Generalized Rational Zeta Series for $\Zeta (2N) $ and $\Zeta (2N+ 1) $

$Generalized Rational Zeta Series for $\Zeta (2N) $ and $\Zeta (2N+ 1) $$

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Later, Euler gave the formula

∞ 1 (−1)k+1B (2π)2k ζ(2k)= = 2k , k ∈ N , n2k 2(2k)! 0 n=1 X

where Bn are the Bernoulli numbers, deﬁned by

z ∞ B = n zn, |z| < 2π. ez − 1 n! n=0 X These Bernoulli numbers also arise in certain power series, namely of the cotangent function

∞ (−1)n22nB ∞ ζ(2n) (1) cot(x)= 2n x2n−1 = −2 x2n−1, |x| < π. (2n)! π2n n=0 n=0 X X The odd arguments of ζ(s) are the interesting as they do not have a closed form, though many mathematicians have studied them in detail. In 1979, Roger Apery [3] proved that ζ(3) is irrational using the fast converging series

5 ∞ (−1)n−1 ζ(3) = . 2 3 2n n=1 n n X It is still unknown whether ζ(5) is irrational however, it is known that at least one of ζ(5), ζ(7), ζ(9), and ζ(11) is irrational (see [18]). The rational zeta series in this paper will involve odd arguments of ζ(s) among other things. Another main function needed is the Clausen function (or Clausen’s integral),

∞ sin(kθ) θ φ (2) Cl2(θ) := 2 = − log 2 sin dφ. k 0 2 Xk=1 Z The Clausen function also has a power series representation which will be used later in the paper. It is given as

∞ Cl (θ) ζ(2n) θ 2n (3) 2 =1 − log |θ| + , |θ| < 2π. θ n(2n + 1) 2π n=1 X There are also higher order Clausen-type function deﬁned as

∞ sin(kθ) ∞ cos(kθ) (4) Cl (θ) := , Cl (θ) := . 2m k2m 2m+1 k2m+1 Xk=1 Xk=1 GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n)AND ζ(2n + 1) 3

The Clausen function is widely studied and has many applications in mathematics and mathematical physics ([5], [6], [9], [12], [14], [15], [17]). We will also discuss the Dirichlet beta function

∞ (−1)n β(s)= , ℜ(s) > 0. (2n + 1)s n=0 X When s = 2,

∞ (−1)n β(2) = G = (2n + 1)2 n=0 X is known as Catalan’s constant. Using this and the Riemann zeta function, we ﬁnd

(4m − 1)ζ(2m + 1) (5) Cl (π)=0, Cl (π)= − , 2m 2m+1 4m

and

(4m − 1)ζ(2m + 1) (6) Cl (π/2) = β(2m), Cl (π/2) = − . 2m 2m+1 24m+1

Also, from (4) we can see

d d (7) Cl (θ) = Cl (θ), Cl (θ)= − Cl (θ), dθ 2m 2m−1 dθ 2m+1 2m

θ θ (8) Cl2m(x) dx = ζ(2m + 1) − Cl2m+1(θ), Cl2m−1(x) dx = Cl2m(θ). Z0 Z0 Using (4) and (7), we ﬁnd

θ (9) Cl (θ)= − log 2 sin , |θ| < 2π. 1 2 Writing others out,

z Cl3(z)= ζ(3) − Cl2(t) dt, Z0 z z x z Cl4(z)= Cl3(x) dx = zζ(3) − Cl2(t) dt dx = zζ(3) − (z − t) Cl2(t) dt, Z0 Z0 Z0 Z0 4 DEREK ORR z 1 1 z Cl (z) = ζ(5) − Cl (x) dx = ζ(5) − z2ζ(3) + (z − t)2 Cl (t) dt, 5 4 2 2 2 Z0 Z0 and by induction, for m ≥ 3,

− ⌊ m 1 ⌋ 2 k m−2k−1 m−1 (−1) z (10) Cl (z)=(−1)⌊ 2 ⌋ ζ(2k + 1) m (m − 2k − 1)! k=1 X − m 1 z (−1)⌊ 2 ⌋ + (z − t)m−3 Cl (t) dt. (m − 3)! 2 Z0 In the latter sections of this paper, we will focus on the polygamma function. Its deﬁnition is given by

dn+1 ψ(n)(z) := log Γ(z), n ∈ N . dzn+1 0

This paper will discuss when n = 0. For n = 0, ψ(0)(z) = ψ(z) is called the digamma function. It has been shown (see [2]) that there is a closed form of z xnψ(x) dx in terms of sums involving Harmonic numbers, Bernoulli numbers 0 Zand Bernoulli polynomials. There are also deﬁnitions for negative order polygamma functions, called negapolygammas, given by

ψ(−1)(z) = log Γ(z),

z ψ(−2)(z)= log Γ(x) dx, Z0 z x z z z ψ(−3)(z)= log Γ(t) dtdx = log Γ(t) dxdt = (z − t) log Γ(t) dt, Z0 Z0 Z0 Zt Z0 and by induction, for n ≥ 2,

1 z (11) ψ(−n)(z)= (z − t)n−2 log Γ(t) dt. (n − 2)! Z0 Note the Taylor series for log Γ(z) is

∞ (−1)kζ(k) (12) log Γ(z)= − log z − γz + zk, |z| < 1, k Xk=2 where γ is the Euler-Mascheroni constant. The last function to introduce is the Hurwitz zeta function, deﬁned by GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n)AND ζ(2n + 1) 5

∞ 1 ζ(s, a)= , a ∈ R\ − N, ℜ(s) > 1. (k + a)s Xk=0 The negapolygammas are related to the derivative of the Hurwitz zeta function with respect to the ﬁrst variable (see [2]).

πz p 1.1. Organization of the Paper. We begin by investigating 0 x cot(x) dx for |z| < 1, which is studied in [15]. We then compute this same integral using the power series for the cotangent and obtain a rational zeta series rRepresentation with ζ(2n) on the numerator for z = 1/2 and z = 1/4. Afterwords, we work on a generalized rational zeta series with ζ(2n) on the numerator and an arbitrary number of monomials on the denominator using the generalized Clausen function Clm(z). As a special case, we immediately prove a conjecture given in 2012 by F.M.S. Lima. After doing this, we go back to the cotangent function and discover a separate class of generalized ζ(2n) series. Lastly, we perform the same analysis but for the digamma function ψ(x) instead of cot(x) and instead of Clm(x), we use the negapolygammas ψ(−m)(x). Using the ζ(2n) sums from earlier, we extract the same rational zeta series representations with ζ(2n + 1) on the numerator for z =1/2 and z =1/4.

Acknowledgements. I would like to thank Cezar Lupu, Tom Hales, and George Sparling for their interest in my paper as well as their advice which led to some improvements of the paper.

2. Rational ζ(2n) series with cot(x)

Theorem 2.1. For p ∈ N and |z| < 1,

(13) p k+3 p πz ⌊ 2 ⌋ 2 p p p!(−1) p!(−1) x cot(x) dx =(πz) Cl (2πz)+δ p p ζ(p+1), k k+1 ⌊ 2 ⌋, 2 p 0 (p − k)!(2πz) 2 Z Xk=0

where δj,k is the Kronecker delta function.

