Question

Highly grateful for your help/steers on the following question (at the end):

Take the infinite product:

$$\displaystyle T(s) = \prod _{n=2}^{\infty } \left( \dfrac{{n}^{s}} {{n}^{s}-1}\right)$$

for $\Re(s) > 1$ it is equal to:

$$\displaystyle \prod _{primes}^{\infty } \left( \dfrac{{p}^{s}} {{p}^{s}-1}\right) * \prod _{composites}^{\infty } \left( \dfrac{{c}^{s}} {{c}^{s}-1}\right)$$

I.e. the Euler-product (equal to $\zeta(s)$) multiplied by its composite "equivalent" ( excluding 1 since that is a bit of a strange composite anyway).

Why my interest? I wanted to learn more about the composite infinite product (and see if it had a 'zeta' like version). Soon became clear to me that the only way to learn more about this product, is to concentrate on $T(s)$ and then divide it by $\zeta(s)$.

I searched the web but there is hardly anything known about $T(s)$. F.i. Wolfram math only shows (formula 20) two different solutions (note: both need to be raised to $^{-1}$ to get $T(s)$ !) for odd and even integers and by reading through some arxiv math pre-prints the best I could find was a single, but still integer only formula that is:

$$\prod _{k=1}^{s-1}\Gamma \left( 2- {{\rm e}} ^{{\frac {2 i \pi k}{s}}} \right), ( \Re(s) > 1, s \in \mathbb{N})$$

I then decided to explore ways to extend the domain for $s$ and derived the following formula:

$$\displaystyle \ln \left( T\left( s \right) \right) = \ln \prod_{n=2}^{\infty } \left( \left( -1+{n}^{-s} \right) ^{-1} \right) = \sum_{n=2}^{\infty } \ln \left( \left( 1-{n}^{-s} \right) ^{-1} \right)$$

$$\displaystyle = \sum_{m=1}^\infty \sum_{n=2}^{\infty } \frac{1}{mn^{ms}} = \sum_{n=1}^\infty \frac{\zeta(n s)-1}{n}$$

And this brings us to:

$$ T(s)={\rm e}^{\left( \displaystyle \sum_{n=1}^\infty \frac{\zeta(n s)-1}{n} \right)}$$

Yep, there's always a $\zeta(s)$ hiding around the corner somewhere...

So, let's see what the plot looks like for $s>0$ ($T(s)$ diverges for $s<0$).

$T(s)=\displaystyle {\rm e}^{\left( \displaystyle \sum_{n=1}^\infty \frac{\zeta(n s)-1}{n} \right)} \text{ blue}$

$\displaystyle \zeta(s) = \prod _{primes}^{\infty } \left( \dfrac{{p}^{s}} {{p}^{s}-1}\right) \text{purple}$

$\displaystyle \frac{T(s)}{\zeta(s)} = \prod _{composites}^{\infty } \left( \dfrac{{c}^{s}} {{c}^{s}-1} \right) \text{ brownish}$

graph

For $s>1$ I could numerically solve the following equation:

$$\displaystyle \prod _{primes}^{\infty } \left( \dfrac{{p}^{s}} {{p}^{s}-1}\right) = \prod _{composites}^{\infty } \left( \dfrac{{c}^{s}} {{c}^{s}-1} \right)$$

giving this interesting number $s = 1.397737620...$ (there is only one for $\Re(s) > 1$ )

I obviously took a deep dive with this number on Google and Plouffe's inverter, but have not found anything 'beautiful' or related to other constants as yet.

Then the domain $0 < s < 1$. It is easy to see in the graph that $T(s)$, and therefore also $\dfrac{T(s)}{\zeta(s)}$, have 'trivial' poles for $s= \dfrac{1}{k}, k \in \mathbb{N}$ that are induced by the fact that for each $s= \dfrac{1}{k}$ there always is a $n s = 1$ that makes at least one term in the infinite sum equal to the pole $\zeta(1)$ (hence the whole sum turns into a pole).

But I'm actually mostly intrigued by what happens under the x-axis and especially where:

$$\zeta(s) = \dfrac{T(s)}{\zeta(s)}$$ or

$$|\zeta(s)| = {\rm e}^{\displaystyle \left(\frac12 \sum_{n=1}^\infty \frac{\zeta(n s)-1}{n} \right)}$$.

