On Ramanujan's formula for ζ(1/2) and ζ(2m + 1)

ON RAMANUJAN’S FORMULA FOR ζ(1/2) AND ζ(2m + 1) arXiv:2106.04797v1 [math.NT] 9 Jun 2021 ANUSHREE GUPTA AND BIBEKANANDA MAJI Abstract. Page 332 of Ramanujan’s Lost Notebook contains a compelling identity for ζ(1/2), which has been studied by many mathematicians over the years. On the same page, Ramanujan also recorded the series, 1r 2r 3r + + +··· , s s exp(1 x) − 1 exp(2 x) − 1 exp(3s x) − 1 where s is a positive integer and r −s is any even integer. Unfortunately, Ramanujan doesn’t give any formula for it. This series was rediscovered by Kanemitsu, Tanigawa, and Yoshimoto, although they studied it only when r − s is a negative even integer. Recently, Dixit and the second author generalized the work of Kanemitsu et al. and obtained a transformation formula for the aforementioned series with r − s is any even integer. While extending the work of Kanemitsu et al., Dixit and the second author obtained a beautiful generalization of Ramanujan’s formula for odd zeta values. In the current paper, we investigate transformation formulas for an infinite series, and interestingly, we derive Ramanujan’s formula for ζ(1/2), Wigert’s formula for ζ(1/k) as well as Ramanujan’s formula for ζ(2m + 1). Furthermore, we obtain a new identity for ζ(−1/2) in the spirit of Ramanujan. 1. Introduction The study of understanding the algebraic nature of the special values of the Riemann zeta function ζ(s) has been a constant source of motivation for mathematicians to produce many beautiful results. In 1735, Euler established the following elegant formula for ζ(2m). For every positive integer m, (2π)2m B2m , (1.1) ζ(2m) = (−1)m+1 2(2m)! where B2m denotes the 2mth Bernoulli number. These are certain rational numbers, defined by the generating function: ∞ X z zn = , B n ez − 1 n=0 n! |z| < 2π. Euler’s formula along with the transcendence of π, proved by F. Lindemann, immediately tells us that all even zeta values are transcendental. However, the arithmetic nature of the odd zeta values is yet to be determined. Surprisingly in 1979, Roger Apéry [2, 3] proved the irrationality of ζ(3). But till today we do not know about the algebraic nature of ζ(3), that is, whether ζ(3) is algebraic or transcendental. In 2001, Rivoal [28] Ball and Rivoal [5] 2010 Mathematics Subject Classification. Primary 11M06; Secondary 11J81 . Keywords and phrases. Riemann zeta function, Odd zeta values, Ramanujan’s formula. 1 2 ANUSHREE GUPTA AND BIBEKANANDA MAJI made a breakthrough by proving that there exist infinitely many odd zeta values which are irrational, but their result does not tells us anything about the algebraic nature of a specific odd zeta value. Around the same time, Zudilin [32] proved that at least one of the members of ζ(5), ζ(7), ζ(9) or ζ(11) is irrational, which is the current best result in this direction. Before going to England, Ramanujan, in his second notebook [26, p. 173, Ch. 14, Entry 21(i)], noted down the following remarkable formula for odd zeta values ζ(2m + 1). Let α and β be two positive numbers such that αβ = π 2 . For any non-zero integer m, 1 (4α)m ∞ X n−2m−1 1 ζ(2m + 1) + 2 (e2nα − 1) n=1 ! (−1)m+1 + (4β)m = m+1 X ∞ X n−2m−1 1 ζ(2m + 1) + 2 (e2nβ − 1) n=1 (−1)k−1 k=0 ! B2k B2m+2−2k αm+1−k β k . (2k)! (2m + 2 − 2k)! (1.2) This formula can also be found in Ramanujan’s Lost Notebook [27, pp. 319-320, formula (28)]. In sharp contrast to Euler’s formula, it falls short of providing an explicit formula for odd zeta values, but it is indeed another wonderful discovery of Ramanujan that has attracted the attention of several mathematicians. The first published proof was given by Malurkar [21] in 1925, who had no idea about its presence in Ramanujan’s Notebook. Lerch, in 1901, proved a particular case of the formula (1.2), namely, corresponding to α = β = π and m an odd positive integer. Replacing m by 2m + 1 in (1.2), it reduces to ζ(4m + 3) + 2 2m+2 ∞ X (−1)j+1 B2j B4m+4−2j X n−4m−3 4m+3 4m+2 = π 2 . (e2πn − 1) (2j)!(4m + 2 − 2j)! n=1 (1.3) j=0 Note that the infinite series present in the left side of (1.3) is a rapidly convergent series, which indicates that ζ(4m + 3) is almost a rational multiple of π 4m+3 . Moreover, one can P n−4m−3 say that atleast one of the expressions ζ(4m + 3) or ∞ n=1 (e2πn −1) must be transcendental since the right side expression is transcendental. A folklore conjecture is that π and odd zeta values do not satisfy any non-zero polynomial with rational coefficients, which implies that all odd zeta values must be transcendental. Ramanujan’s formula (1.2) has a deep connection with the theory of Eisenstien series over SL2 (Z) and their Eichler integrals. To know more about this connection, one can see [12], [18]. In 1977, Bruce Berndt [7] obtained a modular transformation formula for a generalized Eisenstien series from which he was able to derive Euler’s formula for ζ(2m) as well as Ramanujan’s formula for ζ(2m + 1). Over the years, many mathematicians found generalizations of (1.2) in various directions. Readers are encouraged to see the paper of Berndt and Straub [13] for more information on the history of Ramanujan’s formula for odd zeta values. ON RAMANUJAN’S FORMULA FOR ζ(1/2) AND ζ(2m + 1) 3 In 2001, Kanemitsu, Tanigawa and Yoshimoto [20] studied a generalized Lambert series defined by ∞ X nN −2h , exp(nN x) − 1 n=1 (1.4) where h, N ∈ N such that 1 ≤ h ≤ N/2. They established an important modular transformational relation for this generalized Lambert series which enabled them to find a formula for the Riemann zeta function at rational arguments. Recently, Dixit and the second author [15] pointed out that the above