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Library · docs/03-functional-equation.md

Riemann's Functional Equation, Derived Line by Line

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The short version

Riemann's functional equation is not a coincidence you verify afterwards; it is a change of variable. Write Gamma(s/2) pi^(-s/2) n^(-s) as an integral of exp(-pi n^2 x) against x^(s/2-1), sum over n, and zeta(s) becomes the Mellin transform of the theta function theta(x) = sum_{n in Z} exp(-pi n^2 x). Theta has the hidden self-similarity proved in docs/02-theta-heat-and-modularity.md — theta(1/x) = sqrt(x) theta(x) — so if you cut the integral at x = 1 and flip the short half onto the long half by x -> 1/x, you land on a formula whose right-hand side is visibly unchanged by s -> 1-s. That one formula simultaneously (a) analytically continues zeta to the whole plane, (b) proves the functional equation, and (c) shows the only poles are simple ones at s = 0 and s = 1. Everything else below — the completed function xi, the critical line, the trivial zeros, the Hadamard product — is bookkeeping on top of that single line.

Notation. Riemann's memoir (and Edwards) call the half-theta psi(x); here — as in docs/02 and zeta/core.py — it is omega(x), because psi is already taken by the Chebyshev prime-counting function in docs/04-explicit-formula.md and in zeta/explicit.py. Same object.

1. The one input: theta's transformation law

We import exactly one external fact.

THEOREM (Jacobi). For x > 0, with theta(x) = sum_{n = -inf}^{inf} exp(-pi n^2 x),

    theta(1/x)  =  sqrt(x) · theta(x)

It is convenient to drop the n = 0 term and fold the two symmetric tails:

    omega(x)  :=  (theta(x) - 1)/2  =  sum_{n >= 1} exp(-pi n^2 x)

Two properties of omega carry the whole argument.

Fast decay. For x > 0, since n^2 >= n,

    omega(x)  <=  sum_{n >= 1} exp(-pi n x)  =  exp(-pi x) / (1 - exp(-pi x))

so on x >= 1 we get omega(x) <= 1.0452 · exp(-pi x). Numerically omega(1) = 0.04321740560665... (consistent with theta(1) = 1.08643481121330801... quoted in docs/02) and omega(2) = 0.00186744274..., already within 10^(-11) of its single leading term exp(-2 pi).

Jacobi restated for omega. Substitute theta = 2 omega + 1 into theta(1/x) = sqrt(x) theta(x):

    2 omega(1/x) + 1  =  sqrt(x) (2 omega(x) + 1)

    ==>   omega(1/x)  =  sqrt(x) · omega(x)  +  (sqrt(x) - 1)/2                          (*)

That leftover (sqrt(x) - 1)/2 is not debris. It is exactly where the poles of zeta come from. Watch it.

2. Step one: n^(-s) is a Gaussian integral in disguise

Start from the definition Gamma(w) = integral_0^inf t^(w-1) e^(-t) dt, valid for Re w > 0. Put w = s/2 and substitute t = pi n^2 x, so dt = pi n^2 dx and t^(s/2 - 1) = (pi n^2 x)^(s/2 - 1):

    Gamma(s/2)  =  integral_0^inf (pi n^2 x)^(s/2 - 1) exp(-pi n^2 x) · pi n^2 dx
                =  (pi n^2)^(s/2) · integral_0^inf x^(s/2 - 1) exp(-pi n^2 x) dx

Divide by (pi n^2)^(s/2) = pi^(s/2) n^s:

    pi^(-s/2) Gamma(s/2) n^(-s)  =  integral_0^inf x^(s/2 - 1) exp(-pi n^2 x) dx,    Re s > 0    (1)

Why this is the right move, not just a legal one. On the left is one term of the Dirichlet series with a Gamma factor bolted on. On the right is an object that knows about a lattice: exp(-pi n^2 x) is a fixed Gaussian sampled at the integer n. Summing over n converts an arithmetic sum into a geometric one, and geometry is where the symmetry hides. The Mellin transform is the right currency for the reason docs/01-sums-integrals-and-continuation.md §4 gives: it is the Laplace transform of the multiplicative group (0, inf), whose defining involution is x -> 1/x.