Proof. Let f(z) be the left hand side of the equation and let g(z) be the right hand side. Note that f ′(z)= πp+1zp cot(πz). Using (7) and (9),

p k+3 1 p!(−1)⌊ 2 ⌋(p − k)(2π)p−kzp−k−1 (πz)p2π cos(πz) g′(z)= Cl (2πz)+ 2p (p − k)! k+1 2 sin(πz) k=0 X p k+3 1 p!(−1)⌊ 2 ⌋(2πz)p−k + (−1)k+12π Cl (2πz) 2p (p − k)! k Xk=1 n o 6 DEREK ORR

p p! Clk(2πz) k+2 k+3 = πp+1zp cot(πz)+(πz)p (−1)⌊ 2 ⌋ +(−1)⌊ 2 ⌋(−1)k . (p − k)!zk+1(2π)k−1 Xk=1 So indeed, g′(z)= πp+1zp cot(πz)= f ′(z). Clearly, f(0) = 0. For g(z), note that all terms in the sum are zero except when k = p. So we have

p 2 1 ⌊ p+3 ⌋ p!(−1) g(0) = p!(−1) 2 Clp+1(0) + δ p p ζ(p + 1). 2p ⌊ 2 ⌋, 2 2p

From (4), Cl (0) = δ p p ζ(p + 1). So we see g(0) = 0. Since f(0) = g(0) and p+1 ⌊ 2 ⌋, 2 f ′(z)= g′(z), f(z)= g(z).

Using (5), (6), and (9), setting z =1/2 and z =1/4, we ﬁnd

(14) p ⌊ 2 ⌋ p π/2 p k k 2 p π p!(−1) (4 − 1) p!(−1) ζ(p + 1) x cot(x) dx = log 2+ ζ(2k+1) +δ p p , 2k ⌊ 2 ⌋, 2 p 0 2 (p − 2k)!(2π) 2 Z Xk=1 and

⌊ p ⌋ π/4 p 2 k k p 1 π p!(−1) (4 − 1) (15) x cot(x) dx = log 2 + 2k ζ(2k + 1) 0 2 4 (p − 2k)!(2π) Z k=1 p+1 X ⌊ 2 ⌋ p p!(−4)kβ(2k) p!(−1) 2 − + δ p p ζ(p + 1). (p +1 − 2k)!π2k−1 ⌊ 2 ⌋, 2 2p Xk=1

We can also integrate xp cot(x) using (1) and Fubini’s theorem. Doing so, we obtain

πz πz ∞ ζ(2n) ∞ ζ(2n)z2n xp cot(x) dx = −2 x2n−1+p dx = −2(πz)p , π2n 2n + p 0 0 n=0 n=0 Z Z X X and using (13),

∞ p k+3 p ζ(2n)z2n p!(−1)⌊ 2 ⌋ p!(−1) 2 (16) − 2 = Clk+1(2πz)+ δ p p ζ(p + 1). 2n + p (p − k)!(2πz)k ⌊ 2 ⌋, 2 (2πz)p n=0 X Xk=0

So for z =1/2 and z =1/4, we have GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n)AND ζ(2n + 1) 7

(17) p ∞ ⌊ 2 ⌋ p ζ(2n) p!(−1)k(4k − 1)ζ(2k + 1) p!(−1) 2 ζ(p + 1) − 2 = log 2 + + δ p p , (2n + p)4n (p − 2k)!(2π)2k ⌊ 2 ⌋, 2 πp n=0 X Xk=1 and

⌊ p ⌋ ∞ ζ(2n) 1 1 2 p!(−1)k(4k − 1)ζ(2k + 1) (18) − 2 = log 2 + (2n + p)16n 2 2 (p − 2k)!(2π)2k n=0 k=1 X p+1 X ⌊ 2 ⌋ p π p!(−4)kβ(2k) p!(−1) 2 2pζ(p + 1) − + δ p p . 2 (p +1 − 2k)!π2k ⌊ 2 ⌋, 2 πp Xk=1 Plugging in p = 1 into both of the equations yields two nice series representations for log 2,

∞ ζ(2n) (19) − 2 = log 2, (2n + 1)4n n=0 X and

4G ∞ ζ(2n) (20) − − 4 = log 2. π (2n + 1)16n n=0 X Other series from (17) and (18) are

∞ ζ(2n) 7ζ(3) = − log 2 + , (n + 1)4n 2π2 n=0 X ∞ ζ(2n) 1 9ζ(3) = − log 2 + , (2n + 3)4n 2 4π2 n=0 X ∞ ζ(2n) 9ζ(3) 93ζ(5) = − log 2 + − , (n + 2)4n π2 2π4 n=0 X ∞ ζ(2n) 1 35ζ(3) 4G = − log 2 + − , (n + 1)16n 2 4π2 π n=0 X and

∞ ζ(2n) 1 9ζ(3) 3G 24β(4) = − log 2 + − + . (2n + 3)16n 4 8π2 π π3 n=0 X 8 DEREK ORR

3. General ζ(2n) series using Clm(x)

Theorem 3.1. For p ∈ N0, m ∈ N and |z| < 1,

m+p+1 m k 2πz p+m+1−k ⌊ 2 ⌋ ⌊ 2 ⌋ p (2πz) p!(−1) (−1) (21) x Clm(x) dx = − Clk(2πz) 0 (p + m +1 − k)! Z k=Xm+1 ⌊ m ⌋ p+m + δ p+m p+m (−1) 2 p!(−1) 2 ζ(p + m + 1). ⌊ 2 ⌋, 2

Proof. Similar to the proof of (13), we will call the left hand side f(z) and the ′ p+1 p right hand side g(z). Note f(0) = g(0) and f (z)=(2π) z Clm(2πz). Using (7),

m+p m k p!(2π)p+m+1−kzp+m−k(−1)⌊ 2 ⌋+⌊ 2 ⌋ g′(z)= − Cl (2πz) (p + m − k)! k k=Xm+1 m+p+1 m k p!(2πz)p+m+1−k(−1)⌊ 2 ⌋+⌊ 2 ⌋ − (−1)k2π Cl (2πz) (p + m +1 − k)! k−1 k=Xm+1 n o

m+p p+m−k ⌊ m ⌋ m m+1 p!(2πz) (−1) 2 = −(2π)p+1zp(−1)⌊ 2 ⌋+⌊ 2 ⌋(−1)m+1 Cl (2πz) − 2π ∗ m (p + m − k)! k=Xm+1 ⌊ k+1 ⌋ k+1 ⌊ k ⌋ p+1 p Clk(2πz) (−1) 2 (−1) +(−1) 2 = (2π) z Clm(2πz). n o So indeed f ′(z)= g′(z) and f(0) = g(0), which implies f(z)= g(z).

We can see if p = 0, we recover (8). Now letting z =1/2 and z =1/4, we ﬁnd

⌊ p+m ⌋ π 2 k k m (−1) (4 − 1)ζ(2k + 1) p ⌊ 2 ⌋ p+m (22) x Clm(x) dx =(−1) p!π 2k 0 m+1 (p + m − 2k)!(2π) Z k=X⌊ 2 ⌋ ⌊ m ⌋ p+m + δ p+m p+m (−1) 2 p!(−1) 2 ζ(p + m + 1), ⌊ 2 ⌋, 2

and

⌊ p+m ⌋ π/2 2 k k m π p+m 1 (−1) (4 − 1)ζ(2k + 1) p ⌊ 2 ⌋ (23) x Clm(x) dx =(−1) p! 2k 0 2 2 (p + m − 2k)!(2π) k=⌊ m+1 ⌋ Z X2 ⌊ p+m+1 ⌋ 2 k k π (−1) 4 β(2k) ⌊ m ⌋ p+m − + δ p+m p+m (−1) 2 p!(−1) 2 ζ(p + m + 1). 2 (p + m +1 − 2k)!π2k ⌊ 2 ⌋, 2 k=⌊ m+2 ⌋ X2 GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n)AND ζ(2n + 1) 9

