# Euler's Product

Everyone knows Euler’s famous identity, linking the imaginary unit to trigonometric functions:

$$\mathrm{e}^{\mathrm{i} \pi} = -1$$

But another remarkable identity is also due to Euler, this one linking the set of prime numbers to an analytical function:

$$\prod_\text{all primes p} \left( 1 - \frac{1}{p^s} \right)^{-1} = \sum_\text{all positive integers n} \frac{1}{n^s}$$

where $s$ is any complex number.

The expression on the right is the definition of the Riemann Zeta Function:

\begin{eqnarray*} \zeta(s) & = & \sum_{n=1}^{\infty} \frac{1}{n^s} \\
& = & 1 + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + \dots \end{eqnarray*}

It is astounding that a product over the primes (surely a discrete set!) can yield an analytic function! But, as a matter of fact, Euler’s Product follows pretty much by simply multiplying out the product. Consider:

\begin{eqnarray*} f_2(s) = \left( 1 - \frac{1}{2^s} \right) \zeta(s) & = & \zeta(s) - \frac{1}{2^s} \zeta(s) \\
& = & 1 + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + \dots \\
& & \phantom{1} - \frac{1}{2^s} - \frac{1}{4^s} - \frac{1}{6^s} - \dots \\
& = & 1 + \frac{1}{3^s} + \frac{1}{5^s} + \frac{1}{7^s} + \dots \end{eqnarray*}

In other words, all terms with an even integer in the denominator have been eliminated. Now we do it again with the next prime, $3$:

\begin{eqnarray*} f_3(s) = \left( 1 - \frac{1}{3^s} \right) f_2(s) & = & \left( 1 - \frac{1}{3^s} \right) \left( 1 - \frac{1}{2^s} \right) \zeta(s) \\
& = & f_2(s) - \frac{1}{3^s} f_2(s) \\
& = & 1 + \frac{1}{3^s} + \frac{1}{5^s} + \frac{1}{7^s} + \dots \\
& & \phantom{1} - \frac{1}{3^s} - \frac{1}{9^s} - \frac{1}{15^s} - \dots \\
& = & 1 + \frac{1}{5^s} + \frac{1}{7^s} + \frac{1}{11^s} + \dots \end{eqnarray*}

This eliminates any term where the integer in the denominator is divisible by $3$. Continuing like this with all primes, we end up with:

$$\prod_\text{p prime} \left( 1 - \frac{1}{p^s} \right) \zeta(s) = 1$$

which is equivalent to Euler’s Product.

Euler’s Product is one of the more remarkable identities in mathematics. By making a connection between the primes and an analytic function, it is the starting point for all analytic number theory, which investigates the analytic properties of $\zeta(s)$ in order to gain information about the primes.