generalized Lambert series is in fact present on page 332 of Ramanujan’s Lost Notebook [27] with more general conditions on the parameters. At the end of page 332, Ramanujan wrote 2r 3r 1r + + + ··· , exp(1s x) − 1 exp(2s x) − 1 exp(3s x) − 1 (1.5) where s is a positive integer and r − s is any even integer. Unfortunately, Ramanujan does not give any formula for the series (1.5). Comparing (1.4) and (1.5), one can observe that Kanemitsu et al. studied (1.5) with the restriction that r−s is a negative even integer that lies in a restricted domain. This motivated Dixit and the second author to generalize the main result of Kanemitsu et al. [20, Theorem 1.1] and find a formula for (1.5). While extending the main result of Kanemitsu et al., Dixit and the second author [15, Theorem 1.2] obtained the following beautiful generalization of the Ramanujan’s formula for ζ(2m + 1). Let N ≥ 1 be an odd positive integer and α, β be two positive real numbers such that αβ N = π N +1 . Then for any integer non-zero integer m 6= 0, we have ! ∞ −2N m−1 X n 1 − 2Nm ζ(2N m + 1) + α N+1 2 exp ((2n)N α) − 1 n=1 N−1  −m 22m(N −1) 2N = −β N+1 N 2N m m+ N+3 2 2 + (−1) N+3 1 ζ(2m + 1) + (−1) 2 2 +m ⌊ N+1 2N X ⌋ j=0 2 X j= −(N−1) 2 (−1)j ∞ X n=1 n−2m−1   iπj 1 exp (2n) N βe N − 1 2N 2 (m−j) 2j (−1)j B2j BN +1+2N (m−j) α N+1 β N + N+1 . (2j)!(N + 1 + 2N (m − j))! ! (1.6) This formula is of interest as it establishes a relation between two distinct odd zeta values, namely ζ(2m + 1) and ζ(2N m + 1) when N > 1. To derive the above generalization of Ramanujan’s formula for ζ(2m + 1), Dixit and the second author studied the following line integration representation of (1.4): ∞ X n=1 nN −2h 1 = N exp(n x) − 1 2πi where c0 > max(1, (N − 2h + 1)/N ). Z c0 +i∞ c0 −i∞ Γ(s)ζ(s)ζ(N s − (N − 2h))x−s ds, 4 ANUSHREE GUPTA AND BIBEKANANDA MAJI Inspired by the work of Dixit and the second author, in the current paper, we would like to investigate the following line integration: Z c+i∞ 1 Γ(s)ζ(ks)ζ(s − r)x−s ds, 2πi c−i∞ where k ∈ N, r ∈ Z and c > max (1/k, 1 + r). Surprisingly, when k − r is an even integer, we obtain nice transformation formulas which allow us to derive many well-known formulas in the literature, for example, Ramanujan’s formula for ζ(1/2), Wigert’s formula for ζ(1/k) for k ≥ 2 even, and Ramanujan’s formula for odd zeta values. Furthermore, we also obtain a new identity for ζ(−1/2) analogous to Ramanujan’s formula for ζ(1/2). 2. Preliminaries The study of the well-known divisor function d(n) plays an important role in analytic number theory. This function has been generalized in many directions, and one of the important generalizations is defined by X σk (m) := dk , d|m where k ∈ Z. In this paper, we encountered two new divisor functions that are defined by X  n r X  n −r Dk,r (n) := , and S (n) := dk−1 , (2.1) k,r k k d d k k d |n d |n where k ∈ N and r ∈ Z. At this point, it may look artificial to define the above two divisor functions, but in due time we will see that these divisor functions naturally arise with the theory. One can easily observe that D1,r (n) = σr (n), and S1,r = σ−r (n). (2.2) P σk (n) It is well-known that ζ(s)ζ(s−k) = ∞ n=1 ns , absolutely convergent for Re(s) > max{1, k+ 1}. We now look for the Dirichlet series associated to Dk,r (n) and Sk,r (n) in terms of the Riemann zeta function. Using the definition (2.1), we see that    r ∞ ∞ ∞ ∞ X X X X Dk,r (n) 1 X 1 1 n   = ζ(ks)ζ(s − r), (2.3) = = ns ns nk1 nks ns−r k n=1 n=1 n =1 1 n =1 2 n1 |n 1 2
valid for Re(s) > max k1 , 1 + r . Similarly, with the help of the definition (2.1), one can show that, for Re(s) > max(1, 1 − r), ∞ X Sk,r (n) n=1 ns = ζ(1 + ks − k)ζ(s + r). (2.4) P∞ −mx associated to the divisor function We observe that the Lambert series m=1 σk (m)e σk (m) is in fact present in Ramanujan’s formula (1.2) as it can be clearly seen from the ON RAMANUJAN’S FORMULA FOR ζ(1/2) AND ζ(2m + 1) 5 relation that ∞ X m=1 σk (m)e−mx = ∞ X m=1   X d|m  dk  e−mx = ∞ X ∞ X dk e−mdx = m=1 d=1 ∞ X nk . enx − 1 n=1 P −nx associated Let us try to find a similar expression for the Lambert series ∞ n=1 Dk,r (n)e P∞ k to the new divisor function Dk,r (n). We know that, for x > 0, m=1 e−n mx = nk1x . Now e −1 differentiating both sides of this series r times, and then taking sum over n, we obtain   ∞ X dr 1 r k r −nk mx (−1) (n m) e = dxr enk x − 1 m=1   ∞ ∞ X ∞ X X 1 (−1)r dr k mr e−n mx = ⇒ nkr dxr enk x − 1 n=1 n=1 m=1   ∞ ∞ X X (−1)r dr 1 −nx , for r ≥ 0. (2.5) ⇒ Dk,r (n)e = nkr dxr enk x − 1 n=1 n=1 Next, we state another important special function, the Meijer G-function, which is the generalization of many well-known special functions in the literature. Let 0 ≤ m ≤ q, 0 ≤ n ≤ p. Let a1 , · · · , ap and b1 , · · · , bq complex numbers such that ai − bj 6∈ N for 1 ≤ i ≤ n and 1 ≤ j ≤ m. Then the Meijer G-function [24, p. 415, Definition 16.17] is defined by the line integral: ! Qn Z Qm −s 1 j=1 Γ(bj + s) j=1 Γ(1 − aj − s)z m,n a1 , · · · , ap Qq Qp Gp,q z = ds, (2.6) 2πi L j=m+1 Γ(1 − bj − s) j=n+1 Γ(aj + s) b1 , · · · , b q where the vertical line of integration L, going from −i∞ to +i∞, separates the poles of the factors Γ(1 − aj − s) from those of the factors Γ(bj + s). This integral converges if p + q < 2(m + n) and | arg(z)| < (m + n − p+q 2 )π. Now we are ready to state the main results. 3. Main results Theorem 3.1. Let k ≥ 2 and r be an even integers. Let Dk,r (n) and Sk,r (n) be defined as in (2.1). Then for any x > 0, we have     k+r−2 k+1−2r ∞ X (−1) 2 (2π) 2 1 1 1 1 −nx − k1 Dk,r (n)e = − ζ(−r) + Γ − r x + R1+r + ζ 2k−1 2 k k k xk 2 n=1 ! (k−1) ∞ X ′′ X {} k,0 Sk,r (n) G0,k × (k−1) X(j) , 1 r, − , · · · , − k k j=−(k−1) n=1 where ′′ means summation runs over j = −(k − 1), −(k − 3), · · · , (k − 3), (k − 1), and X(j) := Xx,n,k (j) := e− iπj 2 (2π)k+1 n , kk x (3.1) 6 ANUSHREE GUPTA AND BIBEKANANDA MAJI and R1+r =  r!