A Few Remarks on Values of Hurwitz Zeta Function at Natural and Rational Arguments

Mathematica Aeterna, Vol. 5, 2015, no. 2, 383 - 394

A few remarks on values of Hurwitz Zeta function at natural and rational arguments

PawełJ. Szabłowski

Department of Mathematics and Information Sciences, Warsaw University of Technology ul. Koszykowa 75, 00-662 Warsaw, Poland

Abstract We exploit some properties of the Hurwitz zeta function ζ(n, x) in order to 1 n study sums of the form πn j∞= 1/(jk + l) and 1 j n −∞ πn j∞= ( 1) /(jk + l) for 2 n, k N, and integer l k/2. We show −∞ − P ≤ ∈ ≤ that these sums are algebraic numbers. We also show that 1 < n N and P n n ∈ p Q (0, 1) : the numbers (ζ(n, p) + ( 1) ζ(n, 1 p))/π are algebraic. ∈ ∩ − − On the way we find polynomials sm and cm of order respectively 2m + 1 and 2m + 2 such that their n th coeffi cients of sine and cosine Fourier transforms are equal to ( 1)n/n2m+1−and ( 1)n/n2m+2 respectively. − − Mathematics Subject Classification: Primary 11M35, 11M06, Sec- ondary 11M36, 11J72

Keywords: Riemann zeta function, Hurwitz zeta function, Bernoulli numbers, Dirichlet series

1 Introduction

Firstly we ﬁnd polynomials sm and cm of orders respectively 2m+1 and 2m+2 such that their n th coeﬃ cients in respectively sine and cosine Fourier series are of the form (−1)n/n2m+1 and ( 1)n/n2m+2. Secondly using these polynomials we study sums− of the form: −

∞ 1 ∞ ( 1)j S(n, k, l) = , Sˆ(n, k, l) = − . (jk + l)n (jk + l)n j= j= X−∞ X−∞ for n, k, l N such that n 2, k 2 generalizing known result of Euler (re- cently proved∈ differently by≥ Elkies≥ ([6])) stating that the number S(n, 4, 1)/πn 384 PawełJ. Szabłowski is rational. In particular we show that the numbers S(n, k, j)/πn and Sˆ(n, k, j)/πn for N n, k 2, and 1 j < k are algebraic. The3 results≥ of the paper≤ are somewhat in the spirit of [5] in the sense that the sums we study are equal to certain combinations of values of the Hurwitz zeta function calculated for natural and rational arguments. The paper is organized as follows. In the present section we recall basic definitions and a few elementary properties of the Hurwitz and Riemann zeta functions. We present them for completeness of the paper and also to introduce in this way our notation. In Section 2 are our main results. Longer proof are shifted to Section 3. To fix notation let us recall that the following function

ζ(s, x) = 1/(n + x)s, n 0 X≥ defined for all Re(s) > 1 and 1 Re(x) > 0 is called Hurwitz zeta function df ≥ while the function ζ(s, 1) = ζ(s) is called the Riemann zeta function. One shows that functions ζ(s, x) and ζ(s) can be extended to meromorphic functions of s defined for all s = 1. Hurwitz zeta function has been studied from different points of view in e.g.6 [7], [12], [4]. Let us consider slight generalization of the Hurwitz zeta function namely the function; ζˆ(s, x) = ( 1)n/(n + x)s, − n 0 X≥ for the same as before s and x. Using multiplication theorem for ζ(s, x) and little algebra we get:

ζ(s, x) = (ζ(s, x/2)+ζ(s, (1+x)/2))/2s, ζˆ(s, x) = (ζ(s, x/2) ζ(s, (1+x)/2))/2s, − (1)

1 l l S(n, k, l) = (ζ(n, ) + ( 1)nζ(n, 1 )), (2) kn k − − k 1 l l Sˆ(n, k, l) = (ζˆ(n, ) + ( 1)n+1ζˆ(n, 1 )), kn k − − k