If I have done my analysis correctly, this result would imply that there are an infinite number of values for $0 < s < 1$, where the (analytically continued) infinite products of primes and composites are equal (since $\zeta(s)$ remains negative between $0 < s < 1$ and there are an infinite number of poles separating the intersection points). And that would imply/reveal an infinite amount of tiny bits of information about how the primes 'grow like weed between the composites'.

Of course I checked $T(s)$ also for $s \in \mathbb{C}$, however, any graph I've produced sofar for $s=a+bi$ of $T(s)={\rm e}^{\left(\displaystyle \sum_{n=1}^\infty \frac{\zeta(n s)-1}{n} \right)}$ did not reveal any non-trivial zeroes (nope, not even at $a=\frac12$...), although the curves do seem to be trending towards a large number of very chaotically distributed zeroes when $a \rightarrow 0$.

So, apologies for the relatively long intro to my question:

Since $\zeta(s)$ has been analytically continued throughout the entire complex domain, is it allowed to also analytically continue the division of $\dfrac{T(s)}{\zeta(s)}$ into the domain $s<1$? Or do the nominator and denominator each require an individual continuation and does the concept of division get 'lost in continuation'?

Answer 1

I think that T is meromorphic on $\mathbb{C}$ just like $\zeta$, with a single pole at $s=0$. The ratio should be fine everywhere except at $s=1$, the negative integers, and the critical strip (or line, on the RH).

Answer 2

Just to elaborate a bit on my reaction to David Hansen's valid comment that I actually should have explained in my original question ("my intended response was too large to fit in the margin" ;-) ).

My interest in the infinite product of composites is based on the following idea.

Infinite products of the shape $\prod _{primes}^{\infty } \left( \dfrac{{p}^{s}} {{p}^{s}-1}\right)$ are only defined for $R_e >1$. The domain can be extended via analytical continuation as Riemann showed in his 1859 paper for the Euler product. He named the new function $\zeta(s)$ and proved it be valid for all $s \in \mathbb{C}$ with the only exception a pole at $s=1$. He found $\zeta(s)$ had zeroes (trivial ones at -2, -4, ... and non-trivial ones that appear to all be on the line $s = 0.5 + bi$). And via further transformations he also established a direct connection between the zeroes and the prime counting function $\pi(s)$.

If we take the logarithmic version of the prime-counting function $\psi(s)$ (i.e. the sum of all prime powers less than x, weighted by a natural logarithm of the power e.g.:

$\psi(10) = 3 \log(2) + 2 \log(3) + 1 \log(5) + 1 \log(7)$

then the exact prime counting function is ($\rho_k$ is a non-trivial zero):

$\psi(x) = x - \log(2\pi) - \frac12 \log(1- \frac{1}{x^2}) - \sum_{\rho} \dfrac{x^{\rho}}{\rho}$

Guess this is pretty standard stuff for the readers of this board and I also fully appreciate that the prime numbers are the atoms of the composites (what's in a name), but I wondered whether there could exist a Composite-counting-function that might be derived from $\prod _{composites}^{\infty } \left( \dfrac{{c}^{s}} {{c}^{s}-1}\right)$ in a similar fashion as Riemann did for the prime product. If so, one could use it as a sort of "detour" to approach the Riemann hypothesis from the other side. Let's just try a very small step backwards from the end result:

$C(x)$ = number of composities < x.

$\psi(x) = x - C(x)$

$C(x) = (\log(2\pi) + \frac12 \log(1- \frac{1}{x^2}) + \sum_{\rho} \dfrac{x^{\rho}}{\rho})$

A Composite-counting-function will therefore also be dependent on the non-trivial zeros. Since I couldn't find any way to obtain a zeta-like $C(x)$ function for the composite infinite product, I started exploring the route via $T(s)$ and got some success (I hope) by getting it expressed fully into $\zeta(s)$'s as:

$C(s) = \dfrac{e^{\sum_{n=1}^\infty \frac{\zeta(n s)-1}{n}}}{\zeta(s)}$

And before I even start dreaming about analytic continuation with contour integrals or further steps with Fourier/Mellin transforms, I'd been keen to understand whether the ratio can indeed be continued into $s \in \mathbb{C}$. If Charles' very encouraging response is indeed confirmed, then this would imply that the division $\dfrac{T(s)}{\zeta(s)}$ would induce a peak in $C(s)$ for each $s=\rho_k$. So, I'd need to count peaks rather than zeroes to compute the infinite sum of $\rho$'s in the Composite-counting-function.

