Euler product

Convergence

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

${\displaystyle \mathrm{Re}(s)>C,}$

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 GL_m.

Examples

The following examples will use the notation ${\displaystyle \mathbb{P}}$ for the set of all primes, that is:

${\displaystyle \mathbb{P}=\{p\in \mathbb{N}\mid p\text{ is prime}\}.}$

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

${\displaystyle \begin{array}{cc}\prod _{p\in \mathbb{P}}\left(\frac{1}{1-\frac{1}{p^s}}\right)& =\prod _{p\in \mathbb{P}}\left(\sum _{k=0}^{\mathrm{\infty }}\frac{1}{p^{ks}}\right)\\ & =\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^s}=\zeta (s).\end{array}}$

while for the Liouville function λ(n) = (−1)^Ω(n), it is

${\displaystyle \prod _{p\in \mathbb{P}}\left(\frac{1}{1+\frac{1}{p^s}}\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{\lambda (n)}{n^s}=\frac{\zeta (2s)}{\zeta (s)}.}$

Using their reciprocals, two Euler products for the Möbius function μ(n) are

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

and

${\displaystyle \prod _{p\in \mathbb{P}}\left(1+\frac{1}{p^s}\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{|\mu (n)|}{n^s}=\frac{\zeta (s)}{\zeta (2s)}.}$

Taking the ratio of these two gives

${\displaystyle \prod _{p\in \mathbb{P}}\left(\frac{1+\frac{1}{p^s}}{1-\frac{1}{p^s}}\right)=\prod _{p\in \mathbb{P}}\left(\frac{p^s+1}{p^s-1}\right)=\frac{\zeta (s{)}^2}{\zeta (2s)}.}$

Since for even values of s the Riemann zeta function ζ(s) has an analytic expression in terms of a rational multiple of π^s, 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

${\displaystyle \begin{array}{cccc}\prod _{p\in \mathbb{P}}\left(\frac{p^2+1}{p^2-1}\right)& =\frac{5}{3}\cdot \frac{10}{8}\cdot \frac{26}{24}\cdot \frac{50}{48}\cdot \frac{122}{120}\cdots & =\frac{\zeta (2{)}^2}{\zeta (4)}& =\frac{5}{2},\\ \prod _{p\in \mathbb{P}}\left(\frac{p^4+1}{p^4-1}\right)& =\frac{17}{15}\cdot \frac{82}{80}\cdot \frac{626}{624}\cdot \frac{2402}{2400}\cdots & =\frac{\zeta (4{)}^2}{\zeta (8)}& =\frac{7}{6},\end{array}}$

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

${\displaystyle \prod _{p\in \mathbb{P}}\left(1+\frac{2}{p^s}+\frac{2}{p^{2s}}+\cdots \right)=\sum _{n=1}^{\mathrm{\infty }}\frac{2^{\omega (n)}}{n^s}=\frac{\zeta (s{)}^2}{\zeta (2s)},}$

where ω(n) counts the number of distinct prime factors of n, and 2^ω(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

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

Here it is convenient to omit the primes p dividing the conductor N from the product.

Notable constants

Many well known constants have Euler product expansions.

The Leibniz formula for π

${\displaystyle \frac{\pi }{4}=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{2n+1}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots }$

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):

${\displaystyle \frac{\pi }{4}=\left(\prod _{p\equiv 1(\mathrm{mod}4)}\frac{p}{p-1}\right)\left(\prod _{p\equiv 3(\mathrm{mod}4)}\frac{p}{p+1}\right)=\frac{3}{4}\cdot \frac{5}{4}\cdot \frac{7}{8}\cdot \frac{11}{12}\cdot \frac{13}{12}\cdots ,}$

where each numerator is a prime number and each denominator is the nearest multiple of 4.

