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Vanishing of the Integral of the Hurwitz Zeta Function Draft 14Th June 2001

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Vanishing of the Integral of the Hurwitz Zeta Function Draft 14Th June 2001 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.
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