Analytic continuation of $zeta$

It is known that$$\frac{1}{s-1} - \frac{\zeta(s)}{s} = \int_1^\infty \frac{x - [x]}{x^{s+1}} dx$$ for $Re(s) > 0.$

The last integration can provide the analytic continuation for $\zeta.$

I encouter one generalization of the situation :

Let $f$ be a continuous function on $[0,1]$. If $$F(s) = s \int_1^\infty f(\{t\})t^{-s-1} dt, s = \sigma + it, \sigma >0$$ then $F(s)$ has the analytic continuation to the entire $\mathbb{C},$ where$ \{a\} = a - [a].$

This general case is for Riemann zeta if $f(x) = x.$

I want to prove this theorem, and it states just set $$A = \int_0^1 f(u) du , f_1(u) = \int_0^u (f(\{v\})-A) dv.$$
Then induction and integration by parts will yields the result, but I do not understand. At least, $F$ has different integration $(\int_1^\infty = \sum_{n=1}^\infty \int_n^{n+1})$ from $A$ and $f_1.$

It sounds like it is a well-known theorem, but most of the time I notice that $\zeta$ analytic continuation method is done using gamma function $\Gamma$ or Bernoulli polynomial.

I cannot find a reference with good detail to follow the steps in the proof.

Any help please ?