Analytic Continuation of ζ(s)

The series \zeta(s) = \sum_{n\ge1} n^{-s} is a well-behaved citizen only in the half-plane \Re(s) > 1. Step even slightly to the left — say to s = \tfrac12, or s = 0, or s = -1 — and the sum simply stops making sense: its terms no longer shrink, and it diverges. And yet mathematicians cheerfully write down \zeta(0) = -\tfrac12 and even the notorious \zeta(-1) = -\tfrac{1}{12}. How can a divergent sum have a value?

The answer is analytic continuation: the astonishing rigidity of complex-analytic functions. If two analytic functions agree on any little patch of the plane, they agree everywhere both are defined. So there is at most one analytic function that equals \sum n^{-s} on the half-plane where that sum converges — and once we find a formula for it that keeps working further left, that formula is zeta there, whether the original series cooperates or not. On this page we build the extension explicitly, twice, and read off the famous special values it forces on us.

Why continuation is even allowed

The whole enterprise rests on one theorem you should keep in mind throughout.

Consequence: an analytic function's values on a tiny region already determine it everywhere it can be continued. There is nothing to "choose". The continued \zeta is forced — unique — the moment we know it on \Re(s) > 1. Our job is only to find a formula for it that survives past that wall.

First extension: peeling the sum with Euler–Maclaurin

The trick is to compare the sum \sum n^{-s} with the integral \int_1^\infty t^{-s}\,dt — they behave almost identically, and the tiny difference between them is far better behaved than either. The clean way to bottle that difference is Abel summation (partial summation), which for a smooth f gives the exact identity

\sum_{n=1}^{N} f(n) = \int_1^{N} f(t)\,dt + \tfrac12\bigl(f(1)+f(N)\bigr) + \int_1^{N}\Bigl(\{t\}-\tfrac12\Bigr) f'(t)\,dt,

where \{t\} = t - \lfloor t\rfloor is the fractional part. Take f(t) = t^{-s}, so f'(t) = -s\,t^{-s-1}, and let N \to \infty. For \Re(s) > 1 the boundary term f(N) \to 0 and everything converges. Doing the bookkeeping (adding back the n=1 term and evaluating \int_1^\infty t^{-s}\,dt = \tfrac1{s-1}) collapses to the star of the show:

\zeta(s) = \frac{s}{s-1} - s\int_1^{\infty}\frac{\{t\}}{t^{s+1}}\,dt,\qquad \Re(s) > 0,\ s\neq 1.

Here is the magic. On the left, the series only made sense for \Re(s) > 1. On the right, the integral converges whenever \Re(s) > 0: the fractional part \{t\} is bounded between 0 and 1, so \int_1^\infty \{t\}\,t^{-s-1}\,dt is dominated by \int_1^\infty t^{-\Re(s)-1}\,dt, finite for \Re(s) > 0. The two sides agree on the overlap \Re(s) > 1, so by rigidity the right-hand side is \zeta on the whole strip 0 < \Re(s) \le 1 too. We have moved the wall.

The single simple pole at s = 1

Look hard at the formula. The integral term is a perfectly finite, analytic function everywhere on \Re(s) > 0. Every trace of bad behaviour is quarantined in the one explicit fraction \dfrac{s}{s-1}, which blows up only at s = 1. So:

Near the pole, \zeta(s) = \dfrac{1}{s-1} + \gamma + O(s-1), where the constant term \gamma \approx 0.5772 is the Euler–Mascheroni constant — the exact same \gamma that measures the gap between the harmonic series and \ln N. That is no coincidence: the pole is the ghost of the diverging harmonic series \zeta(1), and its leftover finite part is \gamma.

Worked example: confirming the pole and residue

Let us watch the formula do its job right next to the danger point. Write s = 1 + \varepsilon for a small real \varepsilon > 0 and read off both terms of \zeta(s) = \dfrac{s}{s-1} - s\displaystyle\int_1^\infty \{t\}\,t^{-s-1}\,dt.

The explicit fraction. With s - 1 = \varepsilon,

\frac{s}{s-1} = \frac{1+\varepsilon}{\varepsilon} = \frac{1}{\varepsilon} + 1.

The integral. As \varepsilon \to 0 it tends to the finite constant \int_1^\infty \{t\}\,t^{-2}\,dt = 1 - \gamma \approx 0.4228, and s \to 1, so the whole term approaches -(1-\gamma). Putting the pieces together,

\zeta(1+\varepsilon) = \frac{1}{\varepsilon} + 1 - (1-\gamma) + O(\varepsilon) = \frac{1}{\varepsilon} + \gamma + O(\varepsilon).

Multiply by (s-1) = \varepsilon and let \varepsilon \to 0: the product is 1 + \gamma\varepsilon + \cdots \to 1. The residue is 1, on the nose, and the constant term is \gamma — exactly as claimed. We can also check numerically that a computed \zeta(1.001) is close to 1000 + \gamma \approx 1000.577:

// Zeta just right of the pole via the continued formula: // ζ(s) = s/(s-1) − s·∫₁^∞ {t} t^(−s−1) dt function fracPart(t: number): number { return t - Math.floor(t); } function zetaContinued(s: number): number { // numerically integrate the {t}·t^(−s−1) tail out to a large cutoff let integral = 0; const dt = 0.001; for (let t = 1; t < 4000; t += dt) { integral += fracPart(t + dt / 2) * Math.pow(t + dt / 2, -s - 1) * dt; } return s / (s - 1) - s * integral; } const s = 1.001; console.log("ζ(1.001) via formula ≈", zetaContinued(s).toFixed(4)); console.log("1/(s−1) + γ ≈", (1 / (s - 1) + 0.5772157).toFixed(4)); console.log("(s−1)·ζ(s) ≈", ((s - 1) * zetaContinued(s)).toFixed(4), " → residue 1");

The last line is the residue check: (s-1)\zeta(s) sits right on top of 1. The pole is real, it is simple, and its residue is 1.

