RIEMANN’S FUNCTIONAL EQUATION KEVIN ZHU Abstract. This paper investigates the functional equation of the Riemann zeta function ζ (s). The functional equation is useful for a few reasons. It allows us to immediately conclude that the function has zeros at the negative even integers. In the process of a proof we shall follow, the analytic continuation of the zeta function also falls out as a consequence. Furthermore, having established the functional equation, we can find formulas for the derivatives of the zeta function. The functional equation has close ties with the Riemann hypothesis, playing a role in empirically searching for zeros on the critical line. 1. Deriving the Functional Equation The aim of the first paper we shall review is to present a short proof of Riemann’s functional equation, [3] Γ(s) πs (1.1) ζ (1 − s) = 2 cos ζ (s) : (2π)s 2 In fact, the method the paper undertakes is to prove the slightly more general functional equation on the Hurwitz zeta function. Riemann’s functional equation follows directly from this relation. The proof is based upon the Lipschitz summation formula, which itself is proved using Poisson summation, a technique from Fourier analysis. The meromorphic continuation of both the periodized zeta function (an- other component of the proofs) and the Hurwitz zeta function to the whole complex plane is a corollary of the results in this paper. Riemann’s original proofs for the relation, on the other hand, use either the theta function and its Mellin transform or contour integration. Comparatively, the proof we shall follow mostly gets away with manipulating infinite series and reasoning about the analytic functions that pop out. In the words of the authors, the paper is intended to provide guidance to readers “unfamiliar with the circle of ideas” related to ζ (s). Now let us set the stage for establishing the functional equation. Recall that the P1 −s Riemann zeta function is defined as the continuation of ζ (s) = n=1 n , initially defined on < (s) > 1, where the series converges absolutely by comparison to P n−p for p > 1. The Hurwitz zeta function is an extension of the Riemann zeta function, 1 RIEMANN’S FUNCTIONAL EQUATION 2 similarly defined as 1 X 1 ζ (s; a) = : (n + a)s n=0 Here we take 0 < a ≤ 1 and < (s) > 1. We have ζ (s; 1) = ζ (s). Another relevant extension of the Riemann zeta function is what we call the periodized zeta function 1 X e2πina F (s; a) = ; ns n=1 again defined initially for < (s) > 1, a 2 R. We also define the Fourier transform ^ ^ R 1 2πixm f of f to be f (m) = −∞ f (x) e dx, so that we can use Poisson summation, which says X X f (n) = f^(m) n2Z m2Z for “sufficiently nice” f. The specifics of this are not terribly important to the result we want. Poisson summation is the main tool in the first theorem we state, the Lipschitz summation formula. Theorem 1. 1 2πiαm X s−1 Γ(s) X e (1.2) (n − α) e2πiτ(n−α) = ; (−2πi)s (τ + m)s n=1 m2Z where < (s) > 1, = (τ) > 0, and 0 ≤ α < 1. The proof is to use the Poisson summation formula on the left-hand side of (1.2), defining the summand function f (x) = (x − α)s−1 e2πiτ(x−α) to be 0 for x ≤ α. This function is nice enough to use Poisson summation on, and we end up with Z 1 2πimα Z X 2πimα s−1 2πi(τ+m)x 1 X e s−1 −y e x e dx = s s y e dy; 0 (−2πi) (τ + m) R m2Z m2Z where R is the complex ray fy : y = −2πi (m + τ) x and x ≥ 0g. Then we employ a familiar integration technique using Cauchy’s integral theorem. Consider the integral along a piece of pie formed by going along the real axis for some radius, then following a circular arc counterclockwise about the origin, then coming back to the origin along R. Since ys−1e−y is analytic, this integral is 0, and as the radius grows, the integral along the arc decays to 0. Therefore, the improper integral may be taken along the real axis, and so the theorem follows by the definition of Γ(s). The next steps involve some results