3. Step two: sum over n

Take Re s > 1 and sum (1) over n >= 1:

    pi^(-s/2) Gamma(s/2) zeta(s)  =  sum_{n>=1} integral_0^inf x^(s/2-1) exp(-pi n^2 x) dx

                                  =  integral_0^inf x^(s/2 - 1) omega(x) dx                  (2)

Why the interchange is legal. For real s > 1 every term is positive, so Tonelli applies with no work. For complex s, run the same argument on |x^(s/2-1)| = x^(Re s/2 - 1) to get absolute convergence, then Fubini.

Where (2) converges. At the top end, omega decays like e^(-pi x), so there is no constraint. At the bottom end, Jacobi gives theta(x) = x^(-1/2) theta(1/x) and theta(1/x) -> 1 as x -> 0+, so omega(x) ~ (1/2) x^(-1/2). Then integral_0 x^(Re s/2 - 1) x^(-1/2) dx converges precisely when Re s > 1. The obstruction to continuation is entirely at x = 0, and it is a single explicit power of x. That is a solvable problem.

The left-hand side of (2) deserves its own name: it is zeta completed by its archimedean factor pi^(-s/2) Gamma(s/2). The Euler product supplies one factor per prime; the Gamma factor is the missing factor at the "prime at infinity". Nothing about zeta looks symmetric until you put it in.

4. Step three: split at x = 1 and fold

Split (2) at the fixed point of x -> 1/x:

    pi^(-s/2) Gamma(s/2) zeta(s)  =  integral_0^1 x^(s/2-1) omega(x) dx
                                   + integral_1^inf x^(s/2-1) omega(x) dx

The second integral is already harmless. Because omega(x) <= 1.05 e^(-pi x) on [1, inf), it converges for every complex s and defines an entire function of s (differentiate under the integral sign; the exponential decay dominates the polynomial-in-log x growth of x^(s/2-1) locally uniformly in s).

Now attack the first integral with x = 1/u, dx = -du/u^2:

    integral_0^1 x^(s/2-1) omega(x) dx  =  integral_1^inf u^(-s/2+1) omega(1/u) u^(-2) du
                                        =  integral_1^inf u^(-s/2-1) omega(1/u) du

and substitute the restated Jacobi identity (*), omega(1/u) = sqrt(u) omega(u) + (sqrt(u)-1)/2:

    =  integral_1^inf u^(-s/2 - 1/2) omega(u) du
     + (1/2) integral_1^inf ( u^(-s/2 - 1/2) - u^(-s/2 - 1) ) du

The first piece is integral_1^inf u^((1-s)/2 - 1) omega(u) du — the same shape as the tail integral we already have, with s replaced by 1-s. That is the whole trick, and it is worth pausing on: the inversion x -> 1/x on the theta side has become the reflection s -> 1-s on the zeta side, because the Mellin kernel x^s turns multiplicative inversion into additive negation, and the sqrt(x) in Jacobi's law shifts the negation's centre from 0 to 1/2.

The second piece is elementary. For Re s > 1 both exponents are < -1, so

    integral_1^inf u^(-s/2-1/2) du = 2/(s-1),        integral_1^inf u^(-s/2-1) du = 2/s

giving (1/2)(2/(s-1) - 2/s) = 1/(s-1) - 1/s = 1/(s(s-1)). Assemble everything:

    pi^(-s/2) Gamma(s/2) zeta(s)  =  1/(s(s-1))
                                   + integral_1^inf ( x^(s/2 - 1) + x^((1-s)/2 - 1) ) omega(x) dx   (3)

We derived (3) assuming Re s > 1. Nothing on the right-hand side cares.

5. Step four: read three theorems off one formula

Stare at (3). Do not compute anything.

Continuation (THEOREM). The integral converges for all s in C and is entire. So the whole right-hand side is meromorphic on C and agrees with the left-hand side on Re s > 1. By the identity theorem the two continuations coincide, so (3) is the continuation — a closed form, not an algorithm. Compare docs/01, where Euler–Maclaurin got us the same continuation by a completely different and far more laborious route.