P.S. After I plotted the graph for $T(s)$ with $s=0.5 + xi$, the $C(s)$-peaks and the $\zeta(s)$-zeroes at $s=\rho_k$ do not fully balance out and keep hovering between 0 and 1.

Answer 3

i'st easy to see that:

$$\ln T(s)=\sum_{n=1}^{\infty}\frac{\zeta(ns)-1}{n}$$

using the integral definition of the zeta function, one can show that:

$$\ln T(s)=s\int_{0}^{\infty}\frac{E_{s}(x^{s})-1}{xe^{x}(e^{x}-1)}dx$$

where : $E_{\alpha}(z)$ is the mittag-leffler fuction .

now, following Riemann's trick, here is what i did :

start with contour integral :

$$I(s)=-s\oint_{c}\frac{E_{s}((-x)^{s})-1}{xe^{x}(e^{x}-1)}dx$$

the contour is the usual Hankel contour. consider $I(-s)$ :

$$I(-s)=s\oint_{c}\frac{E_{-s}((-x)^{-s})-1}{xe^{x}(e^{x}-1)}dx=-s\oint_{c}\frac{E_{s}((-x)^{s})}{xe^{x}(e^{x}-1)}dx$$

• the Mittag-Leffler function admits the beautiful continuation : $E_{\alpha}(z^{-1})=1-E_{-\alpha}(z)$ -

or

$$I(s)-I(-s)=s\oint_{c}\frac{dx}{xe^{x}(e^{x}-1)}=s\oint_{c}(-x)^{-1}e^{-x}dx-s\oint_{c}\frac{(-x)^{-1}dx}{e^{x}-1}$$

now :$$\oint_{c}(-x)^{-1}e^{-x}dx=\frac{-2\pi i}{\Gamma(1)}=-2\pi i$$ and the second integral could be thought of as:

$$\oint_{c}\frac{(-x)^{-1}dx}{e^{x}-1}=\lim_{z\rightarrow 0}\oint_{c}\frac{(-x)^{z-1}dx}{e^{x}-1}=-2i\lim_{z\rightarrow 0}\sin(\pi z)\Gamma(z)\zeta(z)=i\pi$$

or :

$$I(s)-I(-s)=-3\pi is$$

lets go back to the 1st integral, and expand the Mittag-leffler function :

$$I(s)=-s\oint_{c}\frac{E_{s}((-x)^{s})-1}{xe^{x}(e^{x}-1)}dx=-s\sum_{n=1}^{\infty}\frac{1}{\Gamma(1+ns)}\oint_{c}\frac{(-x)^{sk-1}dx}{e^{x}(e^{x}-1)}$$ $$=s\sum_{n=1}^{\infty}\frac{2i \sin(k\pi s)\Gamma(ks)}{\Gamma(1+ns)}\left(\zeta(ks)-1\right)=2i\sum_{n=1}^{\infty}\sin(k\pi s)\frac{\zeta(ks)-1}{k}$$

now the problem becomes finding a function of the variable s -lets call it $A(s)$- such that:

$$\sum_{n=1}^{\infty}\sin(k\pi s)\frac{\zeta(ks)-1}{k}=A(s)\sum_{n=1}^{\infty}\frac{\zeta(ks)-1}{k}$$

if we define :

$$k(s)=\sum_{n=1}^{\infty}\frac{\zeta(ks)-1}{k}$$

then :

$$A(s)k(s)-A(-s)k(-s)=-\frac{3}{2}\pi s$$ and the problem becomes proving the existence of $A(s)$ for all s, and of course, finding it !!

Answer 4

I am seeking a proof of \lim_{m \rigtharrow \infty} \frac{Gamma(m)^m}{prod_{j=0}^{n-1}(m) that you need to use in thrid formula for show that \prod_{k=2}^{\infty}\frac{1}{1-k^{-n}}=\prod_{j=0}^{n-1}\Gamma(2-e^{\frac{2\pi j}{n}i})

You talk about arxiv , do you remember the url?

Thanks.

Fortuna