$f_n(s)=\int_1^n\{x\}x^{-s-1}\mathrm{dx}$$ is converging uniformly as a function in $s$. And it is no problem to see that this is true on any compact sets on the right half-plane. If I can prove now that each of these $f_n(s)$ are holomorphic function, combined with the compact convergence this implies that $$\int_1^\infty \{x\}x^{-s-1}\mathrm{dx}$$ is a holomorphic function. However I was not able to prove that these $f_n(s)$ are holomorphic.
After I failed to apply any theorems on this, I tried the definition $$\lim_{s\to s_0}\frac{f_n(s)-f_n(s_0)}{s-s_0}= \lim_{s\to s_0}\int_1^n\{x\} \frac{x^{-s-1}-x^{-s_0-1}}{s-s_0}\mathrm{dx}$$
At this point if I were able to interchange the Integral and the limit, I would get
$$\lim_{s\to s_0}\frac{f_n(s)-f_n(s_0)}{s-s_0} =\int_1^n\{x\} \lim_{s\to s_0}\frac{x^{-s-1}-x^{-s_0-1}}{s-s_0}\mathrm{dx} = \int_1^n\{x\} (-x^{-s_0-1}\ln(x))\mathrm{dx}$$
and since $0\leq\{x\}\leq 1$ this last integral exists what implies that the limit on the left side exists what finishes the prove.
However, I am unable to acutally prove that this interchange of the limit is valid.
I would appreciate any help on this. Or maybe anyone has an other "better" argument why this integral in the first equation is a holomorphic function, maybe I dont even need all this reasoning.