A well known form of Euler product is

$ \prod_p^\infty \big(1- \frac{1}{ p^{s} } \big) = \frac{1}{ \zeta (s) } $

for all primes $p$.

Are there any other variations of above product, such as :

$ \prod_p^\infty \big(1- \frac{k}{ p^{s} } \big)$ for any integer $ k > 1$


$ \prod_p^\infty \big(1- \frac{f (p)}{ p^{s} } \big)$ where $f(p)$ is any non zero polynomial of prime $p$.

Few of them are discussed on the Wikipedia page for Euler product

But the page only includes numerical values of such products, and none of them are represented in terms of other well known constants like $\pi$ or in terms of other Zeta functions like $\zeta(s)$.

Kindly let me about any such publication which discussed such variation in Euler product in detail.

  • $\begingroup$ There are a lot of variations on the theme...the last product you wrote is always possible once you have a totally multiplicative function $f(n)$. In general, take a look here: en.wikipedia.org/wiki/Dirichlet_series. There a different examples. $\endgroup$
    Feb 16, 2016 at 8:47
  • $\begingroup$ quiet right! But I always wonder why it is kind of difficult if $f(p)$ is some constant or an integer greater than 1. The Wikipedia link provided in question deals with some of them. Even thought the product converges, no exact value is know for them, all are approximated. Any ideas on recent publications that discussed Euler product with constants ? $\endgroup$ Feb 20, 2016 at 8:42
  • $\begingroup$ If $|f(n)|=c$ for some $c>1$, then the sum $\sum_{n\geq 1}c/n^s$ converges absolutely only if ${\rm Re}(s)\geq c$ and only in this case you can perform an Euler product. $\endgroup$
    Feb 20, 2016 at 9:39
  • $\begingroup$ Indeed, but I wonder it converges to what? Even for the $f(n) = 1$ we do not know the exact value of $zeta(s)$ for any odd $s$ greater than $1$. However I can't find any latest work on Exact values of zeta function for Odd $s$.Even exact value of $zeta(3)$ has so far not known, if you have any recent resource that addressed this issue, kindly share. Just out of curiosity though, is there a way to represent $zeta(2k+1)$ where $k>1$, in terms of $zeta(3)$ ? $\endgroup$ Feb 20, 2016 at 10:22
  • $\begingroup$ Did you see this link en.wikipedia.org/wiki/… ? $\endgroup$
    Feb 20, 2016 at 10:56

1 Answer 1


You are looking for Dirichlet series, L-functions and modular forms, which you can conveniently study on the website "L-functions and Modular Forms DataBase" at www.lmfdb.org.


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