ζ(k(1 + r))x−(1+r) ,  (−1)1+r kζ ′ (k(1 (−(1+r))! + r))x−(1+r) , if r ≥ 0, if r < 0. (3.2) As an immediate consequence of the above identity, we obtain Ramanujan’s formula for ζ( 12 ). Corollary 3.2. Let α and β be two positive numbers such that αβ = 4π 3 .Then √ √   √ X √ √ ∞ 1 1 π2 β β cos( mβ) − sin( mβ) − e− mβ √ √ . = + + ζ + √ n2 α − 1 4 6α 4π 2 4π m(cosh( mβ) − cos( mβ)) e n=1 m=1 ∞ X 1 (3.3) Ramanujan1 recorded this formula on the page 332 of his Lost Notebook [27], where Dixit and the second author encountered the series (1.5). On the same page, Ramanujan mentioned another form of this identity, which can also be found in Ramanujan’s second notebook [26, Entry 8, Chapter 15] and in [8, p. 314]. The formula (3.3) has been generalized by many mathematicians, for more information readers can see [4, pp. 191-193], [11, p. 859, Theorem 10.1]. More generally, substituting r = 0 in Theorem 3.1, we obtain the following Wigert’s formula [31] for ζ( k1 ), for k ≥ 2 even. Corollary 3.3. For any x > 0 and k ≥ 2 even, we have     1  1 k ζ(k) 1 1 1 1 k 1 (−1) 2 −1 2π k = + + + Γ ζ x k k k x 4 k x enk x − 1 n=1 k ! −1 (  1 2 X iπ(2j+1)(k−1) 2π k − iπ(2j+1)) 2k 2k e × e L̄k 2π x j=0 !)  1 k iπ(2j+1) iπ(2j+1)(k−1) 2π 2k , L̄k 2π e 2k + e− x ∞ X 1 (3.4) where L̄k (x) := ∞ X n=1 1 n k −1 1 exp(xn k ) − 1 . This formula has been generalized by Dixit and the second author [15, Theorem 1.5] and further a two-variable generalization obtained by Dixit, Gupta, Kumar, and Maji [17, Theorem 2.12]. 1Ramanujan [27, p. 332] missed the factor 1/√m on the right side series expression of (3.3). ON RAMANUJAN’S FORMULA FOR ζ(1/2) AND ζ(2m + 1) 7 Theorem 3.4. Let k ≥ 2 be an even integer and r 6= −1 be an odd integer. Let Dk,r (n) and Sk,r (n) be defined as in (2.1). Then for any x > 0, we have     2k+r−1 k+1−2r ∞ X 1 1 1 1 (−1) 2 (2π) 2 − k1 −nx − r x + R1+r + ζ Dk,r (n)e = − ζ(−r) + Γ 2k−1 2 k k k xk 2 n=1 ! (k−1) ∞ X ′′ X {} k,0 × X(j) , Sk,r (n)G0,k ij r, − k1 , · · · , − (k−1) k n=1 j=−(k−1) where X(j) and R1+r are defined as in (3.1) and (3.2) respectively. Substituting k = 2 and r = 1 in the above result, we derive an interesting identity for
ζ − 12 . Corollary 3.5. For α, β > 0 such that αβ = 4π 3 , √     ∞ X 1 1 1 d 1 π4 β + ζ − = + 2 n2 dα 1 − en α 24 4π 2 90α2 n=1 ( ! √ √ √ ∞ cos( mβ) + sin( mβ) − e− mβ π X 1
√ √ √ − 4α m cosh( mβ mβ) − cos( mβ) m=1 ) √ √ 2 sin( mβ) sinh( mβ) √ √ √ . + (cos(mβ) cosh( mβ) − 1)2 + (sin( mβ) sinh( mβ))2 Note that, Theorem 3.4 does not hold for r = −1, so we present the next result corresponding to r = −1. Theorem 3.6. Let k ≥ 2 be an even integer. Then for x > 0, we have       ∞ X 1 x 1 1 1 1 Dk,−1 (n)e−nx = log Γ + ζ 1 + x− k k 2 (2π) k k k n=1 (k−1) k + (−1) X j=−(k−1) ′′ iπj e ∞ X m=1   − iπj 2k log 1 − exp −e 1+ k1 (2π)  m  1  k x . Till now, we have stated identities when k ≥ 2 is even. From the next result onwards we dealt with k ≥ 1 odd. It turns out that, Ramanujan’s formula for ζ(2m + 1) can be derived as a special case of our result when k is an odd integer. Theorem 3.7. Let k ≥ 1 be an odd integer and r 6= −1 be an odd integer. Let Dk,r (n) and Sk,r (n) be defined as in (2.1). Then for x > 0, we have     2k+r−1 k+1−2r ∞ X 1 1 1 1 (−1) 2 (2π) 2 1 − r x− k + R1+r + R + ζ Dk,r (n)e−nx = − ζ(−r) + Γ 2k−1 2 k k k xk 2 n=1 ! (k−1) ∞ X ′′ X {} k,0 X(−j) , (3.5) × ij Sk,r (n)G0,k r, − k1 , · · · , − (k−1) k n=1 j=−(k−1) 8 ANUSHREE GUPTA AND BIBEKANANDA MAJI where X(j) and R1+r are defined as in (3.1) and (3.2) respectively, and  0, if r ≥ 1, 1+r R= −(1+r) i+1
−r P (−1) Bk(2i+1)+1 B−2i−1−r (2π)  (−1) 2 x 2i+1 2 , if r ≤ −3. i=0 2 2π (2i+1)!(k(2i+1)+1) (−(2i+1+r))! The next result provides a relation between two odd zeta values, namely, ζ(2m + 1) and
ζ(2km + 1), and a zeta value at rational argument ζ k1 + 2m + 1 . Corollary 3.8. Let k ≥ 1 be an odd integer and m be a positive integer. For any x > 0, we have     ∞ X 1 1 1 1 1 −nx Dk,−(2m+1) (n)e = − ζ(2m + 1) + Γ + 2m + 1 x− k ζ 2 k k k n=1 2m  x (−1)m k(2km)!ζ(2km + 1) + 2 (2m)! (2π)k m  x 2i+1 (−1)i+1 Bk(2i+1)+1 B2m−2i (−1)m (2π)2m+1 X + 2 (2i + 1)!(k(2i + 1) + 1)(2m − 2i)! 2π i=0 + (−1)m (2π) k+1 +2m+1 2 (k−1) 1 x kk− 2 X j=−(k−1) k,0 × G0,k ′′ j i ∞ X Sk,−(2m+1) (n) n=1 ! {} X(−j) . −(2m + 1), − k1 , · · · , − (k−1) k Substituting k = 1 in Corollary 3.8, and simplifying, we derive Ramanujan’s formula for odd zeta values. Corollary 3.9. Let m be a positive integer. For any x > 0, ! ∞ ∞  x 2m ζ(2m + 1) X X 2 1 −nx − 4πx n m σ−(2m+1) (n)e + ζ(2m + 1) = (−1) σ−(2m+1) (n)e + 2 2π 2 n=1 n=1   m+1 1 X (−1)i+1 B2i B2m+2−2i x2m+1 2π 2i . (3.6) + 2 (2i)!(2m + 2 − 2i)! x i=0 One can easily show that (3.6) is equivalent to (1.2) under the substitution x = 2α and αβ = π 2 . Again, letting k = 1 and r = 2m + 1, for m ≥ 1, in Theorem 3.7, one can obtain the following interesting identity. Corollary 3.10. Let m be a positive integer. For α, β > 0 with αβ = π 2 , we have αm+1 ∞ ∞ X X B2m+2 n2m+1 n2m+1 m+1 − (−β) = (αm+1 − (−β)m+1 ) . 