S(n, k, l) = S(n, 2k, l) + ( 1)nS(n, 2k, k l), (3) − − Sˆ(n, k, l) = S(n, 2k, l) + ( 1)n+1S(n, 2k, k l), (4) − − S(n, k, k l) = ( 1)nS(n, k, l), Sˆ(n, k, k l) = ( 1)n+1Sˆ(n, k, l). (5) − − − − Hence there is sense to deﬁne sums S and Sˆ for l k/2 only. ≤ In the sequel N and N0 will denote respectively sets of natural and natural plus 0 numbers while Z and Q will denote respectively sets of integer and { } Zeta functions 385

rational numbers. Bk will denote k th Bernoulli number. It is known that − B2n+1 = 0 for n 1 and B1 = 1/2. One knows also that ≥ 2l l+1 (2π) ζ(2l) = ( 1) B2l (6) − 2(2l)! for all l N. For more∈ properties of the Hurwitz zeta function and the Bernoulli numbers see [9], [3], [11] and [2]. Having this and (3) and applying multiplication theorem for ζ(s, x) we can easily generalize result of Euler reminded by Elkies in [6]). Let us recall that since Euler times it is known that.

(22l 1) l+1 1 2(2−l)! ( 1) B2l if n = 2l, S(n, 4, 1) = −l πn ( 1) E2l ( 22−l+2(2l)! if n = 2l + 1.

Apart from this result one can easily show (manipulating multiplication theorem for Hurwitz zeta function) that for for N: ∈

(2l)! 2(2l)! 1 2l l+1 S(2l, 2, 1) = ζ(2l, ) = (2 1)( 1) B2l, π2l (2π)2l 2 − −

2(2l)! 2 2l l+1 2l S(2l, 3, 1) = ( ) ( 1) (3 1)B2l, π2l 3 − − 2(2l)! 2l 2l l+1 S(2l, 6, 1) = (2 1)(1 3− )( 1) B2l. π2l − − − Notice also that from (1) it follows that S(2l + 1, 2, 1) = 0 for l N and from (1) and (2) we get: Sˆ(2l, 2, 1) = 0 and Sˆ(2l + 1, 2, 1) = 2S(2l + 1,∈4, 1). Results concerning S(2l, 6, 1) were also obtained in [5](16b).

2 Main results

To get relationships between particular values of ζ(n, q) for n N and q Q we will use some elementary properties of the Fourier transforms.∈ ∈ Let us denote π π s s 1 c c 1 n(f(.)) = n = f(x) sin(nx)dx, n(f(.)) = n = f(x) cos(nx)dx, F F π π F F π π Z− Z− s c s c for n N0. We will abbreviate n(f(.)) and n(f(.)) to n and n if the function∈ f is given. We have: F F F F 1 f(x) = c + c cos(nx) + s sin(nx) 2F0 Fn Fn n 1 n 1 X≥ X≥ 386 PawełJ. Szabłowski

+ for all points of continuity of f and (f(x ) + f (x−)) /2 if at x f has jump discontinuity as a result of well known properties of Fourier series. Let us deﬁne two families of polynomials:

2 m+1 m x ( 1) 2m + 2 2(m j) 2j 2j 1 cm(x) = − x − π B2j(2 − 1), (7) (2m + 2)! 2j − j=0 m Xm x ( 1) 2m + 1 2(m j) 2j 2j 1 s (x) = − x − π B (2 − 1), (8) m (2m + 1)! 2j 2j j=0 − X for all m N0. Let us consider also the following polynomials: ∈ 2m+1 m 2(2π) x chm(x) = ( 1) B2m+2( ), − (2m + 2)! 2π 2m m 1 2(2π) x shm(x) = ( 1) − B2m+1( ). − (2m + 1)! 2π

Basing partially on [8] we have the following auxiliary lemma.

Lemma 1 For n N, m N0 ∈ ∈ n n s ( 1) c ( 1) (sm(.)) = − , (cm(.)) = − , (9) Fn n2m+1 Fn n2m+2 s 1 c 1 (shm(.)) = , (chm(.)) = , (10) Fn n2m+1 Fn n2m+2 2m+1 c 2m+2 m 2(2 1) 2m 1 (cm(.)) = π ( 1) B2m+2 − = 2ζ(2m + 2)(1 2− − ), F0 − (2m + 2)! − c (chm(.)) = 0. (11) F0 Proof. Is shifted to Section 3.