Now, we will integrate the left hand side of (21) using (3) and (10), giving us

m−1 2πz 2πz ⌊ 2 ⌋ m−1 (−1)⌊ 2 ⌋+kxm−2k−1 xp Cl (x) dx = xp ζ(2k + 1) m (m − 2k − 1)! Z0 Z0 k=1 X − ⌊ m 1 ⌋ x (−1) 2 + (x − t)m−3 Cl (t) dt dx (m − 3)! 2 Z0

m−1 ⌊ 2 ⌋ m−1 m−1 ⌊ 2 ⌋+k m+p−2k ⌊ 2 ⌋ 2πz x (−1) (2πz) (−1) p m−3 = ζ(2k+1)+ x (x−t) Cl2(t) dtdx (m − 1 − 2k)!(m + p − 2k) (m − 3)! 0 0 Xk=1 Z Z

m−1 ⌊ 2 ⌋ m−1 m−1 (−1)⌊ 2 ⌋+k(2πz)m+p−2k (−1)⌊ 2 ⌋ = ζ(2k +1)+ ∗ (m − 1 − 2k)!(m + p − 2k) (m − 3)! Xk=1 2πz x ∞ ζ(2n)t2n+1 xp(x − t)m−3 t − t log t + dtdx n(2n + 1)(2π)2n 0 0 n=1 Z Z X

m−1 ⌊ 2 ⌋ m−1 m−1 2πz (−1)⌊ 2 ⌋+k(2πz)m+p−2k (−1)⌊ 2 ⌋ xm−1(H − log x) = ζ(2k+1)+ xp m−1 (m − 1 − 2k)!(m + p − 2k) (m − 3)! 0 (m − 1)(m − 2) Xk=1 Z ∞ 2ζ(2n)Γ(m − 2)xm+2n−1 + dx, 2n(2n + 1)(2n + 2) ... (2n + m − 1)(2π)2n n=1 X

where Hk is the k-th harmonic number and H0 := 0. Integrating again and simplifying, we arrive at

m−1 2πz p+m ⌊ 2 ⌋ p (2πz) (−1) Hm−1 − log(2πz) 1 (24) x Clm(x) dx = + 2 0 (m − 1)! (p + m) (p + m) − Z ⌊ m 1 ⌋ 2 (m − 1)!(−1)kζ(2k + 1) ∞ (m − 1)!ζ(2n)z2n + + . (m − 1 − 2k)!(p + m − 2k)(2πz)2k n(2n + 1) ... (2n + m − 1)(2n + p + m) n=1 Xk=1 X

Using (21), we can rearrange this and ﬁnd 10 DEREK ORR

∞ ζ(2n)z2n log(2πz) − H (25) = m−1 n(2n + 1) ... (2n + m − 1)(2n + m + p) (m − 1)!(p + m) n=1 m+pX k+1 p+m (−1)mp!(−1)⌊ 2 ⌋ (−1)m+1p!(−1) 2 + Clk+1(2πz)+ δ p+m p+m ζ(p + m + 1) (p + m − k)!(2πz)k ⌊ 2 ⌋, 2 (2πz)p+m k=m X − ⌊ m 1 ⌋ 1 2 (−1)kζ(2k + 1) − − . (m − 1)!(p + m)2 (m − 1 − 2k)!(m + p − 2k)(2πz)2k Xk=1

Note when p = 0, we have the very special formula

∞ m+1 ζ(2n)z2n (−1)m(−1)⌊ 2 ⌋ (26) = Cl (2πz) n(2n + 1) ... (2n + m) (2πz)m m+1 n=1 X − ⌊ m 1 ⌋ 3m 2 k (−1) 2 (−1) ζ(2k + 1) log(2πz) − Hm − δ m m ζ(m + 1) − + . ⌊ 2 ⌋, 2 (2πz)m (m − 2k)!(2πz)2k m! Xk=1

Letting z =1/2 and z =1/4, we see that

∞ 3m+2 ζ(2n) (−1) 2 (2m+1 − 1)ζ(m + 1) (27) = δ m m n(2n + 1) ... (2n + m)4n ⌊ 2 ⌋, 2 (2π)m n=1 X − ⌊ m 1 ⌋ 2 (−1)kζ(2k + 1) log π − H − + m , (m − 2k)!π2k m! Xk=1

and

∞ 3m+1 ζ(2n) (−1) 2 2mβ(m + 1) (28) = δ m+1 m+1 n(2n + 1) ... (2n + m)16n ⌊ 2 ⌋, 2 πm n=1 X − ⌊ m 1 ⌋ 3m 2m+1 m 2 k 1 (−1) 2 (2 +2 − 1)ζ(m + 1) (−4) ζ(2k + 1) log(π/2) − Hm − δ m m − + , 2 ⌊ 2 ⌋, 2 (2π)m (m − 2k)!π2k m! Xk=1

both of which were conjectured in 2012 and only for m odd (see [13]). For general p, we have GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n)AND ζ(2n + 1) 11

∞ ζ(2n) log π − H (29) = m−1 n(2n + 1) ... (2n + m − 1)(2n + m + p)4n (m − 1)!(p + m) n=1 X p+m p+m ⌊ 2 ⌋ 2 k k m+1 (−1) (−1) (4 − 1)ζ(2k + 1) +(−1) p! δ p+m p+m ζ(p + m +1)+ ⌊ 2 ⌋, 2 πp+m (p + m − 2k)!(2π)2k k=⌊ m+1 ⌋ X2 − ⌊ m 1 ⌋ 2 (−1)kζ(2k + 1) 1 − − , (m − 1 − 2k)!(p + m − 2k)π2k (m − 1)!(p + m)2 Xk=1 and

∞ ζ(2n) log(π/2) − H π (30) = m−1 + ∗ n(2n + 1) ... (2n + m − 1)(2n + m + p)16n (m − 1)!(p + m) 2 n=1 p+m+1X ⌊ 2 ⌋ p+m p!(−1)m+k4kβ(2k) (−1)m+1p!(−1) 2 2p+m p!(−1)m +δ p+m p+m ζ(p+m+1)− ∗ 2k ⌊ 2 ⌋, 2 p+m m+2 (p + m +1 − 2k)!π π 2 k=X⌊ 2 ⌋ − ⌊ p+m ⌋ ⌊ m 1 ⌋ 2 (−1)k(4k − 1)ζ(2k + 1) 2 (−1)k4kζ(2k + 1) 1 − − . (p + m − 2k)!(2π)2k (m − 1 − 2k)!(p + m − 2k)π2k (m − 1)!(p + m)2 m+1 k=1 k=X⌊ 2 ⌋ X

Remark. When m = 1 and p = 0, one has

∞ ζ(2n) = log π − 1, n(2n + 1)4n n=1 X and

∞ ζ(2n) 2G π = − 1 + log , n(2n + 1)16n π 2 n=1 X the ﬁrst of which is a famous series (see [16]) and both are immediate from (3) with θ = π and θ = π/2, respectively. Below we compute other sums for certain m and p.

∞ ζ(2n) 3ζ(3) 1 1 = − + log π − n(2n + 3)4n 2π2 3 9 n=1 X ∞ ζ(2n) 7ζ(3) 3 = − log π − n(2n + 1)(n + 1)4n 2π2 2 n=1 X 12 DEREK ORR

∞ ζ(2n) 2ζ(3) 1 11 = + log π − n(2n + 1)(n + 1)(2n + 3)4n π2 3 18 n=1 X ∞ ζ(2n) 2ζ(3) 31ζ(5) 1 7 = + + log π − n(2n + 1)(n + 1)(n + 2)4n π2 4π4 2 8 n=1 X ∞ ζ(2n) ζ(3) ζ(5) 1 137 = − + log π − n(2n + 1) ... (2n + 5)4n 6π2 π4 120 7200 n=1 X ∞ ζ(2n) 35ζ(3) 4G π 1 = − + + log − n(n + 1)16n 4π2 π 2 2 n=1 X ∞ ζ(2n) 35ζ(3) π 3 = + log − n(2n + 1)(n + 1)16n 4π2 2 2 n=1 X ∞ ζ(2n) 3ζ(3) 8β(4) 1 π 4 = + + log − n(2n + 1)(2n + 3)16n 8π2 π3 3 2 9 n=1 X ∞ ζ(2n) 2ζ(3) 527ζ(5) 1 π 25 = − + log − n(2n + 1) ... (2n + 4)16n π2 32π4 24 2 288 n=1 X Now, we will revisit the cotangent function and investigate a general zeta series using its power series.