2nα 2nβ e −1 e −1 4m + 4 n=1 (3.7) n=1 This formula is a special case of (1.2) and it can be found in [26, Vol. 1, p. 259]. Note that Theorem 3.7 is not valid for r = −1. Corresponding to r = −1 and k ≥ 1 odd, we obtain the next result. ON RAMANUJAN’S FORMULA FOR ζ(1/2) AND ζ(2m + 1) 9 Theorem 3.11. Let k ≥ 1 be an odd integer. Then for x > 0, we have       ∞ X 1 x 1 1 1 ζ(−k)x 1 −nx + Γ ζ 1+ x− k + Dk,−1 (n)e = log k 2 (2π) k k k 2 n=1 (k−1) − X ′′ j=−(k−1) ∞ X m=1   log 1 − exp −e iπj 2k 1+ k1 (2π)  m  1  k x . (3.8) Plugging k = 1 in the above identity, we obtain an interesting identity. Corollary 3.12. For α, β > 0 and αβ = π 2 , ∞ X n=1 1 n(e2nα − 1) − ∞ X n=1 1 n(e2nβ β−α 1 = − log − 1) 12 4   α . β (3.9) Remark 1. This identity is popularly known as the transformation formula for the logarithm πiz Q 2nπiz ), which is an important of the Dedekind eta function η(z), defined as e 12 ∞ n=1 (1 − e example of a half-integral weight modular form. Ramanujan recorded this identity twice in his second notebook [26, Ch. 14, Sec. 8, Cor. (ii) and Ch. 16, Entry 27(iii)]. For more information, one can see [8, p. 256], [9, p. 43], [27, p. 320, Formula (29)]. In the next chapter we mention a few important results which will be useful throughout the manuscript. We mention the proof of those results which are not well-known. 4. Well-known results An asymmetric form of the functional equation of ζ(s) can be written as  πs  . ζ(s) = 2s π s−1 ζ(1 − s)Γ(1 − s) sin 2 (4.1) We know that ζ(s) has a simple pole at s = 1 with residue 1 and the Laurent series expansion is ζ(s) = ∞ X 1 (−1)n γn (s − 1)n +γ+ , s−1 n! n=1 where γn are called the Stieltjes constants and defined by   i n+1 n X (log i) (log j) , − γn = lim  i→∞ j n+1 (4.2) j=1 and γ is the well-known Euler–Mascheroni constant. The Laurent series expansion of Γ(s) at s = 0 is     1 1 π2 1 π2 2 3 + 2ζ(3) s2 + · · · . (4.3) Γ(s) = − γ + γ + s− γ +γ s 2 6 6 2 More generally, Γ(s) has Laurent series expansion at every negative integer. 10 ANUSHREE GUPTA AND BIBEKANANDA MAJI Lemma 4.1. For r ≤ −2, we have a−1 + a0 + a1 (s − (1 + r)) + · · · , Γ(s) = s − (1 + r)  P−(1+r)  (−1)r+1 (−1)r+1 −γ + k=1 k1 . and a0 = (−(1+r))! where a−1 = (−(1+r))! Euler proved that Γ(s) satisfy the following beautiful reflection formula: π Γ(s)Γ(1 − s) = , ∀s ∈ C − Z. sin(πs) (4.4) The multiplication formula for the gamma function stated as Lemma 4.2. For any positive integer k, kks Γ(ks) = √ k−1 k(2π) 2 In particular, substituting k = 2, we have k−1 Y l=1  l Γ(s)Γ s + k  . (4.5)   1 22s−1 Γ(s)Γ s + . Γ(2s) = π 2 This is known as duplication formula. The next result gives an important information about the asymptotic expansion of the gamma function, popularly known as Stirling’s formula. Lemma 4.3. For s = σ + iT in a vertical strip α ≤ σ ≤ β,    √ 1 σ−1/2 − 12 π|T | , 1+O e |Γ(σ + iT )| = 2π|T | |T | (4.6) as |T | → ∞. Proof. The proof of this result can be found in [19, p. 151].  Lemma 4.4. For σ ≥ σ0 , there exist a constant C(σ0 ), such that |ζ(σ + iT )| ≪ |T |C(σ0 ) as |T | → ∞. Proof. One can find the proof in [30, p. 95].  Next, we shall mention a few important special cases of the Meijer G-function. An immediate special case is the following result. Putting m = q = 1 and n = p = 0 in (2.6), one can see that ! Z c+i∞ 1 1,0 {} G0,1 z = Γ(b + s)z −s ds = e−z z b , (4.7) 2πi c−i∞ b provided Re(s) = c > − Re(b) and Re(z) > 0. Again, plugging m = q = 2 and n = p = 0 in the definition (2.6) of the Meijer G-function, we will have the following result. ON RAMANUJAN’S FORMULA FOR ζ(1/2) AND ζ(2m + 1) 11 Lemma 4.5. Assuming the real part of the line of integration is bigger than max(−Re(b1 ), − Re(b2 )). For | arg(z)| < π, we have ! √ 1 {} 2,0 G0,2 z = 2z 2 (b1 +b2 ) Kb1 −b2 (2 z), b1 , b2 where Kν (z) is the modified Bessel function. Proof. This can be verified from [23, p. 115, Equn (11.1)]. Lemma 4.6. Let k ≥ 2 be a positive integer. Then k,0 G0,k provided | arg(z)| < {} 2 1 b, b + k , b + k , · · · , b + (k−1) k z !  k−1 (2π) 2 b −kz 1/k = √ z e , k kπ 2 . Proof. With the help of the definition (2.6) of the Meijer-G function, we write !   Z Y k {} 1 l k,0 z = G0,k z −s ds, Γ(s + b)Γ s + b + 2πi k b, b + k1 , · · · , b + (k−1) L k l=1 provided the poles of each of these gamma factor lie on the left of the line of integration. Replacing s by s + b in the multiplication formula (4.5) for Γ(s), we have   k−1 kk(s+b) Y l . Γ(k(s + b)) = √ Γ(s + b)Γ s + b + k−1 k k(2π) 2 l=1 Plugging this expression, one can see that ! Z √ k−1 1 {} Γ(k(s + b)) −s k,0 2 G0,k k(2π) z = z ds (k−1) 1 2πi L kk(s+b) b, b + k , · · · , b + k Z √ s1 k−1 1 ds1 2 Γ(s1 )z − k +b s1 +1 = k(2π) 2πi L′ k k−1 Z 1 (2π) 2 1 Γ(s1 )(kz k )−s1 z b ds1 = √ k 2πi L′ k−1 (2π) 2 z b −kz k1 √ , = e k the last step is possible only when | arg(z)| < kπ 2 . Dixit and Maji [15, Lemma 3.1] used the below lemma to prove (1.6). Lemma 4.7. Let a, u, v be three real numbers. Then   cos(a sin(u) + uv) − e−a cos(u) cos(uv) eiuv . = 2 Re exp(ae−iu ) − 1 cosh(a cos(u)) − cos(a sin(u)) In a similar vein, we have the following lemma.  12 ANUSHREE GUPTA AND BIBEKANANDA MAJI Lemma 4.8. Let a, u, v be three real numbers. Then   ieiuv − sin(uv + a sin(u)) + e−a cos(u) sin(uv) . 2 Re = exp(ae−iu ) − 1 cosh(a cos(u)) − cos(a sin(u)) Proof. The left hand side expression can be written as     − sin(uv) + i cos(uv) ieiuv = 2 Re a cos(u) 2 Re , exp(ae−iu ) − 1 e (cos(a sin(u)) − i sin(a sin(u))) − 1 now multiplying the numerator and the denominator by the conjugate of the denominator reduces to ! (− sin(uv) + i cos(uv))(ea cos(u) cos(a sin(u)) + iea cos(u) sin(a sin(u))) − 1) 2 Re e2a cos(u) − 2ea cos(u) (cos(a sin(u))) + 1 ! −ea cos(u) (cos(a sin(u)) sin(uv)) + sin(uv) − ea cos(u) (sin(a sin(u)) cos(uv)) =2 e2a cos(u) − 2ea cos(u) (cos(a sin(u))) + 1 ! −ea cos(u) (sin(uv + a sin(u)) + sin(uv) , =2 e2a cos(u) − 2ea cos(u) (cos(a sin(u))) + 1 = − sin(uv + a sin(u)) + e−a cos(u) sin(uv) . cosh(a cos(u)) − cos(a sin(u)) In the final step we divided by ea cos(u) on the numerator as well as on the denominator.  