Functions $f(x)=\lfloor x\rfloor$ and $g(s)=\frac{\zeta (s)}{s}$ are related by Mellin inversion theorem, for $c>1$, $\Re(s)>1$.


$$\tag{1.1}f(x)=\lfloor x\rfloor =\frac{1}{2 \pi i} \int_{c-i \infty }^{c+i \infty } \frac{\zeta (s)}{s} x^s \, ds$$

$$\tag{1.2}g(s)=\frac{\zeta (s)}{s}=\int_0^{\infty } \lfloor x\rfloor x^{-s-1} \, dx$$

Functions $f(x)=x-\lfloor x\rfloor$ and $g(s)=\frac{-\zeta (s)}{s}$ are related by Mellin inversion theorem, for $0<c<1$, $0<\Re(s)<1$. $$\mathcal{M}_x(f(x))(s)=\mathcal{M}_s^{-1}(g(s))(x)$$

$$\tag{2.1}f(x)=x-\lfloor x\rfloor =\frac{1}{2 \pi i} \int_{c-i \infty }^{c+i \infty } \frac{-\zeta (s)}{s} x^s \, ds$$

$$\tag{2.2}g(s)=\frac{-\zeta (s)}{s}=\int_0^{\infty } (x-\lfloor x\rfloor) x^{-s-1} \, dx$$

Relations in $(1.1)$ and $(1.2)$ can be derived using Abel's summation formula, as is described on that Wikipedia page.

How relations $(2.1)$ and $(2.2)$ can be derived?

It is interesting that $(1.1)$ and $(1.2)$ are valid for $\Re(s)>1$ while $(2.1)$ and $(2.2)$ for $0<\Re(s)<1$.

Are there any other similar formulas that involve $\lfloor x\rfloor$, $\zeta (s)$ and Mellin inversion theorem that are valid at least in some portion of $\Re(s)<0$.

  • $\begingroup$ For $\Re(s)\in (-1,0)$ it is $\zeta(s)=s\int_0^\infty (\lfloor x\rfloor-x+1/2) x^{-s-1}dx$. The Melin inversion "integral" doesn't converge and needs to be interpreted in the sense of distributions. $\endgroup$
    – reuns
    May 6 at 22:36
  • $\begingroup$ How did you get this formula? $\endgroup$ May 6 at 22:43
  • $\begingroup$ The proof is the same as for $\Re(s)\in (0,1)$. Substract $\frac12 1_{x <1}$ and use analytic continuation. $\endgroup$
    – reuns
    May 6 at 22:46

1 Answer 1


For $\Re(s)\in (-1,0)$ we have $$\zeta(s)=s\int_0^\infty (\lfloor x\rfloor-x+1/2) x^{-s-1}dx$$ whence for $\sigma\in (-1,0)$ $$\lfloor x\rfloor-x+1/2 = \lim_{k \to \infty}\frac1{2i\pi} \int_{(\sigma)} \frac{\zeta(s)}{s} x^s e^{s^2/k} ds\tag{1}$$ By the Cauchy integral theorem and the polynomial growth of $\zeta(s)$ on vertical strips, the RHS integrals stay the same for all $\sigma < 0$, so that $(1)$ stays valid for all $\sigma < 0$.

That is to say we have the inverse Fourier/Mellin transform in the sense of distributions $$\forall \sigma < 0, \qquad \mathcal{M}^{-1}\left[\frac{\zeta(s)}s\right]_{\Re(s)=\sigma}= \lfloor x\rfloor-x+1/2$$

  • $\begingroup$ So is the integral in the inverse transform in $(1.1)$ and $(2.1)$ only convergent for $\Re(s)\in (-1,\infty)$? Why we can not go beyond $\Re(s)=-1$? $\endgroup$ May 7 at 12:52
  • $\begingroup$ Convergent in what sense? $\zeta(s)/s$ is not $L^1$ on any vertical line. The functional equation and Stirling formula gives that it is not $L^2$ for $\sigma <0$. $\endgroup$
    – reuns
    May 10 at 19:27
  • $\begingroup$ Is there some other function that its Mellin transform is convergent even for $\Re(s)<-1$ and that can be used to create analytic continuation of zeta function in that region although in a different form than $\zeta(s)/s$. $\endgroup$ May 10 at 19:47
  • $\begingroup$ $\zeta(s)/(s(s+1)(s+2)\ldots (s+n))$ $\endgroup$
    – reuns
    May 10 at 19:48
  • $\begingroup$ Does it still has any sense even in limit as $n\to\infty$ to create analytic continuation in whole complex plane? $\endgroup$ May 11 at 11:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.