Poles (THEOREM). The only non-analytic ingredient is 1/(s(s-1)): simple poles at s = 0 and s = 1, nowhere else. Now split the credit. Gamma(s/2) has its own simple poles at s = 0, -2, -4, .... The pole of the left side at s = 0 is therefore supplied by Gamma, so zeta need not be singular there — and is not. The pole at s = 1 is not supplied by Gamma, so it must belong to zeta, and it is simple with residue 1. (pi^(-1/2) Gamma(1/2) = 1, so the residue of the left side at s=1 is the residue of zeta, and the right side has residue 1 there.) That pole is the primes' fingerprint, as docs/01 §1 argues.

THEOREM (Riemann's functional equation). Replace s by 1-s throughout (3). Inside the integral, x^(s/2-1) and x^((1-s)/2-1) simply swap places, so the integral is unchanged. And 1/(s(s-1)) is unchanged, because

    (1-s)((1-s) - 1)  =  (1-s)(-s)  =  s(s-1)

The right-hand side is therefore identically the same function of s. Hence

    pi^(-s/2) Gamma(s/2) zeta(s)  =  pi^(-(1-s)/2) Gamma((1-s)/2) zeta(1-s)

No computation was performed. The symmetry is a typographical property of (3), inherited straight from the x <-> 1/x symmetry of the lattice Z inside R. This is why Riemann's second proof (the theta proof, 1859) is the one everyone teaches.

6. The completed zeta function

Multiply through by s(s-1) to clear both poles at once, and normalise:

    xi(s)  :=  (1/2) s (s-1) pi^(-s/2) Gamma(s/2) zeta(s)

THEOREM. xi is entire, satisfies xi(s) = xi(1-s), and has order 1.

Entirety and the symmetry are immediate from (3), which also gives the completely explicit form

    xi(s)  =  1/2  +  (1/2) s(s-1) · integral_1^inf ( x^(s/2-1) + x^((1-s)/2-1) ) omega(x) dx

and hence xi(0) = xi(1) = 1/2 for free. (Numerically xi(1/2) = 0.497120778188314....)

"Order 1" means |xi(s)| <= exp(C |s|^(1+eps)) for every eps > 0 and no smaller exponent works; the true growth along the real axis is exp(c |s| log|s|), driven by Gamma(s/2). So xi has order exactly 1 but infinite type — it is not of finite exponential type. That distinction is not pedantry: it is what forces the convergence factors in §9.

One further consequence, small to state and enormous in practice. zeta is real on the real axis, so Schwarz reflection gives zeta(conj s) = conj(zeta(s)), and the same for xi. Combine with the functional equation:

    conj( xi(1/2 + i t) )  =  xi(1/2 - i t)  =  xi( 1 - (1/2 - i t) )  =  xi(1/2 + i t)

so xi(1/2 + it) is real for real t (verified numerically to 25 digits: the imaginary part at t = 1, 5, 30 is below 10^(-26)). Hunting zeros on the critical line therefore reduces to finding sign changes of a real function of a real variable. That is exactly what Hardy's Z(t) does; see zeta.zeros.Z, zeta.zeros.rs_theta, and zeta.zeros.zeros_by_sign_change.

7. Why the critical line is Re(s) = 1/2

Be precise here, because the usual one-liner is slightly off. The map sigma: s -> 1-s is rotation by 180 degrees about the point 1/2. Its only fixed point is s = 1/2; by itself it does not distinguish a line. The line appears when you use both of xi's symmetries together. xi is invariant under the group generated by

    sigma : s -> 1 - s          (the functional equation)
    kappa : s -> conj(s)        (reality on the real axis)

which is a Klein four-group { id, sigma, kappa, sigma·kappa }. The composite sigma·kappa : s -> 1 - conj(s) is a genuine mirror reflection of the plane, and its fixed set is the entire vertical line Re(s) = 1/2. That is the exact sense in which the critical line is the mirror axis of xi. (docs/02 §5 phrases this as "the fixed line of s -> 1-s"; the statement is the same once reality is folded in, but the reflection doing the work is s -> 1 - conj(s).)

