How to get the floor function as a Mellin inverse of the Hadamard product of the Riemann zeta function?

Question

The floor function is given - by Perron's formula - as a Mellin inverse of the zeta function. namely : $$\left \lfloor x \right \rfloor=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\zeta(s)\frac{x^{s}}{s}ds\;\;\;(c>1)$$ This is easily proven using the Dirichlet series rep. of the zeta function : $\zeta(s)=\sum_{n=1}^{\infty}n^{-s}$. i was wondering if one can obtain the same result using the Hadamard product rep. : $$\zeta(s)=\pi^{s/2}\frac{\Pi_{\rho}\left(1-\frac{s}{\rho}\right)}{2(s-1)\Gamma\left(1+\frac{s}{2}\right)}$$

Answer 1

Well, you can, somewhat indirectly. Riemann originally developed the intermediary step function $J(x)$, with steps of $1/j$ at each prime power $p^j$, to establish the prime number theorem. The function $J(x)$ is directly derived from $\ln \zeta(s)$, where he decomposed the Hadamard product into a sum under the natural logarithm (see "Riemann's Zeta Function," H.M. Edwards; I have the Dover edition, 2001).

In that context, the whole number staircase function (sometimes called the integer floor function, or natural number counting function) may be expressed as, $$ N(w) = \int_{r=1^-}^{w} \exp^*(dJ(e^v))[\ln r] dr, $$ where $\ln r$ is the independent variable of the convolution exponential, and the argument of the c.e. is just Riemann's $J(x)$: $$ dJ(e^u) = \left( \frac{1}{u} - \sum_{\rho} \frac{e^{(\rho-1)u}}{u} - \frac{1}{e^{u}(e^{2u}-1)u} \right), $$ where the $\rho$ are the non-trivial zeros of the Riemann zeta function.

To show it: $$ N(w) = \frac{1}{2 \pi i} \int_{x-i\infty}^{x+i\infty} \zeta(z) \frac{w^z}{z} dz \Rightarrow N(e^u) = \frac{1}{2 \pi i} \int_{x-i\infty}^{x+i\infty} \zeta(z) \frac{e^{uz}}{z} dz. $$ We'll drop the $1/(2\pi i)$ factor in some of what follows. Then rewrite, $$ N(e^u) = \int_{x-i\infty}^{x+i\infty} e^{\ln \zeta(z)} \frac{e^{uz}}{z} dz = \int_{x-i\infty}^{x+i\infty} \sum_{k=0}^{\infty} \frac{(\ln \zeta(z))^k}{k!} \frac{e^{uz}}{z} dz. $$ If we can integrate termwise (*), we've $$ = \sum_{k=0}^{\infty} \int_{x-i\infty}^{x+i\infty} \frac{(\ln \zeta(z))^k}{k!} \frac{e^{uz}}{z} dz. $$ Note, each integral is just Perron's formula applied to the Dirichlet series, $$ D^k = (\ln \zeta(z))^k = \left( \sum_i p_i^{-z} + (1/2)p_i^{-2z} + (1/3)p_i^{-3z} + ... \right)^k $$ where $i$ indexes the primes. We can take the derivative of the Perron step function to obtain a distribution of point masses of weight $c/(j_1\cdots j_k)$ at each $p_{i_1}^{j_1} \cdots p_{i_k}^{j_k}$. We can apply the Laplace transform convolution formula (in the sense of distributions): $$ \int_{0^-}^{u} \int_{x-i\infty}^{x+i\infty} \frac{D^k}{k!} e^{tz} dz dt = \int_{0^-}^{u} (1/k!) (\mathcal{L}^{-1}(D))^{*k} [t] dt, $$ where $t$ is the post-convolution independent variable. The whole series can then be expressed as a convolution exponential, $$ N(e^u) = \int_{0^-}^{u} \exp^*(\mathcal{L}^{-1}(D))[t] dt, $$ where the $u$-integral swap with the exponential series sum is justified by noting $\exp^*(.)$ is just a point measure on $[0,\infty)$, with finitely many delta masses in $[0,u]$.