Remark 1 Notice that we have just summed the two following Dirichlet series: j cos(jx) j sin(jx) DC(s, x) = j 1( 1) js and DS(s, x) = j 1( 1) js , for particular values of s. Namely≥ − we have showed that ≥ − P P 2m+1 2m+2 m+1 (2 1) DC(2m + 2, x) = cm(x) + π ( 1) B2m+2 − , − (2m + 2)!

DS(2m + 3, x) = sm+1(x) and for x π, m N0. What are the values of these series for s = 1, 2, 3,...? | | ≤ ∈ 6 πl 2m+2 πl 2m+1 Remark 2 Notice that cm( k )/π and sm( k )/π for all k N, m N0 and l k, are rational numbers. ∈ ∈ ≤ Zeta functions 387

As a consequence we get the following theorem that allows calculating values of S(n, k, l) and Sˆ(n, k, l) for N n, k 2 and l k. 3 ≥ ≤

Theorem 2 For m N0, i N, i) ∈ ∈

m 2m+2 (2l 1)π ( 1) B2m+2π 2m+1 1 cm( − ) = − (2 (1 + ) 1) (12) 2i + 1 (2m + 2)! (2i + 1)2m+2 − i j(2l 1)π + ( 1)j cos( − )S(2m + 2, 2i + 1, j), 2i + 1 j=1 − X for l = 1, . . . , i 1. ii) −

2m+2 m 2m+1 (2l 1)π π ( 1) B2m+2(2 1) 2m 2 cm( − ) = − − (1 (2i)− − ) (13) 2i (2m + 2)! − i 1 − (2l 1)jπ + ( 1)j cos( − )Sˆ(2m + 2, 2i, j), 2i j=1 − X for l = 1, . . . , i. iii)

2m+2 m 2m+1 2lπ π ( 1) B2m+2(2 1) 2m 2 cm( ) = − − (1 (2i + 1)− − ) (14) 2i + 1 (2m + 2)! − i 2ljπ + ( 1)j cos( )Sˆ(2m + 2, 2i + 1, j), 2i + 1 j=1 − X for l = 1, . . . , i. iv) For m, k N: ∈ (2l 1)π i (2l 1)jπ s ( − ) = ( 1)j sin( − )S(2m + 1, 2i + 1, j), (15) m 2i + 1 2i + 1 j=1 − X for l = 1, . . . i 1. v) −

(2l 1)π i+l+1 1 sm( − ) = ( 1) S(2m + 1, 4, 1) (16) 2i − i2m+1 i 1 − (2l 1)jπ + ( 1)j sin( − )Sˆ(2m + 1, 2i, j), 2i j=1 − X 388 PawełJ. Szabłowski for l = 1, . . . i. vi) 2lπ i 2ljπ s ( ) = ( 1)j sin( )Sˆ(2m + 1, 2i + 1, j), (17) m 2i + 1 2i + 1 j=1 − X for l = 1, . . . , i.

Proof. Is shifted to Section 3. As a simple corollary we get some particular values S(n, k, l)

Corollary 3 i) For m = 0, 1, 2,...,, such that 2m + 1 > 1:

1 √2 2m 1 2m 2 Sˆ(2m+2, 4, 1) = (cm(π/4) ζ(2m+2)(1 2− − )(1 4− − )), π2m+2 −π2m+2 − − −

1 1 m 4m+3 Sˆ(2m + 1, 4, 1) = √2( sm(π/4) ( 1) E2m/(2 (2m)!)), π2m+1 − π2m+1 − − in particular Sˆ(2, 4, 1)/π2 = √2/16, Sˆ(3, 4, 1)/π3 = 3√2/128 ii)

2 √2 2m 1 2m 2 S(2m+2, 8, 1) = (cm(π/4) ζ(2m+2)(1 2− − )(1 4− − )) π2m+2 −π2m+2 − − −

1 2m 2 + ζ(2m + 2)(1 2− − ), π2m+2 −

2 √2 2m 1 2m 2 S(2m + 2, 8, 3) = (cm(π/4) ζ(2m + 2)(1 2− − )(1 4− − )) π2m+2 π2m+2 − − −