A second route: the Dirichlet eta function

There is a slicker (if less quantitative) way to reach \Re(s) > 0, and it makes the disappearance of the pole vivid. Alternate the signs of the zeta series to form the Dirichlet eta function:

\eta(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}} = 1 - \frac{1}{2^{s}} + \frac{1}{3^{s}} - \frac{1}{4^{s}} + \cdots

Because the signs alternate, this series converges (conditionally) for all \Re(s) > 0 by the alternating-series/Dirichlet test — no wall at s = 1 at all. And it relates to \zeta by a short computation: strip out the even terms, which are 2\sum (2m)^{-s} = 2^{1-s}\zeta(s),

\eta(s) = \sum \frac{1}{n^{s}} - 2\sum_{n\ \text{even}} \frac{1}{n^{s}} = \zeta(s) - 2\cdot 2^{-s}\zeta(s) = \bigl(1 - 2^{1-s}\bigr)\zeta(s). \eta(s) = \bigl(1 - 2^{1-s}\bigr)\,\zeta(s)\qquad\Longrightarrow\qquad \zeta(s) = \frac{\eta(s)}{1 - 2^{1-s}}.

Solving for \zeta gives a formula valid on all of \Re(s) > 0: the numerator \eta(s) is a convergent (hence analytic) series there, and we divide by the explicit factor 1 - 2^{1-s}. That factor vanishes exactly when 2^{1-s} = 1, i.e. 1 - s = \tfrac{2\pi i k}{\ln 2} for integer k. At k = 0 that is s = 1 — our pole, reappearing as a zero of the denominator. (The other k would-be poles are cancelled by zeros of \eta; only s = 1 survives, matching the previous section.) Two completely different constructions, one and the same function — that is rigidity in action.

All the way: the functional equation closes the plane

Both routes so far only reach \Re(s) > 0. To cover the rest — the whole left half-plane — we use the crowning symmetry, the functional equation, which Riemann proved links s to its mirror image 1 - s:

\zeta(s) = 2^{s}\pi^{s-1}\sin\!\left(\frac{\pi s}{2}\right)\Gamma(1-s)\,\zeta(1-s).

Read it as a machine. To learn \zeta at some point with \Re(s) < 0, feed the equation the value at 1 - s, which has \Re(1-s) > 1 — where the plain old series already works. The Gamma factor \Gamma(1-s) supplies the analytic glue. So knowing \zeta on the right half-plane, the functional equation reflects it onto the left, and \zeta now lives on all of \mathbb{C} except the single pole at s = 1. The continuation is complete.

Seeing the continued zeta on the real line

Now that \zeta is defined for (almost) every real s, we can plot it. Below is the continued function across -4 < s < 4, computed from the eta relation on the right and the functional equation on the left (clipped near the pole so the picture stays finite). Three landmarks jump out:

Between s = 0 and the pole the curve dips down through \zeta(-1) = -\tfrac{1}{12} and \zeta(0) = -\tfrac12, then for s < -1 it oscillates with growing swings whose zeros land precisely on the negative even integers. Everything you see to the left of s = 1 is continuation, not summation — the raw series diverges at every one of these points.

The famous special values

Plugging the functional equation (and known Gamma / Bernoulli data) into specific integers gives the celebrity values of zeta. State them and keep them:

The trivial zeros come for free from the functional equation: the factor \sin(\pi s/2) is zero at every even integer. At the positive evens this is harmless (the pole of \Gamma(1-s) cancels it), but at the negative evens s = -2, -4, \dots nothing cancels it, so \zeta must vanish. And \zeta(-1) = -\tfrac1{12} drops out of \zeta(-1) = -B_2/2 = -(1/6)/2. Which brings us to the most misunderstood equation in mathematics.

You have surely seen the clickbait: "1 + 2 + 3 + 4 + \cdots = -\tfrac{1}{12}". Taken literally, that is false. The series 1 + 2 + 3 + \cdots genuinely diverges — its partial sums are 1, 3, 6, 10, 15, \dots, marching off to +\infty and never settling anywhere near a small negative fraction.

What is true is a subtler statement. The formal object \sum n = \sum n^{-(-1)} is "what the zeta series would be at s = -1", and the analytic continuation of zeta — the unique analytic function agreeing with the sum where it converges — takes the value -\tfrac1{12} at s = -1. So -\tfrac1{12} is the regularised value assigned to that divergent sum by continuation, not a literal total. Analytic continuation extends the function; it does not make the series converge. The number -\tfrac1{12} earns its keep because it is the value physics and number theory actually need (it appears, correctly, in the Casimir effect and string theory) — but write "\zeta(-1) = -\tfrac1{12}", not "1+2+3+\cdots = -\tfrac1{12}", if you want to be right.

No — and that is the beautiful part. You might worry that our integral formula and the eta formula and the functional equation are three different "extensions" that could disagree. Rigidity forbids it. Any analytic function equal to \sum n^{-s} on the convergence half-plane must agree with all the others everywhere they overlap, and hence everywhere on the connected domain \mathbb{C}\setminus\{1\}. There is exactly one continued zeta. The different formulas are not competing definitions; they are three windows onto the same, unique object, each valid on a different region and all stitched together into one seamless function.

Where this is heading

We now have \zeta as a genuine function on the whole plane bar one point — the object on which the entire theory rests. Its zeros are the prize: the trivial ones at -2, -4, \dots we have already located, and the non-trivial ones, hiding in the critical strip 0 < \Re(s) < 1, are the subject of the Riemann Hypothesis. Continuation was the price of admission; the deep structure — and its link to the primes — is what it buys.