4. General ζ(2n) series using cot(x)

Similar to above, we will investigate the double integral

πz x xp(x − t)mt cot(t) dt dx. Z0 Z0 Using the binomial theorem, (13), and a change of variables among other things,

πz x m m πz x xp(x − t)mt cot(t) dt dx = (−1)j xp+m−j tj+1 cot(t) dt dx j 0 0 j=0 0 0 Z Z X Z Z

m j+1 k+3 πz m (−1)j(j + 1)!(−1)⌊ 2 ⌋ = xp+m+1−k Cl (2x) dx j (j +1 − k)!2k k+1 j=0 0 X Xk=0 Z m j+1 j 2 πz (−1) (j + 1)!(−1) m p+m−j + δ j+1 j+1 ζ(j + 2) x dx ⌊ 2 ⌋, 2 2j+1 j j=0 0 X Z GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n)AND ζ(2n + 1) 13

m 2πz m m k m (−1)j+1 m (−1)j(j + 1)!(−1)⌊ 2 ⌋ = up+m+1 Cl (u) du + ∗ j 2m+p+2 1 j (j − k)!2p+m+2 j=0 0 X Z Xk=0 Xj=k ⌊ m+1 ⌋ 2πz 2 k p+m+2−2k p+m−k (2k)(−1) m!(πz) u Clk+2(u) du − 2k ζ(2k + 1) 0 2 (m +1 − 2k)!(p + m +2 − 2k) Z Xk=1

m 2πz ⌊ 2 ⌋ 2πz δm,0 p+1 (−1) m! p+1 = − p+2 u Cl1(u) du − (1 − δm,0) p+m+2 u Clm+1(u) du 2 0 2 0 Z m+1 Z m+1 ⌊ 2 ⌋ ⌊ 2 ⌋ 2πz k+1 p+m+2−2k (m + 1)!(−1) p (2k)(−1) m!(πz) ζ(2k + 1) + p+m+2 u Clm+2(u) du+ 2k . 2 0 2 (m +1 − 2k)!(p + m +2 − 2k) Z Xk=1

where we’ve evaluated the sum on j and k. The terms with δm,0 cancel each other. We can use (21) to evaluate the other integrals. Rearranging and simplifying, we will arrive at

πz x (31) xp(x − t)mt cot(t) dt dx =(p + m + 2) 2(−1)m(πz)m+p+3p!m!∗ Z0 Z0 p+m+3 ⌊ k ⌋ p+m m (−1) 2 Clk(2πz) p!(−1) 2 (−1) m!ζ(p + m + 3) + δ p+m p+m (p + m +3 − k)!(2πz)k ⌊ 2 ⌋, 2 2m+p+2 k=m+3 X m+1 m+1 ⌊ 2 ⌋ (−1)⌊ 2 ⌋m!(πz)p+1 Cl (2πz) (2k)(−1)k+1m!(πz)p+m+2−2kζ(2k + 1) − m+2 + . 2m+1 22k(m +1 − 2k)!(p + m +2 − 2k) Xk=1

Another way to evaluate this double integral is to use (1) and Fubini’s theorem, similar to the ﬁrst section. Doing so, we see

πz x πz ∞ ζ(2n) x xp(x − t)mt cot(t) dt dx = −2 xp (x − t)mt2n dt dx π2n 0 0 0 n=0 0 Z Z Z X Z ∞ ζ(2n)m!Γ(2n + 1) πz = −2 x2n+m+p+1 dx π2nΓ(m +2n + 2) n=0 0 X Z ∞ ζ(2n)m!(πz)m+p+2z2n = −2 . (2n + 1) ... (2n + m + 1)(2n + m + p + 2) n=0 X

Putting this and (31) together, we ﬁnd 14 DEREK ORR

∞ ζ(2n)z2n (32) =(p + m + 2) (−1)m+1p!∗ (2n + 1) ... (2n + m + 1)(2n + m + p + 2) n=0 X p+m+3 ⌊ k ⌋ p+m m πz(−1) 2 Clk(2πz) p!(−1) 2 (−1) ζ(p + m + 3) − δ p+m p+m (p + m +3 − k)!(2πz)k ⌊ 2 ⌋, 2 2(2πz)m+p+2 k=m+3 X m+1 m+1 ⌊ 2 ⌋ (−1)⌊ 2 ⌋ Cl (2πz) k(−1)kζ(2k + 1) + m+2 + . 2(2πz)m+1 (2πz)2k(m +1 − 2k)!(m + p +2 − 2k) Xk=1

When p = 0, this formula simpliﬁes to

∞ m+1 ζ(2n)z2n (−1)⌊ 2 ⌋ Cl (2πz) (33) = m+2 (2n + 1) ... (2n + m + 2) 2(2πz)m+1 n=0 X m+1 m ⌊ 2 ⌋ ⌊ 2 ⌋ m m k (−1) Clm+3(2πz) − δ⌊ 2 ⌋, 2 ζ(m + 3) (m + 2) k(−1) ζ(2k + 1) + m+2 + 2k . 2(2πz) (2πz) (m +2 − 2k)! Xk=1

Setting z =1/2 and z =1/4, we obtain

∞ m+1 ζ(2n) (−1) 2 (2m+1 − 1)ζ(m + 2) (34) = −δ m+1 m+1 (2n + 1) ... (2n + m + 2)4n ⌊ 2 ⌋, 2 2(2π)m+1 n=0 X m+1 m ⌊ 2 ⌋ (−1) 2 (2m+3 − 1)(m + 2)ζ(m + 3) k(−1)kζ(2k + 1) − δ m m + , ⌊ 2 ⌋, 2 2(2π)m+2 (2πz)2k(m +2 − 2k)! Xk=1 and

∞ ζ(2n) m (35) =(−1)⌊ 2 ⌋(m + 2)∗ (2n + 1) ... (2n + m + 2)16n n=0 X m+1 2m+5 m+2 2 β(m + 2) (2 +2 − 1)ζ(m + 3) ⌊ m+1 ⌋ δ m+1 m+1 − δ⌊ m ⌋, m +(−1) 2 ∗ ⌊ 2 ⌋, 2 πm+2 2 2 4(2π)m+2 ⌊ m+1 ⌋ 2mβ(m + 2) (2m+1 − 1)ζ(m + 2) 2 k(−4)kζ(2k + 1) δ⌊ m ⌋, m − δ m+1 m+1 + . 2 2 πm+1 ⌊ 2 ⌋, 2 4(2π)m+1 π2k(m +2 − 2k)! Xk=1

For general p, GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n)AND ζ(2n + 1) 15

∞ ζ(2n) (−1)mp!(m + p + 2) (36) = ∗ (2n + 1) ... (2n + m + 1)(2n + m + p + 2)4n 2 n=0 Xp+m+2 ⌊ 2 ⌋ p+m (−1)k(4k − 1)ζ(2k + 1) (−1) 2 ζ(p + m + 3) − δ p+m p+m (p + m +2 − 2k)!(2π)2k ⌊ 2 ⌋, 2 πm+p+2 m+3 ! k=X⌊ 2 ⌋ m+1 m+1 ⌊ 2 ⌋ (−1) 2 (2m+1 − 1)ζ(m + 2) k(−1)kζ(2k + 1) −δ m+1 m+1 + , ⌊ 2 ⌋, 2 2(2π)m+1 π2k(m +1 − 2k)!(m + p +2 − 2k) Xk=1 and

