Vanishing of the integral of the Hurwitz zeta function Draft 14th June 2001 Kevin A. Broughan University of Waikato, Hamilton, New Zealand E-mail: [email protected] A proof is given that the improper Riemann integral of ³(s; a) with respect to the real parameter a, taken over the interval (0; 1], vanishes for all complex s with <(s) < 1. The integral does not exist (as a finite real number) when <(s) ¸ 1. Key Words: Hurwitz zeta function, functional equation, improper Riemann integral. MSC2000 11M35, 30E99. 1. INTRODUCTION A number of authors have considered mean values of powers of the mod- ulus of the Hurwitz zeta function ³(s; a), see [3, 4, 5, 6, 7]. In this paper, the mean of the function itself is considered. First a functional equation relating the Riemann zeta function to sums of the values of the Hurwitz zeta function at rational values of a is de- rived. This functional equation underlies the vanishing of the integral of the Hurwitz zeta function. Consider the values of the function at negative integers: B (a) ³(¡n; a) = ¡ n+1 ; n ¸ 0 n + 1 where Bn(a) is the n’th Bernoulli polynomial. The integral of the right hand side expression between 0 and 1 is zero for every n. This appears to be a side-effect of the properties of Bernoulli polynomials (namely for 0 n ¸ 2, Bn(0) = Bn(1) and Bn(x) = nBn¡1(x)), and nothing particularly intrinsic to the zeta function. However, as the theorem below will show, the integral vanishes at every value of the complex variable s to the left of 1 2 BROUGHAN the line <(s) = 1. The integral does not exist (as a finite real number), on or to the right of this line. 2. THE VANISHING THEOREM The theorem is proved through developing a number of lemmas. The first is a fundamental, yet easy to derive, functional equation. See also, for example, [2]. Lemma 2.1. For all integers k ¸ 1 and all s 2 C ¡ f1g Xk j ks³(s) = ³(s; ): k j=1 Proof. Consider the functional equation for the Hurwitz zeta function [1]: h 2Γ(s) Xk ¼s 2¼jh j ³(1 ¡ s; ) = cos( ¡ )³(s; ) k (2¼k)s 2 k k j=1 This formula holds for all s and all integers h; k with 1 · h · k. Set h = k and obtain 2Γ(s) ¼s Xk j ³(1 ¡ s) = ³(1 ¡ s; 1) = cos( ) ³(s; ) (2¼k)s 2 k j=1 Using the functional equation for the zeta function to write the left hand side in terms of ³(s): ¼s 2Γ(s) ¼s Xk j 2(2¼)¡sΓ(s) cos( )³(s) = cos( ) ³(s; ) 2 (2¼k)s 2 k j=1 so the formula follows for all points except zeros of cos(¼s=2) and poles of Γ(s). But then it must hold at these points also since each side represents an analytic function, except for s = 1. Corollary 2.1. If ³(s0) = 0 then for all integers k ¸ 1 X j ³(s ; ) = 0: 0 k 1·j·k;(j;k)=1 HURWITZ ZETA FUNCTION 3 Proof. Let ³(s0) = 0. If k = 1 then ³(s0; 1=1) = ³(s0) = 0 so assume it is true for all m < k. By the Lemma Xk j ³(s ; ) = 0: 0 k j=1 Divide the sum on the left up into groups of terms corresponding to indices (j; k) having the same gcd. By the inductive hypothesis, each of the groups with a common gcd greater than 1 will sum to zero. Omitting these terms we obtain the result of the corollary. Observation: It follows easily from the corollary that the sums of the values of the Hurwitz zeta function over the Farey fractions of a given order, other than zero, at a zero of zeta function, are all zero. Pn j 1 Lemma 2.2. If <(s) < 1 then limn!1 j=1 ³(s; n ) n = 0. Proof. By Lemma 2.1 Xn j 1 ns¡1³(s) = ³(s; ) : n n j=1 Hence Xn j 1 nσ¡1j³(s)j = j ³(s; ) j: n n j=1 σ¡1 So if σ < 1, limn!1 n j³(s)j = 0, and the lemma follows directly. Lemma 2.3. Let f : (0; 1] ! R be a bounded C1 function. Extend f to a Riemann integrable function on [0; 1] with f(0) = 0. If Xn j 1 lim f( ) = 0 n!1 n n j=1 R 1 then 0 f = 0, because, in this case, the integral is the limit of the given Riemann sums. Lemma 2.4. If σ = <(s) < 0 there exists a positive real number B = B(s) such that for all a 2 (0; 1], j³(s; a)j · B(s). 