# Euler product

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

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

$\prod _{p}P(p,s)\quad {\text{for }}\operatorname {Re} (s)>1.$ where the product is taken over prime numbers p, and P(p, s) is the sum

$\sum _{k=0}^{\infty }{\frac {a(p^{k})}{p^{ks}}}=1+{\frac {a(p)}{p^{s}}}+{\frac {a(p^{2})}{p^{2s}}}+{\frac {a(p^{3})}{p^{3s}}}+\cdots$ 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(pk) whenever n factors as the product of the powers pk 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

$P(p,s)={\frac {1}{1-{\frac {a(p)}{p^{s}}}}},$ 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 following examples will use the notation $\mathbb {P}$ for the set of all primes, that is:

$\mathbb {P} =\{p\in N\,|\,p{\text{ is prime}}\}.$ The Euler product attached to the Riemann zeta function ζ(s), also using the sum of the geometric series, is

{\begin{aligned}\prod _{p\,\in \,\mathbb {P} }\left({\frac {1}{1-{\frac {1}{p^{s}}}}}\right)&=\prod _{p\ \in \ \mathbb {P} }\left(\sum _{k=0}^{\infty }{\frac {1}{p^{ks}}}\right)\\&=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}=\zeta (s).\end{aligned}} while for the Liouville function λ(n) = (−1)ω(n), it is

$\prod _{p\,\in \,\mathbb {P} }\left({\frac {1}{1+{\frac {1}{p^{s}}}}}\right)=\sum _{n=1}^{\infty }{\frac {\lambda (n)}{n^{s}}}={\frac {\zeta (2s)}{\zeta (s)}}.$ Using their reciprocals, two Euler products for the Möbius function μ(n) are

$\prod _{p\,\in \,\mathbb {P} }\left(1-{\frac {1}{p^{s}}}\right)=\sum _{n=1}^{\infty }{\frac {\mu (n)}{n^{s}}}={\frac {1}{\zeta (s)}}$ and

$\prod _{p\,\in \,\mathbb {P} }\left(1+{\frac {1}{p^{s}}}\right)=\sum _{n=1}^{\infty }{\frac {|\mu (n)|}{n^{s}}}={\frac {\zeta (s)}{\zeta (2s)}}.$ Taking the ratio of these two gives

$\prod _{p\,\in \,\mathbb {P} }\left({\frac {1+{\frac {1}{p^{s}}}}{1-{\frac {1}{p^{s}}}}}\right)=\prod _{p\,\in \,\mathbb {P} }\left({\frac {p^{s}+1}{p^{s}-1}}\right)={\frac {\zeta (s)^{2}}{\zeta (2s)}}.$ Since for even values of s the Riemann zeta function ζ(s) has an analytic expression in terms of a rational multiple of πs, then for even exponents, this infinite product evaluates to a rational number. For example, since ζ(2) = π2/6, ζ(4) = π4/90, and ζ(8) = π8/9450, then

{\begin{aligned}\prod _{p\,\in \,\mathbb {P} }\left({\frac {p^{2}+1}{p^{2}-1}}\right)&={\frac {5}{3}}\cdot {\frac {10}{8}}\cdot {\frac {26}{24}}\cdot {\frac {50}{48}}\cdot {\frac {122}{120}}\cdots &={\frac {\zeta (2)^{2}}{\zeta (4)}}&={\frac {5}{2}},\\[6pt]\prod _{p\,\in \,\mathbb {P} }\left({\frac {p^{4}+1}{p^{4}-1}}\right)&={\frac {17}{15}}\cdot {\frac {82}{80}}\cdot {\frac {626}{624}}\cdot {\frac {2402}{2400}}\cdots &={\frac {\zeta (4)^{2}}{\zeta (8)}}&={\frac {7}{6}},\end{aligned}} and so on, with the first result known by Ramanujan. This family of infinite products is also equivalent to

$\prod _{p\,\in \,\mathbb {P} }\left(1+{\frac {2}{p^{s}}}+{\frac {2}{p^{2s}}}+\cdots \right)=\sum _{n=1}^{\infty }{\frac {2^{\omega (n)}}{n^{s}}}={\frac {\zeta (s)^{2}}{\zeta (2s)}},$ where ω(n) counts the number of distinct prime factors of n, and 2ω(n) is the number of square-free divisors.

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

$\prod _{p\,\in \,\mathbb {P} }{\frac {1}{1-{\frac {\chi (p)}{p^{s}}}}}=\sum _{n=1}^{\infty }{\frac {\chi (n)}{n^{s}}}.$ 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

$\prod _{p\,\in \,\mathbb {P} }\left(x-{\frac {1}{p^{s}}}\right)\approx {\frac {1}{\operatorname {Li} _{s}(x)}}$ for s > 1 where Lis(x) is the polylogarithm. For x = 1 the product above is just 1/ζ(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 (fractions where numerator and denominator differ by 1):

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

Other Euler products for known constants include:

$\prod _{p>2}\left(1-{\frac {1}{\left(p-1\right)^{2}}}\right)=0.660161...$ {\begin{aligned}{\frac {\pi }{4}}\prod _{p\equiv 1{\pmod {4}}}\left(1-{\frac {1}{p^{2}}}\right)^{\frac {1}{2}}&=0.764223...\\[6pt]{\frac {1}{\sqrt {2}}}\prod _{p\equiv 3{\pmod {4}}}\left(1-{\frac {1}{p^{2}}}\right)^{-{\frac {1}{2}}}&=0.764223...\end{aligned}} $\prod _{p}\left(1+{\frac {1}{\left(p-1\right)^{2}}}\right)=2.826419...$ $\prod _{p}\left(1-{\frac {1}{\left(p+1\right)^{2}}}\right)=0.775883...$ $\prod _{p}\left(1-{\frac {1}{p(p-1)}}\right)=0.373955...$ $\prod _{p}\left(1+{\frac {1}{p(p-1)}}\right)={\frac {315}{2\pi ^{4}}}\zeta (3)=1.943596...$ $\prod _{p}\left(1-{\frac {1}{p(p+1)}}\right)=0.704442...$ and its reciprocal :
$\prod _{p}\left(1+{\frac {1}{p^{2}+p-1}}\right)=1.419562...$ ${\frac {1}{2}}+{\frac {1}{2}}\prod _{p}\left(1-{\frac {2}{p^{2}}}\right)=0.661317...$ $\prod _{p}\left(1-{\frac {1}{p^{2}(p+1)}}\right)=0.881513...$ $\prod _{p}\left(1+{\frac {1}{p^{2}(p-1)}}\right)=1.339784...$ $\prod _{p>2}\left(1-{\frac {p+2}{p^{3}}}\right)=0.723648...$ $\prod _{p}\left(1-{\frac {2p-1}{p^{3}}}\right)=0.428249...$ $\prod _{p}\left(1-{\frac {3p-2}{p^{3}}}\right)=0.286747...$ $\prod _{p}\left(1-{\frac {p}{p^{3}-1}}\right)=0.575959...$ $\prod _{p}\left(1+{\frac {3p^{2}-1}{p(p+1)\left(p^{2}-1\right)}}\right)=2.596536...$ $\prod _{p}\left(1-{\frac {3}{p^{3}}}+{\frac {2}{p^{4}}}+{\frac {1}{p^{5}}}-{\frac {1}{p^{6}}}\right)=0.678234...$ $\prod _{p}\left(1-{\frac {1}{p}}\right)^{7}\left(1+{\frac {7p+1}{p^{2}}}\right)=0.0013176...$ 