∞ ζ(2n) (−1)mp!(m + p + 2) (37) = ∗ (2n + 1) ... (2n + m + 1)(2n + m + p + 2)16n 4 n=0 X ⌊ p+m+2 ⌋ ⌊ p+m+3 ⌋ 2 (−1)k(4k − 1)ζ(2k + 1) 2 (−1)kβ(2k)4k − (p + m +2 − 2k)!(2π)2k (p + m +3 − 2k)!π2k−1 m+3 m+4 k=X⌊ 2 ⌋ k=X⌊ 2 ⌋ p+m m+1 (−1) 2 ζ(p + m + 3)2p+m+3 (−1)⌊ 2 ⌋2mβ(m + 2) − δ p+m p+m + δ m m ⌊ ⌋, m+p+2 ⌊ 2 ⌋, 2 m+1 2 2 π ! π m+1 m+1 ⌊ 2 ⌋ (−1) 2 (2m+1 − 1)ζ(m + 2) k(−1)kζ(2k + 1)4k −δ m+1 m+1 + . ⌊ 2 ⌋, 2 4(2π)m+1 π2k(m +1 − 2k)!(m + p +2 − 2k) Xk=1 Remark. For m = 0 and p = 0, we have

∞ ζ(2n) 7ζ(3) = − , (2n + 1)(n + 1)4n 2π2 n=0 X and

∞ ζ(2n) 2G 35ζ(3) = − , (2n + 1)(n + 1)16n π 4π2 n=0 X the ﬁrst of which was rediscovered by Ewell (see [11]). Below we compute other sums for certain m and p.

∞ ζ(2n) 9ζ(3) = − (2n + 1)(2n + 3)4n 8π2 n=0 X ∞ ζ(2n) 3ζ(3) 31ζ(5) = − + (2n + 1)(n + 2)4n π2 2π4 n=0 X 16 DEREK ORR

∞ ζ(2n) 5ζ(3) = − (2n + 1)(2n + 2)(2n + 3)4n 8π2 n=0 X ∞ ζ(2n) ζ(3) 31ζ(5) = − − (2n + 1)(n + 1)(n + 2)4n 2π2 2π4 n=0 X ∞ ζ(2n) ζ(3) 49ζ(5) = − + (2n + 1) ... (2n + 5)4n 6π2 32π4 n=0 X ∞ ζ(2n) 9ζ(3) G 12β(4) = − + − (2n + 1)(2n + 3)16n 16π2 π π3 n=0 X ∞ ζ(2n) 61ζ(3) 12β(4) = − + (2n + 1)(2n + 2)(2n + 3)16n 16π2 π3 n=0 X ∞ ζ(2n) 2ζ(3) 527ζ(5) 4β(4) = − + − (2n + 1) ... (2n + 4)16n π2 16π4 π3 n=0 X Now, we will use these results to establish the same families of general rational zeta series but for ζ(2n + 1).

5. Rational ζ(2n + 1) series using ψ(x)

Theorem 5.1. For p ∈ N and |z| < 1,

p z p!(−1)kψ(−k−1)(z) (38) xpψ(x) dx = zp−k. 0 (p − k)! Z Xk=0 Proof. Let f(z) be the left hand side and g(z) be the right hand side. It is clear that f(0) = g(0) = 0 and f ′(z)= zpψ(z). Taking the derivative of g(z), we ﬁnd

p p−1 p!(−1)kψ(−k)(z) p!(−1)kψ(−k−1)(z) g′(z)= zp−k + zp−k−1 (p − k)! (p − k − 1)! Xk=0 Xk=0 p p−1 p!(−1)kψ(−k)(z) p!(−1)kψ(−k−1)(z) = zpψ(z)+ zp−k + zp−k−1. (p − k)! (p − k − 1)! Xk=1 Xk=0 Reindexing the ﬁrst sum, one can see the two sums cancel each other out. So, g′(z)= f ′(z) for all z. Since they are equal at z = 0, then f(z)= g(z).

We can also compute this integral using (12). So we will have

z z 1 ∞ xpψ(x) dx = xp − γ − + (−1)kζ(k)xk−1 dx 0 0 x Z Z Xk=2 GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n)AND ζ(2n + 1) 17

γ zp ∞ (−1)kζ(k) = − zp+1 − + zk+p. p +1 p k + p Xk=2 Setting these two results equal to each other, one has

p ∞ (−1)kζ(k) 1 γz p!(−1)kψ(−k−1)(z) (39) zk = + + . k + p p p +1 (p − k)!zk Xk=2 Xk=0

Now we can split this up as follows:

∞ (−1)kζ(k)zk ∞ ζ(2n)z2n ∞ ζ(2n + 1)z2n+1 = − . k + p 2n + p 2n + p +1 n=1 n=1 Xk=2 X X Using (13) and rearranging, we see

∞ p k+1 ζ(2n + 1)z2n 1 γ p!(−1)⌊ 2 ⌋π (40) = − − + Cl (2πz) 2n + p +1 2zp p +1 (p − k)!(2πz)k+1 k+1 n=1 k=0 X Xp p p!(−1) 2 π p!(−1)kψ(−k−1)(z) − δ p p ζ(p + 1) − . ⌊ 2 ⌋, 2 (2πz)p+1 (p − k)!zk+1 Xk=0

For z =1/2 and z =1/4, we ﬁnd

⌊ p ⌋ ∞ ζ(2n + 1) 1 γ 2 p!(−1)k(4k − 1)ζ(2k + 1) (41) = − − log 2 − − (2n + p + 1)4n p p +1 (p − 2k)!(2π)2k n=1 k=1 X p X p p!(−1) 2 ζ(p + 1) p!(−2)kψ(−k−1)(1/2) − δ p p − 2 , ⌊ 2 ⌋, 2 πp (p − k)! Xk=0 and

⌊ p ⌋ ∞ ζ(2n + 1) 2 γ 2 p!(−1)k(4k − 1)ζ(2k + 1) (42) = − − log 2 − − (2n + p + 1)16n p p +1 (p − 2k)!(2π)2k n=1 k=1 p+1X X ⌊ 2 ⌋ p p p!(−4)kβ(2k) p!(−1) 2 2p+1ζ(p + 1) p!(−4)kψ(−k−1)(1/4) +π −δ p p −4 . (p +1 − 2k)!π2k ⌊ 2 ⌋, 2 πp (p − k)! Xk=1 Xk=0

Below we compute a few examples for speciﬁc p. Note that A is the Glaisher- 1 ′ Kinkelin constant, deﬁned by log A = 12 − ζ (−1). 18 DEREK ORR

∞ ζ(2n + 1) 1 = −2 − γ + 12 log A − log 2 (n + 1)4n 3 n=1 X ∞ ζ(2n + 1) 1 γ 1 = − − + 4 log A − log 2 (2n + 3)4n 2 3 3 n=1 X ∞ ζ(2n + 1) 199 γ 3 = − − + 8 log A − 56ζ′(−3) − log 2 (2n + 5)4n 180 5 5 n=1 X ∞ ζ(2n + 1) γ 1 = −2 − + 18 log A + log 2π2 − 4 log Γ (2n + 2)16n 2 4 n=1 X ∞ ζ(2n + 1) 1 1 γ 3ζ(3) 1 = −64ζ′ − 2, + + 4 log A + log 2π2 − + − 4 log Γ (2n + 3)16n 4 2 3 2π2 4 n=1 X

6. General ζ(2n + 1) series using ψ(−m)(x)

Theorem 6.1. For p ∈ N0, m ∈ N and |z| < 1,

p z p!(−1)kψ(−k−m−1)(z) (43) xpψ(−m)(x) dx = zp−k. 0 (p − k)! Z Xk=0 Proof. The proof is exactly the same as the proof of (38) as there was no depen- dence on ψ(x)= ψ(0)(x).

