# Euler product

In number theory, an Euler product is an expansion of a Dirichlet series into an infinite product indexed by prime numbers. The name arose from the case of the Riemann zeta-function, where such a product representation was proved by Leonhard Euler.

## Definition

In general, if $a$ is a multiplicative function, then the Dirichlet series

$\sum_{n} a(n)n^{-s}\,$

is equal to

$\prod_{p} P(p, s)\,$

where the product is taken over prime numbers $p$, and $P(p, s)$ is the sum

$1+a(p)p^{-s} + a(p^2)p^{-2s} + \cdots .$

In fact, if we consider these as formal generating functions, the existence of such a formal Euler product expansion is a necessary and sufficient condition that $a(n)$ be multiplicative: this says exactly that $a(n)$ is the product of the $a(p^k)$ whenever $n$ factors as the product of the powers $p^k$ of distinct primes $p$.

An important special case is that in which $a(n)$ is totally multiplicative, so that $P(p, s)$ is a geometric series. Then

$P(p, s)=\frac{1}{1-a(p)p^{-s}},$

as is the case for the Riemann zeta-function, where $a(n) = 1$, and more generally for Dirichlet characters.

## Convergence

In practice all the important cases are such that the infinite series and infinite product expansions are absolutely convergent in some region

Re(s) > C

that is, in some right half-plane in the complex numbers. This already gives some information, since the infinite product, to converge, must give a non-zero value; hence the function given by the infinite series is not zero in such a half-plane.

In the theory of modular forms it is typical to have Euler products with quadratic polynomials in the denominator here. The general Langlands philosophy includes a comparable explanation of the connection of polynomials of degree m, and the representation theory for GLm.

## Examples

The Euler product attached to the Riemann zeta function $\zeta(s)$, using also the sum of the geometric series, is

$\prod_{p} (1-p^{-s})^{-1} = \prod_{p} \Big(\sum_{n=0}^{\infty}p^{-ns}\Big) = \sum_{n=1}^{\infty} \frac{1}{n^{s}} = \zeta(s)$.

while for the Liouville function $\lambda(n) = (-1)^{\Omega(n)}$, it is,

$\prod_{p} (1+p^{-s})^{-1} = \sum_{n=1}^{\infty} \frac{\lambda(n)}{n^{s}} = \frac{\zeta(2s)}{\zeta(s)}$

Using their reciprocals, two Euler products for the Möbius function $\mu(n)$ are,

$\prod_{p} (1-p^{-s}) = \sum_{n=1}^{\infty} \frac{\mu (n)}{n^{s}} = \frac{1}{\zeta(s)}$

and,

$\prod_{p} (1+p^{-s}) = \sum_{n=1}^{\infty} \frac{|\mu(n)|}{n^{s}} = \frac{\zeta(s)}{\zeta(2s)}$

and taking the ratio of these two gives,

$\prod_{p} \Big(\frac{1+p^{-s}}{1-p^{-s}}\Big) = \prod_{p} \Big(\frac{p^{s}+1}{p^{s}-1}\Big) = \frac{\zeta(s)^2}{\zeta(2s)}$

Since for even s the Riemann zeta function $\zeta(s)$ has an analytic expression in terms of a rational multiple of $\pi^{s}$, then for even exponents, this infinite product evaluates to a rational number. For example, since $\zeta(2)=\pi^2/6$, $\zeta(4)=\pi^4/90$, and $\zeta(8)=\pi^8/9450$, then,

$\prod_{p} \Big(\frac{p^{2}+1}{p^{2}-1}\Big) = \frac{5}{2}$
$\prod_{p} \Big(\frac{p^{4}+1}{p^{4}-1}\Big) = \frac{7}{6}$

and so on, with the first result known by Ramanujan. This family of infinite products is also equivalent to,

$\prod_{p} (1+2p^{-s}+2p^{-2s}+\cdots) = \sum_{n=1}^{\infty}2^{\omega(n)} n^{-s} = \frac{\zeta(s)^2}{\zeta(2s)}$

where $\omega(n)$ counts the number of distinct prime factors of n and $2^{\omega(n)}$ the number of square-free divisors.

If $\chi(n)$ is a Dirichlet character of conductor $N$, so that $\chi$ is totally multiplicative and $\chi(n)$ only depends on n modulo N, and $\chi(n) = 0$ if n is not coprime to N then,

$\prod_{p} (1- \chi(p) p^{-s})^{-1} = \sum_{n=1}^{\infty}\chi(n)n^{-s}$.

Here it is convenient to omit the primes p dividing the conductor N from the product. Ramanujan in his notebooks tried to generalize the Euler product for Zeta function in the form:

$\prod_{p} (x-p^{-s})\approx \frac{1}{\operatorname{Li}_{s} (x)}$

for $s > 1$ where $\operatorname{Li}_s(x)$ is the polylogarithm. For $x=1$ the product above is just $1/ \zeta (s).$

## Notable constants

Many well known constants have Euler product expansions.

