advanced28 min studyLesson 584 in Mathematics

Riemann zeta function and analytic continuation

Continue zeta beyond its defining series while keeping poles, gamma factors, and special values distinct.

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What you’ll understand

  • Derive continuation to the positive-real-part half-plane by partial summation.
  • Interpret the functional equation as a meromorphic identity.
  • Distinguish continued special values from divergent ordinary series.

The zeta series converges absolutely to the right of one, but the function it defines has a much larger domain. Continuing the function exposes a pole, a symmetry, and zeros that the original sum cannot directly show. A continued value is determined by agreement on overlapping domains and the identity theorem. It is not permission to evaluate a divergent series by its usual summation rule.

The pole emerges from a floor function

In the original half-plane, partial summation expresses zeta as s∫1∞⌊x⌋x−s−1 dxs\int_1^\infty\lfloor x\rfloor x^{-s-1}\,dx. Split the floor function into the variable and minus its fractional part. The integral of the variable evaluates explicitly, leaving a bounded numerator in the remaining improper integral. This remainder converges absolutely and locally uniformly when Re⁡s>0\operatorname{Re}s>0, so it defines a holomorphic function there. Agreement where the original series converges proves that this is its unique continuation.

The rational term has a simple pole at one with residue one, while the remainder is holomorphic near that point. This identifies both the location and the order of the singularity. It also illustrates the difference between the convergence domain of a representation and the analytic domain of a function: the floor-remainder representation reaches further left than the series, but does not yet reach the entire complex plane. A new identity will extend it further without redefining any previously determined value.

ζ(s)=ss−1−s∫1∞{x}x−s−1 dx(Re⁡s>0, s≠1).\zeta(s)=\frac{s}{s-1}-s\int_1^\infty\{x\}x^{-s-1}\,dx\qquad(\operatorname{Re}s>0,\ s\ne1).
The fractional part is bounded, so the remainder integral continues zeta and isolates its residue at one.↔ Scroll wide equations horizontally.

Symmetry after completing the function

The gamma function and a power of pi supply the factors needed for reflection across the central line. Define the completed expression Λζ(s)=π−s/2Γ(s/2)ζ(s)\Lambda_\zeta(s)=\pi^{-s/2}\Gamma(s/2)\zeta(s); the subscript distinguishes it from the arithmetic von Mangoldt function. It satisfies Λζ(s)=Λζ(1−s)\Lambda_\zeta(s)=\Lambda_\zeta(1-s) as a meromorphic identity and has simple poles at zero and one. Multiplying by the polynomial that cancels those poles yields an entire function with the same reflection symmetry.

One route to this equation takes a Mellin transform of a Gaussian theta sum and uses its transformation under reciprocal scaling. The Gaussian decay permits justified integration in an initial half-plane; splitting the integral and using the transformation reveals the reflected argument. This sketches the mechanism, rather than deriving Poisson summation here. An equivalent form is ζ(s)=2sπs−1sin⁡(πs/2)Γ(1−s)ζ(1−s)\zeta(s)=2^s\pi^{s-1}\sin(\pi s/2)\Gamma(1-s)\zeta(1-s), interpreted meromorphically. At apparent singularities one takes limits, rather than multiplying an undefined infinity by zero.

ξ(s)=12s(s−1)π−s/2Γ(s/2)ζ(s),ξ(s)=ξ(1−s),ξ is entire.\xi(s)=\frac12s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s),\qquad \xi(s)=\xi(1-s),\qquad \xi\text{ is entire}.
The polynomial cancels the two poles of the completed zeta expression; gamma poles and zeta zeros must be combined through limits.↔ Scroll wide equations horizontally.

Special values and the meaning of continuation

At each negative even integer the sine factor in the second functional-equation form vanishes while the remaining factors are finite and nonzero. These are the trivial zeros. At negative odd integers, the reflected zeta value lies in its absolutely convergent domain, so the equation provides a concrete continued value using ordinary gamma values. Such computations depend on an analytic identity whose proof starts in a justified domain, not on an algebraic rearrangement of divergent sums of positive integers.