Now, let us compute this integral using (11) and (12). Doing so, we see

z z xp x ∞ (−1)kζ(k) xpψ(−m)(x) dx = (x−t)m−2 −γt−log t+ tk dt dx 0 0 (m − 2)! 0 k Z Z Z Xk=2 z −xp γxm xm−1(log x − H ) ∞ (−1)kζ(k)Γ(m − 1)xm+k−1 = + m−1 − dx 0 (m − 2)! m(m − 1) m − 1 k(k + 1) ... (k + m − 1) Z Xk=2 γzm+p+1 zm+p(H − log z) zm+p = − + m−1 + m!(m + p + 1) (m − 1)!(m + p) (m − 1)!(m + p)2 ∞ (−1)kζ(k)zm+p+k + . k(k + 1) ... (k + m − 1)(k + m + p) Xk=2

Setting this result equal to (43) and simplifying, one has the nice result GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n)AND ζ(2n + 1) 19

∞ (−1)kζ(k)zk γz (44) = k(k + 1) ... (k + m − 1)(k + m + p) m!(m + p + 1) k=2 X p log z − H 1 p!(−1)kψ(−k−m−1)(z) + m−1 − + . (m − 1)!(m + p) (m − 1)!(m + p)2 (p − k)!zm+k Xk=0 In the special case of p = 0, we have

∞ (−1)kζ(k)zk γz log z − H ψ(−m−1)(z) (45) = + m + . k(k + 1) ... (k + m) (m + 1)! m! zm Xk=2 Again, we can split (44) into

∞ (−1)kζ(k)zk ∞ ζ(2n)z2n = k(k + 1) ... (k + m − 1)(k + m + p) 2n(2n + 1) ... (2n + m − 1)(2n + m + p) n=1 Xk=2 X ∞ ζ(2n + 1)z2n+1 − , (2n + 1) ... (2n + m)(2n +1+ m + p) n=1 X and the ﬁrst sum has been computed earlier. Using (25), we ﬁnd

p ∞ ζ(2n + 1)z2n p!(−1)kψ(−k−m−1)(z) (46) = − (2n + 1) ... (2n + m)(2n + m +1+ p) (p − k)!zm+k+1 n=1 X Xk=0 m+p k+1 p+m (−1)mp! (−1)⌊ 2 ⌋ (−1) 2 ζ(p + m + 1) + Clk+1(2πz)−δ p+m p+m 2z (p + m − k)!(2πz)k ⌊ 2 ⌋, 2 (2πz)p+m k=m X− ⌊ m 1 ⌋ 1 2 (−1)kζ(2k + 1) log(2π/z)+ H − + m−1 2z (m − 1 − 2k)!(m + p − 2k)(2πz)2k 2z(m − 1)!(p + m) Xk=1 1 γ + − . 2z(m − 1)!(p + m)2 m!(p + m + 1)

If p = 0, we have the nice representation

− ⌊ m 1 ⌋ ∞ ζ(2n + 1)z2n ψ(−m−1)(z) 1 2 (−1)kζ(2k + 1) (47) = − − (2n + 1) ... (2n + m + 1) zm+1 2z (m − 2k)!(2πz)2k n=1 k=1 X ⌊ m ⌋ X (−1) 2 (Clm+1(2πz) − δ⌊ m ⌋, m ζ(m + 1)) log(2π/z)+ H γ + 2 2 + m − . 2z(2πz)m 2zm! (m + 1)! 20 DEREK ORR

and for z =1/2 and z =1/4, we have

∞ ζ(2n + 1) log 4π + H γ (48) = m − (2n + 1) ... (2n + m + 1)4n m! (m + 1)! n=1 X m−1 m ⌊ 2 ⌋ 2 m+1 k (−1) (2 − 1)ζ(m + 1) (−1) ζ(2k + 1) m+1 (−m−1) − δ m m − − 2 ψ (1/2), ⌊ 2 ⌋, 2 (2π)m (m − 2k)!π2k Xk=1

and

∞ ζ(2n + 1) 2 log 8π +2H γ (49) = m − (2n + 1) ... (2n + m + 1)16n m! (m + 1)! n=1 X m+1 m (−1) 2 2m+1β(m + 1) (−1) 2 (22m+1 +2m − 1)ζ(m + 1) − δ m+1 m+1 − δ⌊ m ⌋, m ⌊ 2 ⌋, 2 πm 2 2 (2π)m − ⌊ m 1 ⌋ 2 2(−4)kζ(2k + 1) − − 4m+1ψ(−m−1)(1/4). (m − 2k)!π2k Xk=1

For general p,

∞ ζ(2n + 1) log 4π + H (50) = m−1 (2n + 1) ... (2n + m)(2n +1+ m + p)4n (m − 1)!(m + p) n=1 X ⌊ p+m ⌋ 2 k k 1 γ m (−1) (4 − 1)ζ(2k + 1) + 2 − − (−1) p! 2k (m − 1)!(m + p) m!(m + p + 1) m+1 (p + m − 2k)!(2π) k=X⌊ 2 ⌋ p+m p m+1 2 k (−k−m−1) (−1) p!(−1) m+1 p!(−2) ψ (1/2) + δ⌊ p+m ⌋, p+m ζ(p + m + 1) − 2 2 2 πp+m (p − k)! k=0 − X ⌊ m 1 ⌋ 2 (−1)kζ(2k + 1) − , (m − 1 − 2k)!(m + p − 2k)π2k Xk=1

and GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n)AND ζ(2n + 1) 21

∞ ζ(2n + 1) 2 log 8π + H (51) = m−1 (2n + 1) ... (2n + m)(2n +1+ m + p)16n (m − 1)! (m + p) n=1 X ⌊ p+m+1 ⌋ ⌊ p+m ⌋ 1 2 p!(−1)m+k4kβ(2k) 2 p!(−1)k+m(4k − 1)ζ(2k + 1) + + − (m + p)2 (p + m +1 − 2k)!π2k−1 (p + m − 2k)!(2π)2k k=⌊ m+2 ⌋ k=⌊ m+1 ⌋ X2 X2 p+m p!(−1)m(−1) 2 2p+m+1ζ(p + m + 1) γ − δ p+m p+m − ⌊ 2 ⌋, 2 πp+m m!(m + p + 1) m−1 ⌊ ⌋ p 2 2(−4)kζ(2k + 1) p!(−4)kψ(−k−m−1)(1/4) − − 4m+1 . (m − 1 − 2k)!(m + p − 2k)π2k (p − k)! Xk=1 Xk=0 Below we compute some sums for certain m and p.

