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 $s<0$).

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

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

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


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'?

  • 1
    $\begingroup$ Why do you think this "composite product" will tell you anything interesting about anything? $\endgroup$ Jan 27, 2011 at 2:16
  • $\begingroup$ David, Here's the thought. Riemann linked the Euler prime product, via the analytically continued Zeta-function, via its non-trivial zeroes (all allegedly lying on line a=1/2) to the prime counting function. Since the logarithmic prime counting function phi(x) = x - ln(2pi) - infinity sum(x^rho / rho), I wondered whether a Composite-counting function exists as well. Since such function requires the same non-trivial zeroes (i.e. = (ln(2pi) + infinity sum(x^rho / rho)), I conjectured that there should be a link back into the infinite composite product (and the Zeta). Hence the quest. $\endgroup$
    – Agno
    Jan 27, 2011 at 11:07
  • 4
    $\begingroup$ This is a pretty old question, but I just saw it for the first time: I think questions like this can be valuable -- asking your own questions rather than just trying to answer other people's is always valuable. But you might find helpful Tim Gowers's discussion of why the Zeta function is a 'natural' and useful thing to consider: dpmms.cam.ac.uk/~wtg10/zetafunction.ps $\endgroup$ May 6, 2012 at 21:44
  • $\begingroup$ Thanks Brad. A pretty old question indeed (actually my very first ever on MathOverflow. I even remember the excitement as well as the anxiety from throwing a pretty rough idea in front of so many sharp brains). Will check out the link on the 'naturalness' of the Zeta function. $\endgroup$
    – Agno
    May 10, 2012 at 19:22
  • $\begingroup$ ImageShack has turned most of its photos into ads. In the process the link to the graph is no longer available. If you have the original photo, you might want to consider uploading it using the SE interface. $\endgroup$
    – Asaf Karagila
    Aug 18, 2015 at 22:26

3 Answers 3


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).

  • $\begingroup$ Do you have any reasons for thinking that $T$ can be extended to a meromorphic function on $\mathbb C$? $\endgroup$ May 10, 2012 at 22:47
  • $\begingroup$ @Greg: Unfortunately I did not make any notes and in the year-plus since I answered the question I do not recall my reasoning. If you have any contrary thoughts, please give a separate answer! $\endgroup$
    – Charles
    May 11, 2012 at 2:36

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.

  • 1
    $\begingroup$ C(x) and pi(x) are related by C(x) + (pi(x)-2)= x/3 since 2x/3 are multiples of 2 and 3. You see this if you only consider the arithmetic progressions (AP) (6k+1) and (6k-1). The numbers of composites and primes in those two AP add up to x/3. But we still need to add the primes 2 and 3 to the total. $\endgroup$
    – user25406
    Feb 5, 2019 at 22:21

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 :


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


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



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 :


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 !!

  • $\begingroup$ Thanks for your response, Mohammad. I like the angle you took, however also got a bit confused by the interchange between $n$ and $k$. Are these used correctly in all the sums? $\endgroup$
    – Agno
    May 10, 2012 at 19:13
  • $\begingroup$ @Agno ... sorry for the late response ... here is my email : [email protected] , you might be interested in my work on this problem . $\endgroup$ Jul 28, 2012 at 21:19

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