4 BROUGHAN Proof. Consider Hurwitz’ formula for the zeta function in terms of the periodic zeta function [1], namely: Γ(s) ³(1 ¡ s; a) = fe¡¼is=2F (a; s) + e¼is=2F (¡a; s)g (2¼)s where 0 < a · 1, 1 < σ and where X1 e2¼ina F (a; s) = : ns n=1 then Γ(1 ¡ s) ³(s; a) = fe¡¼i(1¡s)=2F (a; 1 ¡ s) + e¼i(1¡s)=2F (¡a; 1 ¡ s)g (2¼)1¡s for σ < 0. Hence jΓ(1 ¡ s)j j³(s; a)j · fe¡¼t=2jF (a; 1 ¡ s)j + e¼t=2jF (¡a; 1 ¡ s)jg (2¼)1¡σ jΓ(1 ¡ s)j X1 1 X1 1 · fe¡¼t=2 + e¼t=2 g (2¼)1¡σ n1¡σ n1¡σ n=1 n=1 jΓ(1 ¡ s)j ¼t = 2 cosh( )³(1 ¡ σ) = B(s) (2¼)1¡σ 2 Lemma 2.5. If 0 < σ < 1, there exists a positive real number B = B(s) such that for all a 2 (0; 1], 1 j³(s; a)j · + B(s): aσ Proof. Consider the following expression for the zeta function [1], valid for 0 < σ < 1 and all integers N ¸ 1, namely Z XN 1 (N + a)1¡s 1 x ¡ [x] ³(s; a) = + ¡ s dx: (n + a)s s ¡ 1 (x + a)s+1 n=0 N Then Z XN 1 (N + a)1¡σ 1 1 j³(s; a)j · + + jsj dx: (n + a)σ js ¡ 1j (x + a)1+σ n=0 N HURWITZ ZETA FUNCTION 5 Let N = 1 to derive the upper bound 1 1 (1 + a)1¡σ jsj j³(s; a)j · + + + aσ (1 + a)σ js ¡ 1j σ 1 = + B(s) aσ where we may take 2 jsj B(s) = 1 + + : js ¡ 1j σ Lemma 2.6. Let f : (0; 1] ! R be a C1 function. Let a positive real number M be such that, for some σ 2 (0; 1) M jf(x)j · xσ for all x. Then f is Riemann integrable (proper if f is bounded). If R Pn j 1 1 limn!1 j=1 f( n ) n = 0, then 0+ f = 0. Proof. Let σ1 be such that σ < σ1 < 1. Then jf(x)j · xσ1¡σM 1=xσ1 so jf(x)j lim = 0: x!0+ 1=xσ1 It follows that f is integrable on [0; 1]. R 1 Let 0+ f = ® and suppose ® is not zero. By replacing f with ¡f if necessary we can assume ® > 0. Since f is integrable there is an N1 in N such that, for all n ¸ N1, Z 1 ® f > 1=n 2 There exists an N2 such that for all l ¸ N2 Z ¯ Xnl j 1 1 ¯ ® ¯ f( ) ¡ f¯ < nl nl 4 j=l 1=n 6 BROUGHAN so Z ® Xnl j 1 1 ¡ < f( ) ¡ f 4 nl nl j=l 1=n Therefore Z ® 1 ® Xnl j 1 < f < + f( ) 2 4 nl nl 1=n j=l so ® Xnl j 1 < f( ) : 4 nl nl j=l By the given hypothesis Xn j 1 lim f( ) = 0 n!1 n n j=1 so there is an N3 such that for all l ¸ N3 ® Xln j 1 ® ¡ < f( ) < 8 ln ln 8 j=1 Therefore ® Xl¡1 j 1 Xln j 1 ® ¡ < f( ) + f( ) < 8 ln ln ln ln 8 j=1 j=l and so ® ® Xl¡1 j 1 < ¡ f( ) 4 8 ln ln j=1 which implies ® Xl¡1 j 1 < jf( )j 8 ln ln j=1 Xl ln 1 < M ( )σ j ln j=1 lσnσ Xl 1 = M ( ) ln jσ j=1 lσnσl1¡σ < 2M ln HURWITZ ZETA FUNCTION 7 which can be made arbitarily small for n sufficiently large. This contra- diction shows we must have ® = 0, so completes the proof of the Lemma. Lemma 2.7. If σ = 0 and jtj ¸ 1 then j³(it; a)j · B(t) for some bound B(t). Proof. This follows directly from the inequality [1] valid for ¡± · σ · ± for ± < 1 and jtj ¸ 1 j³(s; a) ¡ a¡sj · A(±)jtj1+±: Lemma 2.8. If σ = 0 and 0 · t · 1 then j³(it; a)j · B(t): Proof. If t = 0, ³(0; a) = 1=2 ¡ a so we may assume t is not zero. To establish a bound we use two expressions for the Hurwitz zeta function derived with Euler summation and integration by parts [1]: For σ > ¡1 and N ¸ 0 XN 1 (N + a)1¡s ³(s; a) = + (n + a)s s ¡ 1 n=0 s XN 1 ¡ f³(s + 1; a) ¡ g 2! (n + a)s+1 n=0 Z s(s + 1) X1 1 u2 ¡ du 2! (n + a + u)s+2 n=N 0 and if σ > 0 XN 1 (N + a)1¡s ³(s; a) = + (n + a)s s ¡ 1 Zn=0 1 x ¡ [x] ¡ s+1 dx: N (x + a) 8 BROUGHAN Substitute σ = 0 and N = 0 in the first formula to obtain the equation 1 a1¡it ³(it; a) = + ait it ¡ 1 it 1 ¡ f³(it + 1; a) ¡ g 2! a1+it Z it(it + 1) X1 1 u2 ¡ du 2! (n + a + u)it+2 n=1 0 so 1 jtj 1 j³(it; a)j · 1 + + j³(it + 1; a) ¡ j jit ¡ 1j 2! a1+it Z jtj(jtj + 1) X1 1 u2 + du 2! (n + u)2 n=1 0 1 jtj(jtj + 1) jtj · 1 + + (³(2) + 1) + jC(t; a)j jit ¡ 1j 2! 2! where 1 C(t; a) = ³(it + 1; a) ¡ : a1+it In the second formula let N = 1 and s = 1 + it so σ = 1 > 0 giving Z 1 (1 + a)1¡(1+it) 1 x ¡ [x] C(t; a) = 1+it + ¡ (1 + it) 2+it dx (1 + a) 1 ¡ (1 + it) 1 (x + a) so 1 p jC(t; a)j · 1 + + 1 + t2: jtj Theorem 2.1.