∞ ζ(2n + 1) 7 = −12 log A +2 − γ + log 2 (2n + 1)(n + 1)4n 3 n=1 X ∞ ζ(2n + 1) 1 γ 2 = −2 log A + − + log 2 (2n + 1)(2n + 3)4n 4 3 3 n=1 X ∞ ζ(2n + 1) 19 γ 89 = −4 log A + 20ζ′(−3) + − − log 2 (2n + 1)(n + 2)4n 36 2 90 n=1 X ∞ ζ(2n + 1) 3 γ = −8 log A + − + log 2 (2n + 1)(n + 1)(2n + 3)4n 2 3 n=1 X ∞ ζ(2n + 1) 53 γ 121 = −8 log A − 20ζ′(−3) + − + log 2 (2n + 1)(n + 1)(n + 2)4n 36 2 90 n=1 X ∞ ζ(2n + 1) 2 8 551 γ 1 = − log A + ζ′(−3) + − + log 2 (2n + 1) ... (2n + 5)4n 3 3 4320 120 24 n=1 X ∞ ζ(2n + 1) = −36 log A +4 − γ + 8 log 2 (2n + 1)(n + 1)16n n=1 X ∞ ζ(2n + 1) 1 3 1 γ = 32ζ′ − 2, − 2 log A − ζ(3) − − + 2 log 2 (2n + 1)(2n + 3)16n 4 4π2 4 3 n=1 X ∞ ζ(2n + 1) 187 γ 41 = −8 log A + 45ζ′(−3) + − + log 2 (2n + 1) ... (2n + 4)16n 144 24 60 n=1 X To conclude, we will revisit the digamma function and ﬁnd another general ζ(2n+ 1) series. 22 DEREK ORR

7. General ζ(2n + 1) series using ψ(x)

Similar to section 4, we will investigate the double integral

z x xp(x − t)mtψ(t) dt dx. Z0 Z0 Using the binomial theorem and (38) among other things, we ﬁnd

z x m m z x xp(x − t)mtψ(t) dt dx = (−1)j xp+m−j tj+1ψ(t) dt dx j 0 0 j=0 0 0 Z Z X Z Z

j+1 m m (−1)j(j + 1)!(−1)k z = xp+m+1−kψ(−k−1)(x) dx j (j +1 − k)! j=0 0 X Xk=0 Z

m m z = (−1)j xp+m+1ψ(−1)(x) dx j j=0 0 X Z m m m (−1)j(j + 1)!(−1)k+1 z + xp+m−kψ(−k−2)(x) dx j (j − k)! 0 Xk=0 Xj=k Z

z z z p+1 (−1) p+1 (−m−1) p (−m−2) = δm,0 x ψ (x) dx+(1−δm,0)m! x ψ (x) dx−(m+1)! x ψ (x) dx Z0 Z0 Z0

Simplifying and using (43), we ﬁnd

(52) z x p k (−k−m−3) p m p (−m−2) p!(−1) ψ (z) x (x−t) tψ(t) dt dx = m!z zψ (z)−(m+p+2) k . 0 0 (p − k)!z Z Z Xk=0

Now we will evaluate the same integral using the power series for ψ(x) and Fubini’s theorem once again. Doing so, we see

z x z x ∞ xp(x−t)mtψ(t) dt dx = − xp (x−t)m tγ +1− (−1)kζ(k)tk dt dx 0 0 0 0 Z Z Z Z Xk=2 z xm+2γ xm+1 ∞ (−1)kζ(k)m!xk+m+1 = − xp + − dx 0 (m + 1)(m + 2) m +1 (k + 1) ... (k + m + 1)! Z Xk=2 GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n)AND ζ(2n + 1) 23

−zp+m+3γ −zp+m+2 = + (m + 1)(m + 2)(m + p + 3) (m + 1)(m + p + 2) ∞ (−1)kζ(k)m!zp+m+2+k + (k + 1) ... (k + m + 1)(k + m + p + 2) Xk=2 Using this result and (52), we arrive at

∞ (−1)kζ(k)zk γz (53) = (k + 1) ... (k + m + 1)(k + m + p + 2) (m + 2)!(p + m + 3) k=2 X p 1 ψ(−m−2)(z) p!(−1)kψ(−k−m−3)(z) + + − (m + p + 2) . (m + 1)!(p + m + 2) zm+1 (p − k)!zk+m+2 Xk=0 Note when p = 0,

∞ (−1)kζ(k)zk γz + m +3 ψ(−m−2)(z) (m + 2)ψ(−m−3)(z) (54) = + − . (k + 1) ... (k + m + 2) (m + 3)! zm+1 zm+2 Xk=2 As before, we will split the sum in (53) as

∞ (−1)kζ(k)zk ∞ ζ(2n)z2n = (k + 1) ... (k + m + 1)(k + m + p + 2) (2n + 1) ... (2n + m + 1)(2n + m + p + 2) n=1 Xk=2 X ∞ ζ(2n + 1)z2n+1 − (2n + 2) ... (2n + m + 2)(2n + m + p + 3) n=1 X and using (32), we have

∞ ζ(2n + 1)z2n (55) =(p + m + 2) (−1)m+1p!∗ (2n + 2) ... (2n + m + 2)(2n + m + p + 3) n=1 X p+m+3 ⌊ k ⌋ p+m m π(−1) 2 Clk(2πz) p!(−1) 2 (−1) ζ(p + m + 3) − δ p+m p+m (p + m +3 − k)!(2πz)k ⌊ 2 ⌋, 2 2z(2πz)m+p+2 k=m+3 pX p!(−1)kψ(−k−m−3)(z) γ 1 + − − (p − k)!zk+m+3 (m + 2)!(p + m + 3) 2z(m + 1)!(p + m + 2) k=0 X m+1 m+1 ⌊ 2 ⌋ (−1)⌊ 2 ⌋ Cl (2πz) ψ(−m−2)(z) k(−1)kζ(2k + 1) + m+2 − + . 2z(2πz)m+1 zm+2 z(2πz)2k(m +1 − 2k)!(m + p +2 − 2k) Xk=1 If p = 0, this simpliﬁes to 24 DEREK ORR

∞ m+1 ζ(2n + 1)z2n (−1)⌊ 2 ⌋ Cl (2πz) (56) = m+2 (2n + 2) ... (2n + m + 3) 2z(2πz)m+1 n=1 X m+1 m ⌊ ⌋ ⌊ ⌋ 2 k (−1) 2 Clm+3(2πz) − δ⌊ m ⌋, m ζ(m + 3) (m + 2) k(−1) ζ(2k + 1) + 2 2 + 2z(2πz)m+2 z(2πz)2k(m +2 − 2k)! k=1 X (2γz + m + 3) zψ(−m−2)(z) − (m + 2)ψ(−m−3)(z) − − . 2z(m + 3)! zm+3

For z =1/2 and z =1/4, we see

∞ m+1 ζ(2n + 1) (−1) 2 (2m+1 − 1)ζ(m + 2) (57) = −δ m+1 m+1 (2n + 2) ... (2n + m + 3)4n ⌊ 2 ⌋, 2 (2π)m+1 n=1 X m+1 m ⌊ 2 ⌋ (−1) 2 (2m+3 − 1)(m + 2)ζ(m + 3) 2k(−1)kζ(2k + 1) − δ m m + ⌊ 2 ⌋, 2 (2π)m+2 π2k(m +2 − 2k)! Xk=1 γ 1 − 2m+2ψ(−m−2)(1/2)+(m + 2)2m+3ψ(−m−3)(1/2) − − , (m + 3)! (m + 2)!

and

∞ ζ(2n + 1) m (58) =(m + 2) 4m+3ψ(−m−3)(1/4)+(−1)⌊ 2 ⌋∗ (2n + 2) ... (2n + m + 3)16n n=1 X m+3 2m+5 m+2 2 β(m + 3) (2 +2 − 1)ζ(m + 3) ⌊ m+1 ⌋ δ m+1 m+1 − δ⌊ m ⌋, m +(−1) 2 ∗ ⌊ 2 ⌋, 2 πm+2 2 2 (2π)m+2 m+2 m+1 2 β(m + 2) (2 − 1)ζ(m + 2) m+2 (−m−2) δ⌊ m ⌋, m − δ m+1 m+1 − 4 ψ (1/4) 2 2 πm+1 ⌊ 2 ⌋, 2 (2π)m+1 ⌊ m+1 ⌋ 2 k(−4)kζ(2k + 1) γ 2 +4 − − . π2k(m +2 − 2k)! (m + 3)! (m + 2)! Xk=1

