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 ?

  • $\begingroup$ Thank you. I ask there already, but no response. Is this site valid to ask a question related to when I read a textbook and stuck on the proof claim in the book ? If not, I am sorry. $\endgroup$
    – user117375
    Nov 2, 2016 at 9:43
  • $\begingroup$ I didn't find your question on Math.SE, is it called like this one? This site is actually for research-level mathematics. $\endgroup$ Nov 2, 2016 at 9:50
  • 1
    $\begingroup$ Usually you should wait longer for an answer there before reasking here. And you should post links in both places for people to know. $\endgroup$
    – abatkai
    Nov 2, 2016 at 10:03
  • $\begingroup$ okay, sorry about that. $\endgroup$
    – user117375
    Nov 2, 2016 at 10:31
  • $\begingroup$ And this theorem is useful for the Dirichlet L-functions, or the Hurwitz zeta, or for a mix of those, for example $\sum_{n=1}^\infty e^{i n x} n^{-s}$ $\endgroup$
    – reuns
    Nov 2, 2016 at 16:25

1 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

  • $\begingroup$ Note that $\zeta(s) = \sum_{n=1}^\infty n^{-s}$ is already the Mellin transform of a periodic distribution : $\zeta(s)= \int_{1-\epsilon}^\infty \text{I}\Pi(x) x^{-s}dx$ where $\text{I}\Pi(x) = \sum_{n=-\infty}^\infty \delta(x-n)$ is the Dirac comb $\endgroup$
    – reuns
    Nov 2, 2016 at 16:31
  • $\begingroup$ Thank you very much for help. I have never heard about Merlin transformation. Can I think of it as a kind of special integration ? I will try to deduce it, but I wonder if it has a simpler longer, but require not much background knowledge. Or when the book says set $A = ...$ and do induction, it means this whole things. $\endgroup$
    – user117375
    Nov 2, 2016 at 17:14
  • $\begingroup$ @user117375 Here you don't need to look at the properties of the Mellin transform, except that $\int_a^\infty g(x) x^{-s-1}dx$ is analytic for $Re(s )> 0$ if $|g(x)| < C$. And $\int_1^\infty h(x) x^{-s-1}dx = \int_0^\infty h(e^t)e^{-st}dt$ so the Mellin transform of $h(x)1_{x > 1}$ is the Laplace transform of $h(e^t)1_{t > 0}$ $\endgroup$
    – reuns
    Nov 2, 2016 at 17:19
  • $\begingroup$ Of course, since Dirichlet series are Mellin transforms, you need to study it in detail for understanding the theory of $\zeta(s)$ $\endgroup$
    – reuns
    Nov 2, 2016 at 17:20
  • $\begingroup$ okay, thank you very much. I will try to understand that. I far more complicated than I first expect. $\endgroup$
    – user117375
    Nov 2, 2016 at 17:21

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