Dividing the Euler product for ${\displaystyle \zeta (2)=\frac{{\pi }^2}{6}}$ by the previous product, known as Basel problem, one finds that

${\displaystyle \frac{\pi }{2}=\left(\prod _{p\equiv 1(\mathrm{mod}4)}\frac{p}{p+1}\right)\left(\prod _{p\equiv 3(\mathrm{mod}4)}\frac{p}{p-1}\right).}$

Taking the ratio of the previous two products gives

${\displaystyle 2=\left(\prod _{p\equiv 1(\mathrm{mod}4)}\frac{p-1}{p+1}\right)\left(\prod _{p\equiv 3(\mathrm{mod}4)}\frac{p+1}{p-1}\right).}$

The infinite products must be taken in order of increasing primes.

Other Euler products for known constants include:

• The Hardy–Littlewood twin prime constant:

${\displaystyle \prod _{p>2}\left(1-\frac{1}{{\left(p-1\right)}^2}\right)=0.660161\dots }$

• The Landau–Ramanujan constant:

${\displaystyle \begin{array}{cc}\frac{\pi }{4}\prod _{p\equiv 1(\mathrm{mod}4)}{\left(1-\frac{1}{p^2}\right)}^{\frac{1}{2}}& =0.764223\dots \\ \frac{1}{\sqrt{2}}\prod _{p\equiv 3(\mathrm{mod}4)}{\left(1-\frac{1}{p^2}\right)}^{-\frac{1}{2}}& =0.764223\dots \end{array}}$

• Murata's constant (sequence A065485 in the OEIS):

${\displaystyle \prod _p\left(1+\frac{1}{{\left(p-1\right)}^2}\right)=2.826419\dots }$

• The strongly carefree constant ×ζ(2)² OEIS: A065472:

${\displaystyle \prod _p\left(1-\frac{1}{{\left(p+1\right)}^2}\right)=0.775883\dots }$

• Artin's constant OEIS: A005596:

${\displaystyle \prod _p\left(1-\frac{1}{p(p-1)}\right)=0.373955\dots }$

• Landau's totient constant OEIS: A082695:

${\displaystyle \prod _p\left(1+\frac{1}{p(p-1)}\right)=\frac{315}{2{\pi }^4}\zeta (3)=1.943596\dots }$

• The carefree constant ×ζ(2) OEIS: A065463:

${\displaystyle \prod _p\left(1-\frac{1}{p(p+1)}\right)=0.704442\dots }$

and its reciprocal OEIS: A065489:

${\displaystyle \prod _p\left(1+\frac{1}{p^2+p-1}\right)=1.419562\dots }$

• The Feller–Tornier constant OEIS: A065493:

${\displaystyle \frac{1}{2}+\frac{1}{2}\prod _p\left(1-\frac{2}{p^2}\right)=0.661317\dots }$

• The quadratic class number constant OEIS: A065465:

${\displaystyle \prod _p\left(1-\frac{1}{p^2(p+1)}\right)=0.881513\dots }$

• The totient summatory constant OEIS: A065483:

${\displaystyle \prod _p\left(1+\frac{1}{p^2(p-1)}\right)=1.339784\dots }$

• Sarnak's constant OEIS: A065476:

${\displaystyle \prod _{p>2}\left(1-\frac{p+2}{p^3}\right)=0.723648\dots }$

• The carefree constant OEIS: A065464:

${\displaystyle \prod _p\left(1-\frac{2p-1}{p^3}\right)=0.428249\dots }$

• The strongly carefree constant OEIS: A065473:

${\displaystyle \prod _p\left(1-\frac{3p-2}{p^3}\right)=0.286747\dots }$

• Stephens' constant OEIS: A065478:

${\displaystyle \prod _p\left(1-\frac{p}{p^3-1}\right)=0.575959\dots }$

• Barban's constant OEIS: A175640:

${\displaystyle \prod _p\left(1+\frac{3p^2-1}{p(p+1)\left(p^2-1\right)}\right)=2.596536\dots }$

• Taniguchi's constant OEIS: A175639:

${\displaystyle \prod _p\left(1-\frac{3}{p^3}+\frac{2}{p^4}+\frac{1}{p^5}-\frac{1}{p^6}\right)=0.678234\dots }$

• The Heath-Brown and Moroz constant OEIS: A118228:

${\displaystyle \prod _p{\left(1-\frac{1}{p}\right)}^7\left(1+\frac{7p+1}{p^2}\right)=0.0013176\dots }$