The Riemann zeta function \zeta(s) is, on its own, a slightly lopsided object. It has a pole at s = 1, a scatter of trivial zeros at the negative even integers, and a functional equation \zeta(s) = 2^{s}\pi^{s-1}\sin(\tfrac{\pi s}{2})\Gamma(1-s)\zeta(1-s) so cluttered with stray factors that its beautiful underlying symmetry is almost hidden. There is a better way to package the same information.

Riemann's insight was to multiply \zeta by exactly the right extra factors to produce a single, flawlessly symmetric function — the completed zeta function. In its cleanest form it satisfies simply \xi(s) = \xi(1-s): reflect across the critical line and nothing changes. This completed object, not the bare \zeta, is really the "true" zeta — the one that reveals what the functional equation is saying.

Building the completed function

Start by attaching to \zeta(s) a factor built from the Gamma function. Define the completed function (often written \Lambda or \xi, conventions differ — see the warning below):

The piece \pi^{-s/2}\Gamma(s/2) is the Gamma factor. It is the single most important guest at the party: with it attached, the whole product obeys the pristine reflection

There is just one blemish left. \zeta still has its pole at s = 1, and the Gamma factor \Gamma(s/2) has a pole at s = 0. So \Lambda is meromorphic with two simple poles, at s = 0 and s = 1. Riemann's \xi removes even those.

Riemann's ξ — the entire version

The extra prefactor \tfrac12 s(s-1) does two clean-up jobs at once. The factor (s-1) has a zero at s = 1 that cancels the pole of \zeta there; the factor s has a zero at s = 0 that cancels the pole of \Gamma(s/2). Both blemishes vanish, and what remains is an entire function. Notice too that s(s-1) is itself symmetric under s \mapsto 1-s (swap the roots), so multiplying by it does not disturb the reflection \xi(s) = \xi(1-s).

The Gamma factor and the "prime at infinity"

Why \pi^{-s/2}\Gamma(s/2), of all things? The name for it is the archimedean factor, or the local factor at infinity. To see the idea, recall the Euler product: \zeta(s) = \prod_{p} (1 - p^{-s})^{-1}, one factor for each prime p. Each of those is a "local factor" attached to a prime.

The completed function slots in one more local factor — the Gamma factor — as though there were an extra "prime", conventionally called the prime at infinity and written p = \infty. Schematically,

With the point at infinity included, the product becomes symmetric — that is exactly the modern (Tate's-thesis) explanation of why the functional equation holds and why the Gamma factor is the shape it is. The finite primes give \zeta; the prime at infinity gives the Gamma factor; together they give the perfectly balanced \Lambda.

Where the trivial zeros go

Here is the most satisfying piece of bookkeeping in the whole subject. On the negative real axis two things happen at each even integer s = -2, -4, -6, \dots: \zeta(s) has a trivial zero, and the Gamma factor \Gamma(s/2) has a pole. In the product they cancel exactly, leaving \xi finite and nonzero at each of those points.

Worked example — check s = -2. The Gamma factor is \Gamma(s/2) = \Gamma(-1), and \Gamma has a simple pole at -1 (indeed at every non-positive integer). Meanwhile \zeta(-2) = 0 is a simple zero. A simple zero multiplied by a simple pole gives a finite, nonzero value:

The same cancellation runs down the entire negative even axis. So the "trivial" zeros of \zeta are, from the completed point of view, not zeros at all — they are artefacts of the Gamma factor's poles. Strip the Gamma factor away and they reappear; that is literally what "trivial" means. What survives as genuine zeros of \xi are only the non-trivial zeros in the critical strip.

The zeros of ξ are exactly the interesting ones

Let's tally every possible source of a zero of \xi(s) = \tfrac12 s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s):

Every zero except the non-trivial ones is neutralised. Hence:

This is why \xi is the right object to state the Riemann Hypothesis with: the hypothesis becomes simply "all zeros of \xi lie on \Re(s) = \tfrac12", with no trivial-zero caveats to carry around.

Order 1, and a product over the zeros

As an entire function, \xi has a growth order — a measure of how fast |\xi(s)| can grow. One shows

which makes \xi entire of order 1 (the |s|\log|s| growth just fails to be order-1 in the strict sense, so one says "order 1, maximal type" — but order 1 for Hadamard's theorem). That single fact is enormously powerful. Hadamard's factorization theorem then says an order-1 entire function is determined, up to an exponential prefactor, by its zeros:

the product running over the non-trivial zeros \rho (paired so it converges). This product is the bridge from the zeros of \zeta straight to the primes: taking its logarithmic derivative feeds the "explicit formula" that expresses the prime count as a sum over exactly these \rho. The completed function is what makes that clean product possible.

Special values and reality

Two small facts pin the function down and are worth memorising. First, the symmetric endpoints:

Worked example — \xi(1). Near s = 1, \zeta(s) \sim \dfrac{1}{s-1}, so the product (s-1)\zeta(s) \to 1. The remaining factors evaluate at s=1 to \tfrac12\cdot 1\cdot\pi^{-1/2}\Gamma(\tfrac12) = \tfrac12\cdot\pi^{-1/2}\cdot\sqrt{\pi} = \tfrac12, using \Gamma(\tfrac12) = \sqrt\pi. So \xi(1) = \tfrac12, and by \xi(s) = \xi(1-s) also \xi(0) = \tfrac12.

Second, reality. Because \overline{\xi(\bar s)} = \xi(s) (all the building blocks are real on the real axis), \xi takes real values on the real axis. Combined with \xi(s) = \xi(1-s), it also takes real values on the critical line \Re(s) = \tfrac12: writing s = \tfrac12 + it, one checks \xi(\tfrac12 + it) is real for real t. This is precisely what lets us hunt for zeros on the critical line by watching a real function change sign — the practical engine behind every large-scale zero computation.