$\pi/4=\sum_{n=0}^\infty \, \frac{(-1)^n}{2n+1}=1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots,$

can be interpreted as a Dirichlet series using the (unique) Dirichlet character modulo 4, and converted to an Euler product of superparticular ratios

$\pi/4=\left(\prod_{p\equiv 1\pmod 4}\frac{p}{p-1}\right)\cdot\left( \prod_{p\equiv 3\pmod 4}\frac{p}{p+1}\right)=\frac{3}{4} \cdot \frac{5}{4} \cdot \frac{7}{8} \cdot \frac{11}{12} \cdot \frac{13}{12} \cdots,$

where each numerator is a prime number and each denominator is the nearest multiple of four.[1]

Other Euler products for known constants include:

$\prod_{p>2} \Big(1 - \frac{1}{(p-1)^2}\Big) = 0.660161...$
$\frac{\pi}{4} \prod_{p = 1\,\text{mod}\,4} \Big(1 - \frac{1}{p^2}\Big)^{1/2} = 0.764223...$
$\frac{1}{\sqrt{2}} \prod_{p = 3\,\text{mod}\,4} \Big(1 - \frac{1}{p^2}\Big)^{-1/2} = 0.764223...$

Murata's constant (sequence A065485 in OEIS):

$\prod_{p} \Big(1 + \frac{1}{(p-1)^2}\Big) = 2.826419...$

Strongly carefree constant $\times\zeta(2)^2$ :

$\prod_{p} \Big(1 - \frac{1}{(p+1)^2}\Big) = 0.775883...$
$\prod_{p} \Big(1 - \frac{1}{p(p-1)}\Big) = 0.373955...$
$\prod_{p} \Big(1 + \frac{1}{p(p-1)}\Big) = \frac{315}{2\pi^4}\zeta(3) = 1.943596...$

Carefree constant $\times\zeta(2)$ :

$\prod_{p} \Big(1 - \frac{1}{p(p+1)}\Big) = 0.704442...$

(with reciprocal) :

$\prod_{p} \Big(1 + \frac{1}{p^2+p-1}\Big) = 1.419562...$
$\frac{1}{2}+\frac{1}{2} \prod_{p} \Big(1 - \frac{2}{p^2}\Big) = 0.661317...$
$\prod_{p} \Big(1 - \frac{1}{p^2(p+1)}\Big) = 0.881513...$
$\prod_{p} \Big(1 + \frac{1}{p^2(p-1)}\Big) = 1.339784...$
$\prod_{p>2} \Big(1 - \frac{p+2}{p^3}\Big) = 0.723648...$
$\prod_{p} \Big(1 - \frac{2p-1}{p^3}\Big) = 0.428249...$
$\prod_{p} \Big(1 - \frac{3p-2}{p^3}\Big) = 0.286747...$
$\prod_{p} \Big(1 - \frac{p}{p^3-1}\Big) = 0.575959...$
$\prod_{p} \Big(1 + \frac{3p^2-1}{p(p+1)(p^2-1)}\Big) = 2.596536...$
$\prod_{p} \Big(1 - \frac{3}{p^3}+\frac{2}{p^4}+\frac{1}{p^5}-\frac{1}{p^6}\Big) = 0.678234...$
$\prod_{p} \Big(1 - \frac{1}{p}\Big)^7 \Big(1 + \frac{7p+1}{p^2}\Big) = 0.0013176...$

## Notes

1. ^ Debnath, Lokenath (2010), The Legacy of Leonhard Euler: A Tricentennial Tribute, World Scientific, p. 214, ISBN 9781848165267.

## References

• G. Polya, Induction and Analogy in Mathematics Volume 1 Princeton University Press (1954) L.C. Card 53-6388 (A very accessible English translation of Euler's memoir regarding this "Most Extraordinary Law of the Numbers" appears starting on page 91)
• Apostol, Tom M. (1976), Introduction to analytic number theory, Undergraduate Texts in Mathematics, New York-Heidelberg: Springer-Verlag, ISBN 978-0-387-90163-3, MR 0434929, Zbl 0335.10001 (Provides an introductory discussion of the Euler product in the context of classical number theory.)
• G.H. Hardy and E.M. Wright, An introduction to the theory of numbers, 5th ed., Oxford (1979) ISBN 0-19-853171-0 (Chapter 17 gives further examples.)
• George E. Andrews, Bruce C. Berndt, Ramanujan's Lost Notebook: Part I, Springer (2005), ISBN 0-387-25529-X
• G. Niklasch, Some number theoretical constants: 1000-digit values"