# Formulas of Mellin inversion theorem that involve Riemann zeta function $\zeta (s)$ and floor function $\lfloor x\rfloor$

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 $$0, $$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 $$\Re(s)<0$$.

• 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. May 6 at 22:36
• How did you get this formula? May 6 at 22:43
• The proof is the same as for $\Re(s)\in (0,1)$. Substract $\frac12 1_{x <1}$ and use analytic continuation. May 6 at 22:46

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 $$\forall \sigma < 0, \qquad \mathcal{M}^{-1}\left[\frac{\zeta(s)}s\right]_{\Re(s)=\sigma}= \lfloor x\rfloor-x+1/2$$
• 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 $\Re(s)=-1$? May 7 at 12:52
• 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 $\sigma <0$. May 10 at 19:27
• 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 $\zeta(s)/s$. May 10 at 19:47
• $\zeta(s)/(s(s+1)(s+2)\ldots (s+n))$ May 10 at 19:48
• Does it still has any sense even in limit as $n\to\infty$ to create analytic continuation in whole complex plane? May 11 at 11:05