Consequence. If rho is a zero of xi, so are 1-rho, conj(rho), and 1-conj(rho). Zeros come in quadruples

    { rho,  1 - rho,  conj(rho),  1 - conj(rho) }

generically four distinct points forming a rectangle centred at s = 1/2, symmetric about both the critical line and the real axis. The quadruple degenerates to a pair {rho, conj(rho)} exactly when rho already sits on the mirror, i.e. Re(rho) = 1/2.

CONJECTURE (Riemann Hypothesis, 1859). Every quadruple degenerates: all non-trivial zeros have Re(rho) = 1/2.

So RH is not "the zeros happen to prefer a line". It is the assertion that the zeros are as symmetric as the function that produces them — that nothing ever breaks the mirror.

Locating the strip is already a theorem. The Euler product gives zeta(s) != 0 for Re s > 1; the functional equation transports that to Re s < 0 (where the only zeros are the trivial ones of §8); and THEOREM (Hadamard and de la Vallée Poussin, independently, 1896) zeta(1 + it) != 0 for real t, which by the functional equation also closes Re s = 0. Hence every non-trivial zero lies in 0 < Re s < 1. RH asks for the middle of a strip we can already fence.

Counting them is also a theorem, not a conjecture. THEOREM (Riemann–von Mangoldt). The number N(T) of zeros with 0 < Im rho < T satisfies

    N(T)  =  (T/2pi) log(T/2pi)  -  T/2pi  +  7/8  +  S(T)  +  O(1/T)

with S(T) the argument term (typically small, provably unbounded). Sanity check: at T = 100 the elementary terms give 29.0023..., and there are indeed exactly 29 zeros below height 100. Code: zeta.zeros.riemann_von_mangoldt, zeta.zeros.N_of_T, zeta.zeros.S_of_T, and zeta.zeros.verify_rh_up_to.

8. Where the trivial zeros come from, and why they are "trivial"

Solve the definition of xi for zeta:

    zeta(s)  =  2 xi(s) pi^(s/2) / ( s (s-1) Gamma(s/2) )

Gamma(s/2) has simple poles at s/2 = 0, -1, -2, ..., i.e. at s = 0, -2, -4, -6, .... A pole of Gamma(s/2) sitting in the denominator forces a zero of zeta — unless something cancels it. At s = 0 the factor s in the denominator does cancel it, which is exactly why zeta(0) = -1/2 is finite and nonzero. At s = -2, -4, -6, ... nothing cancels, so

    zeta(-2) = zeta(-4) = zeta(-6) = ... = 0

These are the trivial zeros. "Trivial" does not mean unimportant — they contribute a genuine term to the explicit formula in docs/04-explicit-formula.md. It means forced: they are artifacts of the archimedean Gamma factor, predictable without knowing anything whatsoever about the primes. Note that xi has none of them; it was constructed precisely to absorb them. The zeros xi keeps are the interesting ones.

9. The Hadamard product: the bridge to the explicit formula

Because xi is entire of order 1, Hadamard's factorisation theorem applies (Hadamard, commonly dated 1893, in the paper where he studied this very function). The Riemann–von Mangoldt count gives N(T) ~ (T/2pi) log T, from which sum_rho 1/|rho| diverges while sum_rho 1/|rho|^2 converges. Divergence of the first sum is exactly why a bare product prod (1 - s/rho) cannot converge; convergence of the second says genus-1 elementary factors suffice:

    xi(s)  =  e^(A + B s) · prod_rho ( 1 - s/rho ) e^(s/rho)

the product taken over all non-trivial zeros (paired as rho with 1-rho, or equivalently rho with conj(rho), so the product converges). The constants are explicit:

    e^A = xi(0) = 1/2                    ==>  A = -log 2

    B  =  xi'(0)/xi(0)  =  (1/2) log(4 pi) - 1 - gamma/2  =  -0.023095708966121...

with gamma the Euler–Mascheroni constant. B also equals -sum_rho Re(1/rho). I checked both: the closed form matches a numerical xi'(0)/xi(0) to 17 digits, and sum_rho Re(1/rho) over the first 400 zeros gives 0.021766, short of 0.023096 by 0.00133, which is exactly the size the Riemann–von Mangoldt tail estimate predicts.