The next lemma will play a crucial role to obtain our main results. Lemma 4.9. Let s ∈ C and k ∈ N. Then sin(ks) = sin(s) (k−1) X ′′ exp(ijs), j=−(k−1) where ′′ means summation runs over j = −(k − 1), −(k − 3), · · · , (k − 3), (k − 1). Thus, for k even, (k−1) X ′′ k sin(ks) 2 = (−1) ij exp(ijs), (4.8) cos(s) j=−(k−1) and for k odd, k−1 cos(ks) = (−1) 2 cos(s) (k−1) X ′′ ij exp(−ijs). (4.9) j=−(k−1) 5. Proof of main results Proof of Theorem 3.1. It is well-known that e−x is the inverse Mellin transform of the gamma function Γ(s), that is, for any x > 0, Z c+i∞ 1 −x Γ(s)x−s ds, (5.1) e = 2πi c−i∞ ON RAMANUJAN’S FORMULA FOR ζ(1/2) AND ζ(2m + 1) 13 valid for any c > 0. Replacing x by nx in (5.1) and using (2.3), one can easily show that ∞ X Dk,r (n)e−nx = n=1 1 2πi Z c+i∞ c−i∞ Γ(s)ζ(ks)ζ(s − r)x−s ds, (5.2) provided with Re(s) = c > max( k1 , 1 + r). In order to evaluate this line integral we choose a contour C in a way so that all the poles of the integrand function lie inside the contour C, determined by the line segments [c − iT, c + iT ], [c + iT, −λ + iT ], [−λ + iT, −λ − iT ], and [−λ − iT, c − iT ], where we wisely take λ > max(0, −r) and T is some large positive real number. The essence of this choice of λ is justified at the later stage in our proof. Now applying Cauchy’s residue theorem, we have Z c+iT Z −λ+iT Z −λ−iT Z c−iT 1 Γ(s)ζ(ks)ζ(s − r)x−s ds + + + 2πi c−iT −λ−iT −λ+iT c+iT X = Res Γ(s)ζ(ks)ζ(s − r)x−s , (5.3) ρ s=ρ where the sum over ρ runs through all the poles of Γ(s)ζ(ks)ζ(s − r)x−s inside the contour C. First, let us analyse the poles of the integrand function. We know that Γ(s) has simple poles at s = 0 and negative integers. The poles at negative integers are getting cancelled by the trivial zeros of ζ(ks) as we are dealing with k even. Thus, the only simple poles of the integrand function Γ(s)ζ(ks)ζ(s − r)x−s are at s = 0, k1 and 1 + r, due to the simple pole of ζ(ks) at s = 1/k and the pole of ζ(s − r) at s = 1 + r. Here we would like to mention that, as we have considered r is any even integer, so it may happen that 1 + r is a negative integer, and in that case, it seems that 1 + r is a pole of order 2 due to the factor Γ(s)ζ(s − r), but ζ(ks) will have a zero at 1 + r since k is even. This shows that 1 + r remains a simple pole of the integrand function as we are dealing with k ≥ 2 is even and r is even. One can easily evaluate residues at s = 0 and s = k1 . Let Rρ denotes the residue term at ρ. Using residue calculation methods, we have 1 R0 = lim sΓ(s)ζ(ks)ζ(s − r)x−s = ζ(0)ζ(−r) = − ζ(−r). s→0 2 (5.4) Similarly, we obtain R 1 = lim k s→ k1  s− 1 k  Γ(s)ζ(ks)ζ(s − r)x−s = 1 Γ k     1 1 1 − r x− k . ζ k k (5.5) Here we note that while calculating the residue at s = 1 + r, we have to make two different cases. First, if r ≥ 0, then R1+r = lim (s − (1 + r))Γ(s)ζ(ks)ζ(s − r)x−s = Γ(1 + r)ζ(k(1 + r))x−(1+r) s→1+r = r!ζ(k(1 + r))x−(1+r) . (5.6) 14 ANUSHREE GUPTA AND BIBEKANANDA MAJI Second, if r < 0 is a negative even integer, then 1 + r is also a negative integer. Thus, Γ(s) satisfy the Laurent series expansion (4.1) at 1 + r. Therefore, we have lim (s − 1 − r)Γ(s) = s→1+r (−1)1+r . (−(1 + r))! (5.7) Again, ζ(ks) has a trivial zero at s = 1 + r as k is even, so we have ζ(ks) = kζ ′ (k(1 + r)). s→1+r (s − 1 − r) lim (5.8) Combining (5.7) and (5.8) and together with the fact that 1 + r is a simple pole of ζ(s − r), we obtain R1+r = lim (s − (1 + r))Γ(s)ζ(ks)ζ(s − r)x−s = s→(1+r) (−1)1+r kζ ′ (k(1 + r))x−(1+r) . (5.9) (−(1 + r))! We now proceed to show that the horizontal integrals Z −λ+iT 1 H1 (T, x) := Γ(s)ζ(ks)ζ(s − r)x−s ds 2πi c+iT and H2 (T, x) = 1 2πi Z c−iT −λ−iT Γ(s)ζ(ks)ζ(s − r)x−s ds vanish as T → ∞. Replace s by σ + iT in H1 (T, x) to see Z d 1 |H1 (T, x)| = Γ(σ + iT )ζ(kσ + ikT )ζ(σ + iT − r)x−σ+iT dσ 2πi c Z d |Γ(σ + iT )ζ(kσ + ikT )ζ(σ + iT − r)|x−σ dσ ≪ c  π  ≪ |T |A exp − |T | , 2 for some constant A. In the final step, we used Stirling’s formula for Γ(s) i.e., Lemma 4.3 and bound for ζ(s) i.e., Lemma 4.4. This immediately implies that H1 (T, x) vanishes as T → ∞. Similarly, one can show that H2 (T, x) also vanishes as T → ∞. Now letting T → ∞ in (5.3) and collecting all the residual terms, we have Z 1 Γ(s)ζ(ks)ζ(s − r)x−s ds = R0 + R 1 + R1+r k 2πi (c) Z 1 Γ(s)ζ(ks)ζ(s − r)x−s ds. (5.10) + 2πi (−λ) R R c+i∞ Here and throughout the paper we denote (c) by c−i∞ . Now one of our main objectives is to evaluate the following vertical integral Z Γ(s)ζ(ks)ζ(s − r)x−s ds. (5.11) Vk,r (x) := (−λ) ON RAMANUJAN’S FORMULA FOR ζ(1/2) AND ζ(2m + 1) Utilize the asymmetric form (4.1) of the functional equation of ζ(s) to write   πks ks ks−1 ζ(ks) = 2 π ζ(1 − ks)Γ(1 − ks) sin , 2   π(s − r) s−r s−r−1 . ζ(s − r) = 2 π ζ(1 − s + r)Γ(1 − s + r) sin 2 15 (5.12) (5.13) Substituting (5.12) and (5.13) in (5.11), the vertical integral becomes Z 1 1 Γ(s)Γ(1 − ks)Γ(1 − s + r)ζ(1 − ks)ζ(1 − s + r) Vk,r (x) = (2π)r π 2 2πi (−λ)    s  π(s − r) (2π)k+1 πks ds. sin × sin 2 2 x Now to shift the line of integration, we replace s by 1 − s. Then we have Z (2π)k+1−r 1 Γ(1 − s)Γ(1 − k + ks)Γ(s + r)ζ(1 − k + ks)ζ(s + r) Vk,r (x) = π2 x 2πi (1+λ)    −s  π(1 − s − r) (2π)k+1 πk(1 − s) ds. (5.14) sin × sin 2 2 x Since k and r are both even, the following identities hold     k ksπ πk(1 − s) −1 sin = (−1) 2 sin , 2 2    πs  r π(1 − s − r) . = (−1) 2 cos sin 2 2 (5.15) (5.16) Using the above two expressions in (5.14), we see k+r Z (−1) 2 −1 (2π)k+1−r 1 Vk,r (x) = Γ(1 − s)Γ(1 − k + ks)Γ(s + r)ζ(1 − k + ks) π2 x 2πi (1+λ)    πs   (2π)k+1 −s πks ×ζ(s + r) sin cos ds 2 2 x k+r Z (−1) 2 −1 (2π)k+1−r 1 Γ(1 − s)Γ(1 − k + ks)Γ(s + r)ζ(1 − k + ks) = 2π 2 x 2πi (1+λ)
 −s sin πks (2π)k+1 2
sin(πs) ×ζ(s + r) ds x sin πs 2 k+r Z Γ(1 − k + ks)Γ(s + r) (−1) 2 −1 (2π)k+1−r 1 ζ(1 − k + ks) = 2πx 2πi (1+λ) Γ(s)
