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 $a(p^{k})$ 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


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

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 $GL_{m}$.

Examples
The following examples will use the notation $$\mathbb{P}$$ for the set of all primes, that is:
 * $$\mathbb{P}=\{p \in \mathbb{N}\,|\,p\text{ is prime}\}.$$

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


 * $$\begin{align}

\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{align}$$

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 $π^{s}$, then for even exponents, this infinite product evaluates to a rational number. For example, since $ζ(2) = π^{2}⁄6$, $ζ(4) = π^{4}⁄90$, and $ζ(8) = π^{8}⁄9450$, then


 * $$\begin{align}

\prod_{p\, \in\, \mathbb{P}} \left(\frac{p^2+1}{p^2-1}\right) &= \frac53 \cdot \frac{10}{8} \cdot \frac{26}{24} \cdot \frac{50}{48} \cdot \frac{122}{120} \cdots &= \frac{\zeta(2)^2}{\zeta(4)} &= \frac52, \\[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)} &= \frac76, \end{align}$$

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 $2^{ω(n)}$ 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


 * $$ \prod_{p\, \in\, \mathbb{P}} \frac{1}{1- \frac{\chi(p)}{p^s}} = \sum_{n=1}^\infty \frac{\chi(n)}{n^s}.$$

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 $Li_{s}(x)$ is the polylogarithm. For $x = 1$ the product above is just $1⁄ζ(s)$.

Notable constants
Many well known constants have Euler product expansions.

The Leibniz formula for $\pi$


 * $$\frac{\pi}{4} = \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} = 1 - \frac13 + \frac15 - \frac17 + \cdots$$

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):


 * $$\frac{\pi}{4} = \left(\prod_{p\equiv 1\pmod 4}\frac{p}{p-1}\right)\left( \prod_{p\equiv 3\pmod 4}\frac{p}{p+1}\right)=\frac34 \cdot \frac54 \cdot \frac78 \cdot \frac{11}{12} \cdot \frac{13}{12} \cdots,$$

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

Other Euler products for known constants include:


 * The Hardy–Littlewood twin prime constant:


 * $$ \prod_{p>2} \left(1 - \frac{1}{\left(p-1\right)^2}\right) = 0.660161... $$


 * The Landau–Ramanujan constant:


 * $$\begin{align}

\frac{\pi}{4} \prod_{p \equiv 1\pmod 4} \left(1 - \frac{1}{p^2}\right)^\frac12 &= 0.764223... \\[6pt] \frac{1}{\sqrt{2}} \prod_{p \equiv 3\pmod 4} \left(1 - \frac{1}{p^2}\right)^{-\frac12} &= 0.764223... \end{align}$$


 * Murata's constant :


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


 * The strongly carefree constant $×ζ(2)^{2}$ :


 * $$ \prod_{p} \left(1 - \frac{1}{\left(p+1\right)^2}\right) = 0.775883... $$


 * Artin's constant :


 * $$ \prod_{p} \left(1 - \frac{1}{p(p-1)}\right) = 0.373955... $$


 * Landau's totient constant :


 * $$ \prod_{p} \left(1 + \frac{1}{p(p-1)}\right) = \frac{315}{2\pi^4}\zeta(3) = 1.943596... $$


 * The carefree constant $×ζ(2)$ :


 * $$ \prod_{p} \left(1 - \frac{1}{p(p+1)}\right) = 0.704442... $$


 * and its reciprocal :


 * $$ \prod_{p} \left(1 + \frac{1}{p^2+p-1}\right) = 1.419562... $$


 * The Feller–Tornier constant :


 * $$ \frac{1}{2}+\frac{1}{2} \prod_{p} \left(1 - \frac{2}{p^2}\right) = 0.661317... $$


 * The quadratic class number constant :


 * $$ \prod_{p} \left(1 - \frac{1}{p^2(p+1)}\right) = 0.881513... $$


 * The totient summatory constant :


 * $$ \prod_{p} \left(1 + \frac{1}{p^2(p-1)}\right) = 1.339784... $$


 * Sarnak's constant :


 * $$ \prod_{p>2} \left(1 - \frac{p+2}{p^3}\right) = 0.723648... $$


 * The carefree constant :


 * $$ \prod_{p} \left(1 - \frac{2p-1}{p^3}\right) = 0.428249... $$


 * The strongly carefree constant :


 * $$ \prod_{p} \left(1 - \frac{3p-2}{p^3}\right) = 0.286747... $$


 * Stephens' constant :


 * $$ \prod_{p} \left(1 - \frac{p}{p^3-1}\right) = 0.575959... $$


 * Barban's constant :


 * $$ \prod_{p} \left(1 + \frac{3p^2-1}{p(p+1)\left(p^2-1\right)}\right) = 2.596536... $$


 * Taniguchi's constant :


 * $$ \prod_{p} \left(1 - \frac{3}{p^3}+\frac{2}{p^4}+\frac{1}{p^5}-\frac{1}{p^6}\right) = 0.678234... $$


 * The Heath-Brown and Moroz constant :


 * $$ \prod_{p} \left(1 - \frac{1}{p}\right)^7 \left(1 + \frac{7p+1}{p^2}\right) = 0.0013176... $$