For general p, GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n)AND ζ(2n + 1) 25

∞ ζ(2n + 1) (59) =(−1)mp!(m + p + 2)∗ (2n + 2) ... (2n + m + 2)(2n + m + p + 3)4n n=1 p+Xm+2 ⌊ 2 ⌋ p+m k k 2 (−1) (4 − 1)ζ(2k + 1) (−1) ζ(p + m + 3) m − δ p+m p+m + 8(−2) ∗ (p + m +2 − 2k)!(2π)2k ⌊ 2 ⌋, 2 πm+p+2 m+3 k=X⌊ 2 ⌋ p m+1 k (−k−m−3) 2 m+1 (−2) ψ (1/2) (−1) (2 − 1)ζ(m + 2) m+2 (−m−2) m+1 m+1 −δ⌊ ⌋, m+1 −2 ψ (1/2) (p − k)! ! 2 2 (2π) Xk=0 ⌊ m+1 ⌋ 2 2k(−1)kζ(2k + 1) γ 1 + − − , π2k(m +1 − 2k)!(m + p +2 − 2k) (m + 2)!(p + m + 3) (m + 1)!(p + m + 2) Xk=1

and

∞ ζ(2n + 1) (60) =(−1)mp!(m + p + 2)∗ (2n + 2) ... (2n + m + 2)(2n + m + p + 3)16n n=1 X ⌊ p+m+2 ⌋ ⌊ p+m+3 ⌋ 2 (−1)k(4k − 1)ζ(2k + 1) 2 (−1)kβ(2k)4k − + 64(−4)m∗ (p + m +2 − 2k)!(2π)2k (p + m +3 − 2k)!π2k−1 m+3 m+4 k=X⌊ 2 ⌋ k=X⌊ 2 ⌋ p p+m k (−k−m−3) 2 p+m+3 (−4) ψ (1/4) (−1) ζ(p + m + 3)2 m+2 (−m−2) −δ p+m p+m −4 ψ (1/4) (p − k)! ⌊ 2 ⌋, 2 πm+p+2 k=0 ! X m+1 (−1)⌊ 2 ⌋2m+2β(m + 2) γ 2 + δ m m − − ⌊ 2 ⌋, 2 πm+1 (m + 2)!(p + m + 3) (m + 1)!(p + m + 2) m+1 m+1 ⌊ 2 ⌋ (−1) 2 (2m+1 − 1)ζ(m + 2) k(−1)kζ(2k + 1)4k+1 −δ m+1 m+1 + . ⌊ 2 ⌋, 2 (2π)m+1 π2k(m +1 − 2k)!(m + p +2 − 2k) Xk=1

Below we compute some sums for certain m and p.

∞ ζ(2n + 1) γ 1 = 4 log A − 1 − + log 2 (n + 1)(2n + 3)4n 3 3 n=1 X ∞ ζ(2n + 1) 5 γ 19 = 60ζ′(−3) − − + log 2 (n + 1)(n + 2)4n 12 2 30 n=1 X ∞ ζ(2n + 1) 4 112ζ′(−3) 19 γ 13 = − log A + + − + log 2 (n + 1)(2n + 5)4n 3 3 270 5 45 n=1 X 26 DEREK ORR

∞ ζ(2n + 1) 19 γ 1 = 8 log A − 60ζ′(−3) − − + log 2 (n + 1)(2n + 3)(n + 2)4n 12 6 30 n=1 X ∞ ζ(2n + 1) 4 28ζ′(−3) 289 γ 1 = log A − − − + log 2 (2n + 2)(2n + 3)(2n + 5)4n 3 3 1080 30 90 n=1 X ∞ ζ(2n + 1) 2 17ζ′(−3) 277 γ 1 = log A − − − − log 2 (2n + 2) ... (2n + 5)4n 3 3 2160 120 360 n=1 X ∞ ζ(2n + 1) 1 3ζ(3) γ = 128ζ′ − 2, + 28 log A − − 5 − (n + 1)(2n + 3)16n 4 π2 3 n=1 X ∞ ζ(2n + 1) 1 9ζ(3) 41 γ 1 = 384ζ′ −2, +24 log A+540ζ′(−3)− − − + log 2 (n + 1)(n + 2)16n 4 π2 12 2 5 n=1 X ∞ ζ(2n + 1) 1 3ζ(3) 79 γ 1 = −32ζ′ −2, +8 log A−135ζ′(−3)+ − − − log 2 (2n + 2) ... (2n + 4)16n 4 4π2 48 24 20 n=1 X

References [1] M. Abramowitz, I. A. Stegun, Handbook of Mathematical Functions with Formulas, Graphs and Mathematical Tables, New York, NY:Dover, 1972. [2] V. S. Adamchik, Polygamma functions of negative order, J. Comp. Appl. Math. 100 (1998) 191–199. [3] R. Apéry, Irrationalitéde ζ(2) et ζ(3), Astérisque, 61 (1979), 11–13. [4] T. M. Apostol, Another elementary proof of Euler’s formula for ζ(2n), Amer. Math. Monthly, 80 (1973), 425–431. [5] J. Choi, Some integral representations of the Clausen function Cl2(x) and the Catalan constant, East Asian Math. J. 32 (2016), 43–46. [6] J. Choi, H. M. Srivastava, The Clausen function Cl2(x) and its related integrals, Thai J. Math. 12 (2014), 251–264. [7] D. Cvijovic, Closed-form evaluation of some families of cotangent and cosecant integrals, Integral Transf. Spec. Func. 19 (2008), 147–155. [8] D. Cvijovic, J. Klinowski, New rapidly convergent series representations for ζ(2n + 1), Proc. Amer. Math. Soc. 125 (1997), 1263–1271. [9] P. J. de Doeder, On the Clausen integral Cl2(θ) and a related integral, J. Comp. Appl. Math. 11 (1984), 325–330. [10] O. Espinosa, V. H. Moll, A generalized polygamma function, Integral Transf. Spec. Funct. 15 (2004), 101–115. [11] J. A. Ewell, A new series representation for ζ(3), Amer. Math. Monthly 97 (1990), 219–220. [12] C. C. Grosjean, Formulae concerning the computation of the Clausen integral Cl2(θ), J. Comp. Appl. Math. 11 (1984), 331–342. [13] F.M.S. Lima, An Euler-type formula for β(2n) and closed-form expressions for a class of zeta series, Integral Transf. Spec. Funct. (2012), DOI:10.1080/10652469.2011.622274 [14] C. Lupu, D. Orr, Approximations for Apery’s constant ζ(3) and rational series representations involving ζ(2n), (2016) preprint at http://arxiv.org/abs/1605.09541. Bibliography 27

[15] H. M. Srivastava, M. L. Glasser, V. S. Adamchik, Some definite integrals associated with the Riemann zeta function, Z. Anal. Anwendungen 19 (2000), 831–846. [16] D. Tyler, P. R. Chernoff, An old sum reappears-Elementary problem 3103, Amer. Math. Monthly 92 (1985), 507. [17] J. Wu, X. Zhang, D. Liu, An efficient calculation of the Clausen functions, BIT Numer. Math. 50 (2010), 193–206. [18] W. V. Zudilin, One of the numbers ζ(5), ζ(7), ζ(9), ζ(11), is irrational, Uspekhi Mat. Nauk 56 (2001), 149–150.

University of Pittsburgh, Department of Mathematics, 301 Thackeray Hall, Pittsburgh, PA 15260, USA E–mail address: [email protected]