 −s sin πks (2π)k+1 2
×ζ(s + r) ds. (5.17) x sin πs 2 Here in the ultimate step, we have used Euler’s reflection formula (4.4) for Γ(s) and in the
in the denominator as well as in the numerator. penultimate step we multiplied by sin πs 2 Now we notice that Re(s + r) = 1 + λ + r > 1 and Re(1 − k + ks) = 1 + kλ > 1 as we have considered λ > max(0, −r). This explains our choice of λ. Thus, we can express 16 ANUSHREE GUPTA AND BIBEKANANDA MAJI ζ(1−k+ks)ζ(s+r) as an infinite series, mainly, use (2.4) and then interchange the summation and integration in (5.17) to deduce that Vk,r (x) = (−1) k+r −1 2 x ∞ (2π)k−r X Sk,r (n)Ik,r (n, x), (5.18) n=1 where 1 Ik,r (n, x) := 2πi Z (1+λ) Γ(1 − k + ks)Γ(s + r) sin Γ(s) sin πks 2
πs 2
 (2π)k+1 n x −s ds. (5.19) Now our main goal is to simplify this integral. Invoking Lemma 4.9 in (5.19) we see that !−s iπj (k−1) X ′′ 1 Z Γ(1 − k + ks)Γ(s + r) e− 2 (2π)k+1 n ds. (5.20) Ik,r (n, x) = 2πi (1+λ) Γ(s) x j=−(k−1) Employing multiplication formula (4.5) for Γ(s), one can derive that Γ(1 − k + ks) = Γ(s) kks 1 kk− 2 (2π) k−1 2 k−1 Y l=1  l Γ s− k  . (5.21) Substituting (5.21) in (5.20) yields that Ik,r (n, x) = (k−1) 1 1 kk− 2 (2π) k−1 2 X j=−(k−1) ′′ 1 2πi Z Γ(s + r) (1+λ) k−1 Y l=1  l Γ s− k  X(j)−s ds, (5.22) where X(j) = e− iπj 2 (2π)k+1 n kk x (5.23) is exactly same as defined in (3.1). Now we shall try to express, the line integral in (5.22), in terms of the Meijer G-function and for that we have to verify all the necessary conditions for the convergence of the integral. First, one can check that all the poles of the integrand function lie on the left of the line of integration Re(s) = 1 + λ since λ > max(0, −r). Now using the definition (2.6) of the Meijer G-function, with m = q = k and n = p = 0 and b1 = r, b2 = − k1 , b3 = − k2 , · · · , bk = − k−1 k , we find that !  Z k−1 Y  {} 1 l k,0 −s (5.24) Γ(s + r) Γ s− X(j) ds = G0,k (k−1) X(j) , 1 2πi (1+λ) k , · · · , − r, − k k l=1 p+q πk convergent since p + q < 2(m + n) and | arg(X(j))| = | πj 2 | < (m + n − 2 ) = 2 as |j| ≤ k − 1. Substituting (5.24) in (5.22), we arrive at ! (k−1) X ′′ k,0 {} 1 G0,k X(j) . (5.25) Ik,r (n, x) = k−1 1 r, − k1 , · · · , − (k−1) kk− 2 (2π) 2 j=−(k−1) k ON RAMANUJAN’S FORMULA FOR ζ(1/2) AND ζ(2m + 1) 17 Now plugging this final expression (5.25) of Ik,r (n, x) in (5.18), the left vertical integral becomes ! k+1 k+r (k−1) ∞ {} (−1) 2 −1 (2π) 2 −r X ′′ X k,0 X(j) . Vk,r (x) = Sk,r (n)G0,k k− 21 r, − k1 , · · · , − (k−1) k x k j=−(k−1) n=1 (5.26) At this moment, we must show that the infinite series ∞ X k,0 Sk,r (n)G0,k n=1 ! {} X(j) r, − k1 , · · · , − (k−1) k (5.27) is convergent for any fixed k ≥ 1, r ∈ Z, |j| ≤ k − 1. To show the convergence of this series, we shall use the integral representation (5.24) of the Meijer G-function. Employing Stirling’s formula (4.6) on each gamma factor that is present in (5.24), letting s = 1 + λ + iT , we have   π 1 |Γ(s + r)| = |Γ(1 + λ + r + iT ) = O |T |λ+r+ 2 e− 2 |T | ,       1 π l l l Γ s− = Γ 1 + λ − + iT = O |T |λ− k + 2 e− 2 |T | , for 1 ≤ l ≤ k − 1, k k as T → ∞. Using the definition (5.23) of X(j), we note that −s |X(j) |= e− iπj 2 (2π)k+1 n xkk !−1−λ−iT = Ok,x  1 n1+λ e π|j||T | 2  . Utilizing the above bounds for gamma functions and the bound for X(j), and upon simplification, we derive that !   Z ∞ {} 1 k,0 kλ+r+ 12 − π2 (k−|j|)|T | |T | e dT G0,k X(j) ≪ Ok,x n1+λ −∞ r, − k1 , · · · , − (k−1) k   1 ≪ Ok,r,x . (5.28) n1+λ The integral present inside the above big-oh bound is convergent since |j| ≤ k − 1 and kλ + r + 32 is a positive quantity as we have λ > max(0, −r). Plugging (5.28) in (5.27), we can P −(1+λ) see that the infinite series (5.27) is convergent since the Dirichlet series ∞ n=1 Sk,r (n)n is convergent. Finally, combining (5.2), (5.10), (5.11), and (5.26), and together with all the residual terms (5.4), (5.5) (5.6), and (5.9), we complete the proof of Theorem 3.1.  5.1. Recovering Ramanujan’s formula for ζ(1/2). 18 ANUSHREE GUPTA AND BIBEKANANDA MAJI Proof of Corollary 3.2. Substituting k = 2 and r = 0 in Theorem 3.1, one can see that   √ ∞ X π 1 1 π2 −nx D2,0 (n)e = + √ ζ + 4 2 x 2 6x n=1 ! !# " iπ iπ ∞ {} e 2 2nπ 3 {} e− 2 2nπ 3 π 3/2 X 2,0 2,0 + G0,2 . S2,0 (n) G0,2 + x x x 0, − 12 0, − 12 n=1 Observe that the Meijer G-functions that are present in the above equation are conjugate to each other. Thus, letting x = α and αβ = 4π 3 , and using (2.5), we obtain   √ ∞ X 1 1 1 π2 π √ = + + ζ 2 4 2 α 2 6α en α − 1 n=1 !# " ∞ − iπ 2 nβ e 2π 3/2 X {} 2,0 + . (5.29) S2,0 (n) Re G0,2 α 2 0, − 21 n=1 Recall Lemma 4.6 with k = 2 and b = − 12 , to produce ! r   p − iπ 2 nβ 2π iπ e {} 2,0 − iπ 4 exp − 4 . 2nβe G0,2 e = 2 nβ 0, − 21 (5.30) Now utilizing (5.30) in (5.29), and using the definition (2.1) of S2,0 (n), we see that   √ ∞ X π 1 1 1 π2 √ = + ζ + 4 2 α 2 6α en2 α − 1 n=1 r ∞ i h iπ  p iπ 2π X X d √ Re e 4 exp − 2nβe− 4 + . α n n=1 2 (5.31) d |n Next aim is to simplify the infinite series ∞ X i  p h iπ X iπ d √ Re e 4 exp − 2nβe− 4 . n n=1 2 d |n Writing n = d2 m, one can simplify the infinite sum in a following way ∞ X i h iπ  p X iπ d √ Re e 4 exp − 2nβe− 4 n n=1 2 d |n ∞ X ∞ X h iπ i  p iπ 1 √ Re e 4 exp − 2mβde− 4 m m=1 d=1   iπ ∞ X 1 e4   √ √ Re  = − iπ m exp 2mβe 4 − 1 m=1 = ∞ X 1 √ = 2 2m m=1 ! √ √ √ cos( mβ) − sin( mβ) − e− mβ √ √ . (cosh( mβ) − cos( mβ)) In the final step, we have used Lemma 4.7 with u = π/4, v = 1, and a = substituting (5.32) in (5.31), one can finish the proof of (3.3). (5.32) √ 2mβ. Now ON RAMANUJAN’S FORMULA FOR ζ(1/2) AND ζ(2m + 1) 19  5.2. Wigert’s formula for ζ(1/k) for k ≥ 2 even. Proof. First, substituting r = 0 in Theorem 3.1, we find that     ∞ X 1 1 1 1 1 ζ(k) −nx Dk,0 (n)e = + Γ ζ x− k + 4 k k k x n=1 + (−1) k−2 2 k iπj − 2 (2π) 2k−1 2 (k−1) k+1 2 x ′′ X ∞ X ! {} X(j) , 0, − k1 , · · · , − (k−1) k k,0 Sk,0 (n) G0,k n=1 j=−(k−1) (5.33) k+1 (2π) n where X(j) = e . Now we shall try to simplify the last term and to do that we kk x make use of Lemma 4.6, with b = − k−1 k , which gives rise to ! k−2 k+1 {} (−1) 2 (2π) 2 k,0 G0,k X(j) 2k−1 0, − k1 , · · · , − (k−1) k 2 x k   1 k  1  1 (−1) 2 −1 2π k iπj(k−1) 1 −1 − iπj +1 n k 2k k 2k k n (2π) exp −e e . = k x x Implement this simplification of the Meijer G-function in the last term of (5.33), and use (2.5) and the definition (2.1) of Sk,0 (n), then (5.33) becomes      1 k ∞ X 1 1 ζ(k) (−1) 2 −1 2π k 1 1 1 − k1 + Γ + ζ x + = 4 k k k x k x enk x − 1 n=1 (k−1) × X ′′ ∞ X X k−1 d n 1 −1 k e iπj(k−1) 2k n=1 dk |n j=−(k−1)  − iπj 2k exp −e (2π) 1 +1 k n1  k x . (5.34) Now writing n = dk m and upon simplification, the last term reduces to   1 (k−1)  k ∞ ∞  1  1 (−1) 2 −1 2π k X ′′ X 1 −1 iπj(k−1) X +1 m k − iπj k 2k 2k k d e (2π) exp −e m k x x m=1 d=1 j=−(k−1) k (−1) 2 −1 = k k (−1) 2 −1 = k (k−1)  2π x 1  2π x k −1 " 1 X 2 k k ′′ X e iπj(k−1) 2k m=1 j=−(k−1) e ∞ X iπ(2j+1)(k−1) 2k ∞ X m=1 j=0 − +e iπ(2j+1)(k−1) 2k 1 m k −1  iπj 1 exp e− 2k (2π) k +1 m x 1 m k −1
1  iπ(2j+1) 1 exp e− 2k (2π) k +1 ∞ X 1 m k −1  iπ(2j+1) 1 exp e 2k (2π) k +1 m=1 k m x −1
1  m x k −1
1  k −1 # . (5.35) Ultimately, plugging (5.35) in (5.34), we complete the proof of Wigert’s formula (3.4) for ζ( k1 ), k ≥ 2 even. 20 ANUSHREE GUPTA AND BIBEKANANDA MAJI  Proof of Theorem 3.4. The proof goes in accordance with the proof of Theorem 3.1, so we mention those steps where it differs from Theorem 3.1. Note that, in this case, we are dealing with k ≥ 2 even and r 6= −1 is any odd integer. First, we point out that, the analysis of poles will be exactly same as in Theorem 3.1, so won’t repeat it here. The only changes will be in the calculation of the vertical integral of Vk,r (x). One can recall that, while calculating Vk,r (x), the equation (5.14) depends on k and r. Since k is even, the equation (5.15) will remain same. For clarity, we mention it once again. Mainly, the equation (5.15) is     k ksπ πk(1 − s) −1 2 sin sin = (−1) , (5.36) 2 2 whereas the equation (5.16) changes to    πs  r+1 π(1 − s − r) , sin = (−1) 2 sin 2 2 (5.37) as r is odd. In view of (5.36) and (5.37), the vertical integral (5.14) reduces to k+r+1 Z (−1) 2 −1 (2π)k+1−r 1 Γ(1 − s)Γ(1 − k + ks)Γ(s + r)ζ(1 − k + ks) Vk,r (x) = π2 x 2πi (1+λ)    πs   (2π)k+1 −s πks ds. sin ×ζ(s + r) sin 2 2 x (5.38)
Here we remark that the only changes happened in the power of −1 and cos πs got replaced 2

πs πs by sin 2 . At this point, to simplify further, we multiply and divide by cos 2 in (5.38). Then using Euler’s reflection formula and simplifying, we arrive at k+r+1 Z (−1) 2 −1 (2π)k−r 1 Γ(1 − k + ks)Γ(s + r) Vk,r (x) = ζ(1 − k + ks) x 2πi (1+λ) Γ(s)
 −s sin πks (2π)k+1 2
ds. ×ζ(s + r) x cos πs 2 Now we shall make use of (4.8), that is,
sin πks k 2
= (−1) 2 πs cos 2 (k−1) X j=−(k−1) ′′ j i exp  iπjs 2  . This is one of the crucial changes in this proof. Substituting this expression, one can show that 2k+r+1 Z (k−1) Γ(1 − k + ks)Γ(s + r) (−1) 2 −1 (2π)k−r X ′′ j 1 i ζ(1 − k + ks) Vk,r (x) = x 2πi (1+λ) Γ(s) j=−(k−1) ×ζ(s + r) e− iπj 2 (2π)k+1 x !−s ds. ON RAMANUJAN’S FORMULA FOR ζ(1/2) AND ζ(2m + 1) 21 From here onwards, the analysis of simplifying Vk,r (x) is exactly same as in Theorem 3.1, so we avoid reproducing the calculations. If we continue the calculations along the same spirit of Theorem 3.1, then the vertical integral takes the shape of ! k+1−2r 2k+r−1 (k−1) ∞ X ′′ X {} (−1) 2 (2π) 2 k,0 X(j) . Sk,r (n)G0,k ij Vk,r (x) = 2k−1 r, − k1 , · · · , − (k−1) xk 2 k n=1 j=−(k−1) (5.39) Now (5.39) and along with all the residual terms in Theorem 3.1, one can complete the proof of Theorem 3.4.  5.3. A new identity for ζ(−1/2). First, let us state a lemma which will be useful in proving the next corollary. Lemma 5.1. Let b and u be two real numbers. Then " # i exp(beiu ) 1 sin(b sin u) sinh(b cos u) Re . = 2 2 (cos(b sin u) cosh(b cos u) − 1)2 + (sin(b sin u) sinh(b cos u))2 (exp(beiu ) − 1) The proof of this lemma goes along the same vein as the proof of Lemma 4.8, so we left to readers to verify. 4 1 Proof of Corollary 3.5. It is well-known that ζ(−1) = − 12 and ζ(4) = π90 . Letting k = 2 and r = 1, Theorem 3.4 yields that   √ ∞ X 1 π4 π 1 −nx √ + ζ − + D2,1 (n)e = 24 2 x 2 90x2 n=1 " ! !# √ X ∞ {} π {} 2,0 2,0 + S2,1 (n) iG0,2 X(−1) . (5.40) 1 X(1) − iG0,2 2x 1, − 1, − 12 2 n=1 Recall Lemma 4.5, with b1 = 1 and b2 = − 21 , and use [1, p. 444, Equation (10.2.17)] the fact that   √ 1 −2 z √π 1 + √ e √ 2 z K 3 (2 z) = , 1 2 2z 4 to deduce that 2,0 G0,2 !# √ −2√X(j) {} X(j) = πe 1, −1 2 where p 2 X(j) = 2π r 1 1+ p 2 X(j) 2πn − iπj e 4 . x ! , (5.41) 22 ANUSHREE GUPTA AND BIBEKANANDA MAJI From (2.5), we see ∞ X −nx D2,1 (n)e n=1   ∞ X 1 d 1 = . n2 dx 1 − en2 x n=1 (5.42) Simplifying the definition (2.1) of S2,1 (n), one can write X d . S2,1 (n) = m 2 (5.43) n=d m Now plugging (5.41), (5.42), and (5.43) in   ∞ X 1 1 1 d + = 2x 2 n n dx 1 − e 24 n=1 (5.40), we derive that   √ 1 π4 π √ ζ − + 2 x 2 90x2 " #  ∞ ∞ √ π X 1 X 1 −θd m + , 1+ √ Re ide x m=1 m θd m (5.44) d=1 where θ = 2π Let us write q − iπ 4 2π x e . Next aim is to decipher the sum: # " ∞ X √ √ i −θd m −θd m + √ e Re i de θ m d=1 " # √ i eθ m i
. √ = Re √
2 + √ θ m eθ m − 1 eθ m − 1 A(θ) = i eθ eθ √ √ m i √
. and B(θ) = √ θ m eθ m − 1