1 2m 2 + ζ(2m + 2)(1 2− − . π2m+2 − In particular S(2, 8, 1) = π2(1 + √2/2)/16,S(2, 8, 3) = π2(1 √2/2)/16, S(4, 8, 1) = π4(1 + 11√2/16)/192,S(4, 8, 3) = π4(1 11√2/16)/−192. iii) S(3, 8, 1) = π3(1 3√2/4)/32,S(3, 8, 3) = π3−(1+3√2/4)/32,S(5, 8, 1) = 5π5(1 57√2/16)/1536− ,S(5, 8, 1) = 5π5(1 + 57√2/16)/1536. − iv) For m N0 ∈ m 2m + 2 2j 1 (m j)+1 m m 2m+1 B2j(2 − 1)/9 − = B2m+2(4 1 + 4 /3 ), 2j − − − j=0 X m 2m + 2 2j 1 2j 2m+1 2m+2 B (2 − 1)2 = B (2 1)(2 1). 2j 2j 2m+2 j=0 − − − − X Zeta functions 389

v) For m N : ∈ m 2(2m + 1)! m+1 4√3 2m + 1 2(m j) 1 2j+1 S(2m+1, 3, 1) = ( 1) B2(m j)(2 − − 1)/3 . π2m+1 − 3 2j + 1 − − j=0 X (18)

Proof. i) We apply Theorem 2, (13) and (16) with i = 2. ii) We use (3) and (4). iii) We argue in a similar manner. iv) We set i = 1 and l = 1 in (12) and π2m+2 2 2m+2 m+1 2m+2 use the fact that S(2m + 2, 3, 1) = ( ) ( 1) (3 1)B2m+2. 2(2m+2)! 3 − − Then we set i = 1 and l = 1 in (13) and then cancel out π2m+2/(2m + 2)!. v) We set i = 1, and l = 1 in (15) . Based on the above mentioned theorem we can deduce our main result:

Theorem 4 For every N k, n 2, and 1 l k the numbers S(n, k, l)/πn and Sˆ(n, k, l)/πn are algebraic.3 ≥ ≤ ≤

Proof. First of all recall that in the Introdution we have found S(n, 2, 1) and Sˆ(n, 2, 1) (compare equations (16) and (13)). Further analyzing equations (12)-(17) we notice that quantities S(2m + 2, 2i + 1, j), Sˆ(2m + 2, 2i, j), Sˆ(2m+2, 2i+1, j),S(2m+1, 2i+1, j), Sˆ(2m+1, 2i, j) satisfy certain systems j (2l 1)jπ of linear equations with matrices [( 1) cos − ]l,j=1,..., (n 1)/2 , n b − c j (2l 1)jπ − j 2ljπ [( 1) sin −n ]l,j=1,..., (n 1)/2 and [( 1) cos n ]l,j=1,..., (n 1)/2 , − j 2ljπ b − c − b − c [( 1) sin ]l,j=1,..., (n 1)/2 . These matrices are non-singular since their de- − n b − c terminants diﬀer form the determinants of matrices Cn,Sn,Cn∗ and Sn∗ (deﬁned by (21) and (22), below) only possibly in sign. As shown in Lemma 6 matrices

Cn,Sn,Cn∗ and Sn∗ are nonsingular. Moreover notice that all entries of these matrices are algebraic numbers. On the other hand as one can see the numbers 2m+2 2m+1 cm(lπ/k)/π and sm(lπ/k)/π for all N m, k, l < k/2 are rational. Hence solutions of these equations are algebraic3 numbers.b Nowc taking into account that for k = 2 we do not get in fact any equation for Sˆ(n, 2, 1) . Thus we deduce that for n, k, l N such that n 2, k > 2, l k 1 the numbers Sˆ(n, l, k)/πn and for n 2∈, odd k and l k≥ 1 the numbers≤ −S(n, k, l)/πn are algebraic. ≥ ≤ − To show that numbers S(n, l, k)/πn are algebraic also for even l we refer to (3) and (4). First taking l odd we see that S(n, 2l, k)/πn for all k l (the case k = l leads to S(n, 2, 1) considered by the end of Introduction) these≤ numbers are algebraic. Then taking l = 2m with m even we deduce in the similar way that S(n, 4m, k)/πn, k 2m are algebraic. And so on by induction we deduce that indeed for all even≤l and k l/2 numbers S(n, l, k) are algebraic. ≤ As an immediate corollary of Theorem 4 and (3) and (4) we have the following result: 390 PawełJ. Szabłowski

Proposition 5 For 1 < n N and p Q (0, 1) : n∈ ∈ n ∩ n 1 n the numbers (ζ(n, p)+( 1) ζ(n, 1 p))/π and (ζˆ(n, p)+( 1) − ζˆ(n, 1 p))/π are algebraic. − − − −

3 Proofs

Proof. [Proof of Lemma 1](10) and (11) were basically proved in [8]. Only small modiﬃ cation connected with the change of variables was nedded. We will need the following observations: For all m 0, n 1: ≥ ≥ m 2(m k) s 2m+1 n+1 2(2m + 1)! k 2(m k) n − (x ) = ( 1) ( 1) π − , (19) Fn − n2m+1 − (2m + 1 2k)! k=0 − m 1 X − 2(m 1 k) c 2m n 2(2m)! k 2(m 1 k) n − − (x ) = ( 1) ( 1) π − − . (20) Fn − n2m − (2m 1 2k)! k=0 X − −

1 π 2m+1 1 π 2m Let us denote: bm = π π x sin(nx)dx and am = π π x cos(nx)dx. Integrating by parts twice− we obtain the following recursions− R R 2m 1 n+1 π − (2m + 1)2m n 4m 2m 1 2m(2m 1) bm = 2( 1) bm 1, am = ( 1) π − + − am 1 − n − n2 − − n2 n2 − n+1 2 with b0 = ( 1) n , and with a0 = 0 for m 1. Iterating them we get (19) and (20). − ≥ We show (7) and (8) by straightforward computation. First let us consider (7). Making use of (20) and after some algebra including changing the order of summation and setting l = m k j we get: − − m m k k 2(k+1) − c m+n+1 ( 1) n− 2(m k) 2l 1 2m 2k + 1 (cm(.)) = ( 1) − π − B2l(2 − 1) − Fn − (2m 2k + 1)! − 2l k=0 l=0 X − X Now we use [15](12) with x = 0 and [15](4) to show that

m k − 2m 2k + 1 2l 1 0 if k < m − B2l(2 − 1) = 2l − 1/2 if k = m l=0 X − c ( 1)n Hence indeed n(cm(.)) = n−2m+2 . To get (8) we proceed similarly and after some algebra weF get

m m k k 2(m k) − s m+n ( 1) π − 2m 2k + 1 2j 1 (sm(.)) = ( 1) − − B2l(2 − 1) Fn − n2k+1(2m 2k + 1)! 2l − k=0 l=0 X − X ( 1)n+1 = − . n2m+1 Zeta functions 391

1 π To get π π cm(x)dx we proceed as follows: − R π m+1 2m+2 m 1 2( 1) π 2m + 2 1 2(m j) 1 cm(x)dx = − B2(m j)(2 − − 1) π (2m + 2)! 2(m j) (2j + 3) − π j=0 − Z− X − m+1 2m+2 m 2( 1) π 2m + 3 2k 1 = − B2k(2 − 1). (2m + 3)! 2k − k=0 X Now we again manipulate [15] (4) and (12) to get the desired formula. The proof of Theorem 4 is based on the Lemma concerning the following matrices

(2l 1)jπ (2l 1)jπ Cn = [cos − ]l,j=1,..., (n 1)/2 ,Sn = [sin − ]l,j=1,..., (n 1)/2 , n b − c n b − c (21) 2ljπ 2ljπ Cn∗ = [cos ]l,j=1,..., (n 1)/2 ,Sn∗ = [sin ]l,j=1,..., (n 1)/2 . (22) n b − c n b − c