that are mostly bookkeeping to justify the steps we will take in proving the relation on the Hurwitz zeta function. The results are not terribly illuminating, but we state them for use in that proof. They are mainly about being able to interchange limiting operations. RIEMANN’S FUNCTIONAL EQUATION 3 Lemma. (a) Suppose 0 ≤ a < 1 and < (s) < 0. Then 1 1 X s−1 X s−1 lim (n − a) e2πinτ = (n − a) = ζ (1 − s; 1 − a) : τ!0 n=1 n=1 (b) Let 0 ≤ a < 1 and y > 0. Write X 2πiam −s −s −s−1 Sy (s) = e (m + iy) − m + siym ; m6=0 where the sum is taken over all nonzero integers and for the complex powers we use the principle branch of the argument, with range (−π; π]. Then: (i) Sy (s) converges absolutely for < (s) > −1; (ii) Sy (s) is holomorphic in s for < (s) > −1; (iii) if < (s) > −1, then limy!0+ Sy (s) = 0. With these in hand, we can prove the Hurwitz relation, from which Riemann’s functional equation will follow. Theorem 2. e−πis=2F (s; a) + eπis=2F (s; −a) can be continued analytically into < (s) > −1. When −1 < < (s) < 0 and 0 < a ≤ 1, we have the Hurwitz relation Γ(s) (1.3) ζ (1 − s; a) = e−πis=2F (s; a) + eπis=2F (s; −a) : (2π)s Proof. In (1.2), there is a (τ + m)−s factor. The idea is to subtract the first two terms of the binomial expansion of this factor, m−s + τsm−s−1. Moving the Γ(s) = (−2πi)s factor in (1.2) to the other side and then subtracting from (τ + m)−s for nonzero indices of the sum, we get 1 X −s + e2πiαm (τ + m) − m−s + τsm−s−1 τ s m6=0 s 1 2πiαm 2πiαm (−2πi) X s−1 X e X e = (n − α) e2πiτ(n−α) − + τs : Γ(s) ms ms+1 n=1 m6=0 m6=0 for 0 ≤ α < 1 and < (s) > 1. We note that 1 1 X e2πiαm X e2πiαm X e2πiα(−m) = + ms ms (−m)s m6=0 m=1 m=1 1 −2πiαm −s X e = F (s; α) + (−1) ms m=1 = F (s; α) + e−πisF (s; −α) RIEMANN’S FUNCTIONAL EQUATION 4 and take τ = iy, y > 0 to rewrite what we have as 1 X −s (1.4) + e2πiαm (m + iy) − m−s + siym−s−1 (iy)s m6=0 s 1 (−2πi) X s−1 = (n − α) e−2πy(n−α) Γ(s) n=1 − F (s; α) + e−πisF (s; −α) + iys F (s + 1; α) − e−πisF (s + 1; −α) : P1 s−1 −2πy(n−α) The sum n=1 (n − α) e is entire in s the exponential term causes s the series to converge normally on C. If we move the (−2πi) =Γ(s) factor back to the left-hand side, the Lemma tells us that the left-hand side is meromorphic in < (s) > −1 with at most a simple pole at s = 0 from Γ(s). This tells us that the remaining terms Γ(s) iysΓ(s) − F (s; α) + e−πisF (s; −α)+ F (s + 1; α) − e−πisF (s + 1; −α) (−2πi)s (−2πi)s are meromorphic in < (s) > −1 with at most a simple pole at s = 0. Since y > 0 is arbitrary, we find that Γ(s) − F (s; α) + e−πisF (s; −α) (−2πi)s is meromorphic in < (s) > −1 with at most a simple pole at s = 0 by letting y ! 0+, since the other term vanishes uniformly on compact sets. This establishes the first part of the theorem. Now for −1 < < (s) < 0, let y ! 0+ in (1.4). The left-hand side vanishes by the Lemma and the restriction of < (s) and the sum on the right-hand side tends to ζ (1 − s; 1 − α). Hence, for 0 ≤ α < 1 we have Γ(s) ζ (1 − s; 1 − a) − F (s; α) + e−πisF (s; −α) = 0: (−2πi)s in −1 < < (s) < 0. Letting a = 1 − α gives (1.3) since F (s; 1 − a) = F (s; −a) and F (s; −1 + a) = F (s; a). Since F (s; 1) = F (s; −1) = ζ (s), 1.3 yields Γ(s) Γ(s) πs ζ (1 − s) = e−πis=2 + eπis=2 ζ (s) = 2 cos ζ (s) ; (2π)s (2π)s 2 which is the functional equation (1.1) that we want. 2. Using the Functional Equation [3] goes on to show that we can analytically continue ζ (s; a) and F (s; a) using the work done so far. The first supplementary result is that F (0; −a) + F (0; a) = −1, which we obtain by taking s ! 0− in 1.4 and using the Lemma. We also note that RIEMANN’S FUNCTIONAL EQUATION 5 lims!0− F (s; a) is finite for this to work, because a bit of manipulation shows that πis −πis πis −πis lims!0− e − e F (s; a) = 0. Since e − e has a simple zero at s = 0, the order of F (s; a) at s = 0 cannot be negative (because the order of the product is the sum of the orders); i.e.