Conjugation gives ζ(s‾)=ζ(s)‾\zeta(\overline{s})=\overline{\zeta(s)}, since this identity holds in the series domain and continues uniquely. Combined with reflection for the completed function, it organizes the nontrivial zeros in the strip between zero and one into symmetric configurations, counted with multiplicity. The statement that these zeros all lie on the central line is the Riemann hypothesis; results conditional on that statement must be labeled conditional. Neither analytic continuation itself nor the functional equation forces every zero to occupy that line.

Evaluate a negative argument correctly

Calculate the continued value of zeta at minus one from the sine-gamma functional equation, using the ordinary convergent value at two.

  1. Set s=−1s=-1. The elementary factors become 2−12^{-1}, π−2\pi^{-2}, and sin⁡(−π/2)=−1\sin(-\pi/2)=-1. The gamma factor is Γ(2)=1\Gamma(2)=1, which is finite.

  2. The reflected argument is 1−s=21-s=2, where the original zeta series converges and ζ(2)=π2/6\zeta(2)=\pi^2/6. Multiplying the finite factors gives −(1/2)π−2(π2/6)=−1/12-(1/2)\pi^{-2}(\pi^2/6)=-1/12.

  3. The series that would formally substitute minus one into the original definition has positive terms 1,2,3,…1,2,3,\ldots and divergent partial sums. The negative rational number is a value of the continued function, not the ordinary sum of those terms.

The result

The analytic continuation satisfies zeta of minus one equals minus one twelfth; the ordinary positive-integer series remains divergent.

Explore the model

Interactive lab

Interactive lab · heat

Gaussian decay and mode scaling

Pause/Reset and set mode 1 and diffusivity 1. At time 1, reached by two Steps, predict norm sqrt(pi) times exp(-1). For a fresh comparison Pause/Reset and set both mode 2 and diffusivity 0.25; after two Steps predict the same norm because diffusivity times mode squared is unchanged.

Gaussian decay and mode scalingThe exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Current readings: Current amplitude: 1; Spatial L² norm: 1.772; Decay rate · κn²: 3.6 /s. All parameters are adjustable with the labeled controls below.Periodic heat equation mode 3 at time 0: the amplitude is 1.0π/2π3π/22π−1−0.500.51x · periodic intervalTemperature perturbation uExact mode · n = 3
Current temperature modeInitial sineCurrent amplitude
Current amplitude
1
Spatial L² norm
1.772
Decay rate · κn²
3.6 /s
t = 0 s

The exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forQuadratic spectral decay underlies Gaussian and theta constructions. This instrument shows a single exact periodic heat mode, not the theta sum or a numerical evaluation of zeta.

Check your understanding

Which statement about the value zeta of minus one is correct?

Try it yourself

Determine the continued value of zeta at minus two and the value of the entire function xi at one. Account for cancellations at the latter point.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

At minus two, inspect the sine factor and the finiteness of the other factors. At one, use the residue of zeta instead of substituting its pole directly.

Compare with the solution

At s=−2s=-2, the sine is zero, while Γ(3)\Gamma(3) and ζ(3)\zeta(3) are finite, so ζ(−2)=0\zeta(-2)=0. As s→1s\to1, the product (s−1)ζ(s)(s-1)\zeta(s) tends to one. In the definition of xi, the other factors tend to s=1s=1 and π−1/2Γ(1/2)=1\pi^{-1/2}\Gamma(1/2)=1, giving ξ(1)=1/2\xi(1)=1/2. Directly treating the polynomial zero and zeta pole as unrelated values would conceal this removable singularity.

Take it with you

  • Partial summation isolates the simple pole at one before the full functional equation is used.
  • Completed zeta is meromorphic; xi is entire after polynomial cancellation.
  • Continued special values do not change the convergence of the original series.

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