Lemma 6 For every N n 2 matrices Cn,Sn,Cn∗,Sn∗ are nonsingular. 3 ≥ Proof. The proof is based on the following observations. Let i denote unitary unit. Since sin x = (exp(ix) exp( ix))/2i one can easily notice that matrices −j j− Sn and Sn∗ are of the type [xl xl ]l,j=1,..., (n 1)/2 where xl equals respectively b − c (2l 1)iπ 2liπ − exp( −n ) and exp( n ) (n is odd in this case). Now we apply formula (2.3) of [10] and deduce that these matrices are nonsingular provided that quantities xl are nonzero, are all diﬀerent, and xlxk = 1, k, l = 1,..., (n 1)/2 which is the case in both situations. Now notice6 that n 0b : cos(− α) =c n 1 (2l 1)jπ l∀1 ≥ ( 1) − sin((2n 1)π/2 + α). Hence cos −n = ( 1) − sin((2l 1)π/2 + (2−l 1)jπ −l 1 π(j 1/2)(2l 1) l 1 (j 1−/2) (j 1/2)− − ) = ( 1) − sin( − − ) = ( 1) − (x − x− − )/(2i) and n − n − l − l we use formula (2.4) of [10] to deduce that matrix Cn is nonsingular. Now let us consider matrix Cn∗ . Notice that then n = 2m + 1 for some natural 2ljπ j 2ljπ j jπ(2m l+1) 1) m. cos( ) = ( 1) cos(jπ ) = ( 1) cos( − − ). And thus we 2m+1 − − 2m+1 − 2m+1 have reduced this case to the previous one. Hence Cn∗ is also nonsingular. Proof. [Proof of Theorem 2]In the proofs of all assertions we use basically the same tricks concerning properties of trigonometric functions and changing order of summation. That is why we will present proof of only one (out of 6) in detail. All remaining proofs are similar. We start with an obvious identity:

k r(k+l) πl (22m+1 1) jlπ ∞ ( 1) c ( ) = π2m+2( 1)mB − + cos( )( 1)j − m k 2m+2 (2m + 2)! k (rk + j)2m+2 − j=1 − r=0 X X (23) 392 PawełJ. Szabłowski that is true since by cos(nπ + α) = ( 1)n cos α we have: − k ∞ nπl ∞ jπl ( 1)n cos( )/n2m+2 = ( 1)r(k+l)+j cos( )/(rk + j)2m+2, − k − k n 1 j=1 r=0 X≥ X X

Hence we have to consider two cases k + l odd and k + l even. If k + l is even then we have

k ∞ nπl 1 πjl ( 1)n cos( )/n2m+2 = ( 1)j cos( )ζ(2m + 2, j/k). − k k2m+2 − k n 1 j=1 X≥ X

2i+1 j For k = 2i + 1 and l odd we have: j=1 ( 1) cos(jπ/(2i + 1))ζ(2m + 2, j/(2i + 1)). Now recall that for j = 2i + 1 we− get cos(π) = 1 and that for j = 1, . . . , i we have cos((2i + 1 j)lπ/P(2i + 1) = cos(jlπ/−(2i + 1) and 2i+1 j j − − ( 1) − = ( 1) . Thus we get − − − 2i+1 ljπ j ( 1)j cos( )ζ(2m + 2, ) = ζ(2m + 2) 2i + 1 2i + 1 j=1 − X i jlπ j j + ( 1)j cos( )(ζ(2m + 2, ) + ζ(2m + 2, 1 )). 2i + 1 2i + 1 2i + 1 j=1 − − X Now we use (3). Since k + l even and k even leads to triviality we consider two cases leading to k + l odd i.e. k = 2i and l = 2m 1 odd m < i which leads to: −

2i j(2m 1)π ∞ cos( − )( 1)j ( 1)r(2i++2m+1) /(r(2i + 1) + j)2m+2 2i j=1 − r=0 − X X 2m+2 2m+1 m π 2 1 = ( 1) B2m+2 − − (2m + 2)! 22m+2i2m+2 i 1 − jlπ + ( 1)j cos( ))(S(2m + 2, 4i, j) S(2m + 2, 4i, 2i j)). 2i j=1 − − − X We used here (6) (3) and (23). Similarly we consider the case k = 2i + 1 and l = 2m which leads to (14). Note that this case is very similar to the case k = 2i + 1 l = 2m 1. − Assertions iv)-vi) are proved in the similar way. The only diﬀerence is that 1 c we do not have to remember about 2 0 and that we utilize properties of sin function not of cos . F Zeta functions 393

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Received: April, 2015