The argument of the convolution exponential has a common form, $$ \frac{1}{2\pi i} \int_{x-i\infty}^{x+i\infty} \ln(\zeta(z)) e^{uz} dz = dJ(e^u)e^u, $$ where $J(x)$ is from Edwards.

The outer $e^u$ term can be passed through the convolution exponential, and we can change variables back, $u=\ln w$, and let $t=\ln r$ for the variable of integration, $$ N(w) = \int_{t=0^-}^{\ln w} \exp^*(e^v dJ(e^v))[t] dt = \int_{r=1^-}^{w} r \exp^*(dJ(e^v))[\ln r] \frac{dr}{r}. $$

The convolution exponential seems a bit underserved in the literature. I suppose it's assumed things can be done in the frequency domain. It has some of the expected properties, like for functions $f$ and $g$ under convergence conditions, $\exp^*(f+g) = \exp^*(f)*\exp^*(g)$, and if $0<f<g$, then $\exp^*(f)<\exp^*(g)$.

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(*) We need to pass the integral inside the sum in, $$ N(e^u) = \int_{x-i\infty}^{x+i\infty} \sum_{k=0}^{\infty} \frac{(\ln \zeta(z))^k}{k!} \frac{e^{uz}}{z} dz. $$ The justification is indifferent to finitely many terms of the sum. From the Dirichlet sum, $D$, $(\ln \zeta(z))^k$ has the lowest term in its Dirichlet series as $2^{-kz}$ ($2$ being the lowest prime). Take $K = \mbox{ceil}(e^u/2)+1$. We'll show the integral sum swap works for $$ \int_{x-i\infty}^{x+i\infty} \sum_{k=K}^{\infty} \frac{(\ln \zeta(z))^k}{k!} \frac{e^{uz}}{z} dz $$

Note by Perron's formula, we can integrate the relevant Dirichlet series termwise to give, $$ \int_{x-i\infty}^{x+i\infty} \frac{(\ln \zeta(z))^k}{k!} \frac{e^{uz}}{z} dz = 0, $$ for all $k\ge K$. To show the swap, we are then concerned with the rate at which the partial sum of the improper-integrals converge to $0$.

Since $\zeta(z)$ is bounded away from $0$ and $\infty$ along the integration contour ($x>1$), so is $\ln \zeta(z)$. Since on finite vertical contours $[x-ih,x+ih]$ the partial sums of the functions converge uniformly, we can pass this "middle" portion of the contour inside the infinite sum--we need only be concerned with the "tails" of the contours: $$ \int_{x+ih}^{x+i\infty} \frac{(\ln \zeta(z))^k}{k!} \frac{e^{wz}}{z} dz, $$ (likewise for $\rightarrow x-i\infty$). These actually behave nicely.

In fact, from Edwards (Section 3.5), $$ \left| \int_{x+ih}^{x+i\infty} \left( \frac{w}{n} \right)^z \frac{dz}{z} \right| \le c \frac{1}{n^x h} $$ We apply this termwise to the terms in the $(\ln \zeta(z))^k$ Dirichlet series (again, by standard Perron formula results, this swap is OK to do), treating $e^u$ as $w$ to get $$ \left| \int_{x+ih}^{x+i\infty} \frac{(\ln \zeta(z))^k}{k!} \frac{e^{uz}}{z} dz \right| \le \frac{c_2^k}{h k!}, $$ where, $$ \left| \sum_i p_i^{-z} + (1/2)p_i^{-2z} + (1/3)p_i^{-3z} + ... \right| \le \sum_i p_i^{-x} + (1/2)p_i^{-2x} + (1/3)p_i^{-3x} + ... \equiv c_2 $$ since $x>1$ and everything is absolutely convergent. Since the partial sum of the tails converge $\mathcal{O}(1/h$), this justifies the swap.