Now take the logarithmic derivative of the product, substitute into a contour integral of -(zeta'/zeta)(s) · x^s / s, and out falls the explicit formula relating prime powers to zeros. That is docs/04-explicit-formula.md, implemented in zeta/explicit.py (psi_from_zeros, pi_from_zeros, prime_spectrum, convergence_table). The functional equation is what makes the product symmetric; the product is what makes the primes audible.

10. An honest caveat: zeta satisfies no differential equation

It is tempting to hope that something this structured obeys a nice ODE, the way theta obeys the heat equation. It does not.

THEOREM (Hölder, 1887). Gamma satisfies no algebraic differential equation: there is no nonzero polynomial P with P(z, Gamma(z), Gamma'(z), ..., Gamma^(n)(z)) === 0. (O. Hölder, Ueber die Eigenschaft der Gammafunction keiner algebraischen Differentialgleichung zu genügen, Mathematische Annalen 28. Sources differ between 1886 and 1887 for the volume date; 1887 is the usual citation.)

THEOREM (Ostrowski, 1920). The same holds for zeta — it is differentially transcendental (hypertranscendental) over the rational functions. A. Ostrowski, Über Dirichletsche Reihen und algebraische Differentialgleichungen, Mathematische Zeitschrift 8 (1920). Secondary sources routinely describe this as settling an assertion of Hilbert's from around 1900 and credit Mordukhai-Boltovskoi with an independent proof; I have not checked those attribution claims against primary sources, so treat the story as hearsay and the theorem as solid.

So there is no hidden finite-order ODE for zeta waiting to be integrated. Any structural handle has to come from somewhere else — the Euler product, the functional equation, or a spectral interpretation.

CURIOSITY — clearly labelled, NOT a working tool. R. A. Van Gorder, Does the Riemann zeta function satisfy a differential equation? (Journal of Number Theory, 2015), gave an elementary construction of an infinite-order linear differential relation with analytic coefficients,

    sum_{n >= 0}  a_n(s) · zeta^(n)(s)  =  b(s)

which evades Ostrowski because Ostrowski forbids only finite-order algebraic relations. Van Gorder was explicit that his relation was formal and claimed no region or mode of convergence. B. B. Prado and K. Klinger-Logan, Linear Operators, the Hurwitz Zeta Function and Dirichlet L-Functions (Journal of Number Theory 217 (2020), 422–442), subsequently showed that Van Gorder's operator applied to zeta does not converge pointwise at any point of the complex plane, and constructed a modified operator that does converge. Real mathematics, worth knowing about — but do not build anything on the original formal operator, and do not cite it as "zeta satisfies a differential equation" without every one of those qualifications.

The contrast worth internalising. Theta, unlike zeta, genuinely does satisfy a clean PDE. This is the whole content of docs/02: with

    u(z, x)  =  sum_{n in Z} exp(-pi n^2 x) exp(2 pi i n z),          u(0, x) = theta(x)

term-by-term differentiation gives d/dx contributing -pi n^2 and d^2/dz^2 contributing -4 pi^2 n^2, so

    du/dx  =  (1/(4 pi)) · d^2u/dz^2

the heat equation on the circle, with a periodic delta comb as initial data at x -> 0+. All of zeta's good behaviour is borrowed from theta's heat flow through the Mellin transform (2), and none of it survives on the zeta side as a differential equation. Borrowing it back — running heat flow on xi rather than using heat flow to build xi — is the De Bruijn–Newman story in docs/05-de-bruijn-newman.md and zeta/heatflow.py.

11. Check it yourself

The identity to test is (3): the direct product pi^(-s/2) Gamma(s/2) zeta(s) against the theta-integral right-hand side. zeta.core.theta_mellin_xi is the documented entry point — it computes both sides for a given s, so you can compare them anywhere you like, including inside the critical strip and at Re s < 0 where the left-hand side only exists because of the right-hand side.

Running that comparison at 30-digit precision at s = 3, 2 + i, 0.3 + 5i, -1.5 + 2.3i, and 1/2 + 14.134725...i, the two sides agree to within 10^(-31) in every case. The last is the most instructive: both sides evaluate to about 2 · 10^(-12) — the first non-trivial zero announcing itself in a formula that never mentions zeros at all.

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