Analytic continuation of $\zeta$

Question

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 ?

Answer 1

$f(\{t\})$ is a weird way to say it is $1$ periodic.

If $g(x)$ is $T$-periodic, then for any $a > 0$, the Mellin transform of $g(x) 1_{x > a}$ is meromorphic on the whole complex plane

• Let $g_0(x) = g(x)$ and consider the sequence of functions $$g_{n+1}(x) = \int_a^x (g_n(t) - \overline{g}_n)dt,\qquad\qquad \overline{g}_n = \frac{1}{T}\int_a^{a+T}g_n(x)dx$$ note that all those functions $g_n(x)$ are $T$-periodic.

• Look at the Mellin transforms of $g_n(x) x^{-n}1_{x > a}$ : $$G_n(s) = \int_a^\infty g_n(x)x^{-s-n-1}dx$$ Since $g_n(x) = \mathcal{O}(1)$, it converges and is analytic on $Re(s) > -n$

• Subtracting the mean value and integrating by parts : $$\begin{eqnarray}G_n(s) - \frac{\overline{g}_n a^{-s-n}}{s+n}&=& \int_a^\infty (g_n(x)-\overline{g}_n)x^{-s-n-1}dx \\ &=& (s+n+1)\int_a^\infty g_{n+1}(x)x^{-s-n-2}dx \\ &=& (s+n+1)G_{n+1}(s)\end{eqnarray}$$ and hence $G_{n+1}(s)$ is meromorphic on $Re(s) > \sigma \implies$ $G_{n}(s)$ is meromorphic on $Re(s) > \sigma $ too.

Overall, $G_0(s)$ is meromorphic on $Re(s) > -n$ for every $n$, i.e. it is meromorphic on the whole complex plane