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

$$\mathcal{M}_x(f(x))(s)=\mathcal{M}_s^{-1}(g(s))(x)$$

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

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

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 $\Re(s)<0$.

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  • $\begingroup$ 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

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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 $$\forall \sigma < 0, \qquad \mathcal{M}^{-1}\left[\frac{\zeta(s)}s\right]_{\Re(s)=\sigma}= \lfloor x\rfloor-x+1/2$$

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  • $\begingroup$ 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

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