# Euler product

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In number theory, an Euler product is an expansion of a Dirichlet series into an infinite product indexed by prime numbers. The original such product was given for the sum of all positive integers raised to a certain power as proven by Leonhard Euler. This series and its continuation to the entire complex plane would later become known as the Riemann zeta function.

## Definition

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

$\sum _{n}a(n)n^{-s}\,$ is equal to

$\prod _{p}P(p,s)\,$ for Re(s) >1 .

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

$\operatorname {Re} (s)>C,$ 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

$\prod _{p}(1+p^{-s})=\sum _{n=1}^{\infty }{\frac {|\mu (n)|}{n^{s}}}={\frac {\zeta (s)}{\zeta (2s)}}.$ 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 $2^{\omega (n)}$ is 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. In his notebooks, Ramanujan generalized the Euler product for the zeta function as

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

${\frac {\pi }{4}}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}=1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+\cdots ,$ can be interpreted as a Dirichlet series using the (unique) Dirichlet character modulo 4, and converted to an Euler product of superparticular ratios

${\frac {\pi }{4}}=\left(\prod _{p\equiv 1{\pmod {4}}}{\frac {p}{p-1}}\right)\cdot \left(\prod _{p\equiv 3{\pmod {4}}}{\frac {p}{p+1}}\right)={\frac {3}{4}}\cdot {\frac {5}{4}}\cdot {\frac {7}{8}}\cdot {\frac {11}{12}}\cdot {\frac {13}{12}}\cdots ,$ where each numerator is a prime number and each denominator is the nearest multiple of four.

Other Euler products for known constants include:

$\prod _{p>2}\left(1-{\frac {1}{(p-1)^{2}}}\right)=0.660161...$ ${\frac {\pi }{4}}\prod _{p\equiv 1{\pmod {4}}}\left(1-{\frac {1}{p^{2}}}\right)^{1/2}=0.764223...$ ${\frac {1}{\sqrt {2}}}\prod _{p\equiv 3{\pmod {4}}}\left(1-{\frac {1}{p^{2}}}\right)^{-1/2}=0.764223...$ Murata's constant (sequence A065485 in the OEIS):

$\prod _{p}\left(1+{\frac {1}{(p-1)^{2}}}\right)=2.826419...$ Strongly carefree constant $\times \zeta (2)^{2}$ :

$\prod _{p}\left(1-{\frac {1}{(p+1)^{2}}}\right)=0.775883...$ $\prod _{p}\left(1-{\frac {1}{p(p-1)}}\right)=0.373955...$ $\prod _{p}\left(1+{\frac {1}{p(p-1)}}\right)={\frac {315}{2\pi ^{4}}}\zeta (3)=1.943596...$ Carefree constant $\times \zeta (2)$ :

$\prod _{p}\left(1-{\frac {1}{p(p+1)}}\right)=0.704442...$ (with reciprocal) :

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