, m−1 2 √ − iπ 4 , where To extract q real part of A(θ), we need to use Lemma 5.1. We write θ m = be π b = 2π 2πm x . Then, employing Lemma 5.1, with u = − 4 , we get " # − iπ i ebe 4 Re[A(θ)] = Re  2 − iπ ebe 4 − 1     √b √b sin sinh 1 2 2 =−  (5.45)     2     2 . 2 b b cos √2 cosh √2 − 1 + sin √b2 sinh √b2 Again, to find the real part of B(θ), we shall use Lemma 4.8, with u = we have # " iπ ie 4 1 Re[B(θ)] = Re − iπ b ebe 4 − 1     − √b 2 √b √b + cos − e sin 1  2   2   . =− 4 √b cosh √b − cos √b 2 2 2 π 4 and v = 1. Thus, (5.46) ON RAMANUJAN’S FORMULA FOR ζ(1/2) AND ζ(2m + 1) 23 Now in view of (5.45) and (5.46), the equation (5.44) becomes     √ ∞ X 1 1 1 d π4 π 1 √ + = + ζ − n2 dx 1 − en2 x 24 2 x 2 90x2 n=1   "  b  − √b ∞ √ √b sin + cos −e 2 X 1 π 2 2      − 4x m √b cosh √b − cos √b m=1 2 2 2     # 2 sin √b2 sinh √b2 +   2     2   cos √b2 cosh √b2 − 1 + sin √b2 sinh √b2 3 Finally, replace x by α and β = 4πα , then one check that variables, we complete the proof of Corollary (3.5). √b 2 = √ mβ. Plugging all these  Proof of Theorem 3.6. In this theorem, we are concerned with k ≥ 2 even and r = −1, so the integrand function changes to F (s) = Γ(s)ζ(ks)ζ(s + 1)x−s . One can easily show that the only poles of this integrand function are at s = k1 and s = 0. We shall note that s = 0 is a pole of order 2, where as k1 remains a simple pole. Thus, the residue at k1 will be     1 1 1 1 + 1 x− k . R1/k = Γ ζ k k k And the residue at s = 0 will be  d  2 s Γ(s)ζ(ks)ζ(s + 1)x−s . s→0 ds R0 = Res Γ(s)ζ(ks)ζ(s + 1)x−s = lim s=0 From (4.2) and (4.3), we write the Laurent series expansion at s = 0 of each factor of the integrand function:   1 π2 1 2 γ + s + ··· , Γ(s) = − γ + s 2 6 1 ζ(ks) = − + kζ ′ (0)s + · · · , 2 1 ζ(s + 1) = + γ − γ1 s + · · · , s x−s = 1 − log(x)s + · · · . Multiplying these expansions, one can find that the coefficient of s in s2 Γ(s)ζ(ks)ζ(s + 1)x−s is kζ ′ (0) + log(x) 2 . Thus, the residue R0 equals to   x 1 , (5.47) R0 = log 2 (2π)k 24 ANUSHREE GUPTA AND BIBEKANANDA MAJI since ζ ′ (0) = − 12 log(2π). The remaining part of the proof is exactly same as in Theorem 3.4, as we are dealing a particular case of when k is even and r is odd. Hence, considering the above residual terms, Theorem 3.4, with r = −1, yields that       k+3 ∞ X x 1 1 1 (−1)k−1 (2π) 2 1 − k1 −nx +1 x + + Γ ζ Dk,−1 (n)e = log 2k−1 2 (2π)k k k k xk 2 n=1 ! (k−1) ∞ X ′′ X {} k,0 × X(j) . (5.48) Sk,−1 (n)G0,k ij −1, − k1 , · · · , − (k−1) k n=1 j=−(k−1) Here we invoke Lemma 4.6 with b = −1 to simplify that {} k,0 G0,k −1, − k1 , · · · , − (k−1) k ! X(j)   1 exp −kX(j) k k−1 2 (2π) √ k = X(j) . (5.49) Substituting (5.49) in (5.48), the right side infinite sum reduces to   1 (k−1) ∞ k exp −kX(j) k−1 k+1 X X ′′ j (−1) (2π) Sk,−1 (n) i k xk X(j) n=1 j=−(k−1) (k−1) k−1 = (−1) ′′ X j i e iπj 2 (k−1) = (−1) ′′ X iπj e (k−1) X j=−(k−1) ′′ ∞ X ∞ X 1 m=1 d=1 j=−(k−1) = (−1)k n n=1 j=−(k−1) k−1 ∞ X Sk,−1 (n) eiπj d  − iπj 2k exp −e  − iπj 2k exp −e ∞ X (2π) 1+ k1 (2π) 1+ k1 d n1  k x m1  k x    m  1  iπj 1 k . log 1 − exp −e− 2k (2π)1+ k x m=1 (5.50) In the penultimate step, we have used the definition of Sk,−1 (n), and in the final step we P ym used the identity − log(1 − y) = ∞ m=1 m when |y| < 1. Finally, in view (5.48) and (5.50), we complete the proof of Theorem 3.6.  Proof of Theorem 3.7. Note that here we are dealing with k ≥ 1 and r 6= −1 are both as an odd integer. Let us recall that the integrand function is F (s) = Γ(s)ζ(ks)ζ(s − r)x−s . We shall divide the analysis of poles in two cases. Case I: First, we consider r as an odd positive integer. We know that Γ(s) has simple poles at 0, −1, −2, −3, · · · . The poles at even negative integers are getting cancelled by the trivial zeros of ζ(ks), whereas the poles at odd negative integers will be cancelled by ζ(s − r). Therefore, in this case, the only poles of the integrand function are at s = 0, k1 , and 1 + r. These are all simple poles and their corresponding residues have been already calculated in Theorem 3.1, so we won’t repeat it here. ON RAMANUJAN’S FORMULA FOR ζ(1/2) AND ζ(2m + 1) 25 Case II: We consider r 6= −1 as an odd negative integer. One can easily check that 0 and are simple poles in this case too. As r ≤ −3 is odd, so 1 + r is an even negative integer. Thus, 1 + r is a pole of order 2 of the factor Γ(s)ζ(s − r). And we also note that 1 + r is a trivial zero of ζ(ks). Therefore, 1 + r remains as a simple pole of the integrand function. 1 k An important point is to note that, the poles of Γ(s) at s = −1, −3, · · · , r will not get neutralized by ζ(s−r), but the poles at the negative odd integers beyond r, say r−2, r−4, · · · will be neutralized by ζ(s − r) since they are trivial zeros of ζ(s − r). Therefore, in this case, we have to take into account the residual terms coming from the contribution of the poles at s = −1, −3, · · · , r. To calculate the residual term corresponding to s = 1 + r, we have to use the Laurent series expansion of Γ(s) i.e., Lemma 4.1, and the Laurent series expansion (4.2) of ζ(s) with s replace by s − r. Thus, we will have R1+r = lim (s − (1 + r))Γ(s)ζ(ks)ζ(s − r)x−s s→1+r = lim (s − (1 + r))Γ(s) s→1+r = ζ(ks) (s − (1 + r))ζ(s − r)x−s (s − (1 + r)) (−1)1+r kζ ′ (k(1 + r))x−(1+r) . (−(1 + r))! (5.51) Let R be the sum of the residual terms corresponding to the poles at s = −1, −3, · · · , r. Then, we have − 1+r 2 R= X i=0 − 1+r 2 = X i=0 (−1) = 2 Res s=−(2i+1) Γ(s)ζ(ks)ζ(s − r)x−s −1 ζ(−k(2i + 1))ζ(−2i − 1 − r)x2i+1 (2i + 1)! 1+r 2 − 1+r 2 X i=0 (−1)i+1 Bk(2i+1)+1 B−2i−1−r (2π)−r  x 2i+1 . (2i + 1)!(k(2i + 1) + 1) (−(2i + 1 + r))! 2π (5.52) The remaining part of the proof is in the same direction of the proof of Theorem 3.1, so we briefly mention all the steps. Mainly, we will concentrate on the calculation of the vertical integral Vk,r (x). From (5.14), we have Vk,r (x) = (2π)k+1−r 1 π2 x 2πi Z (1+λ) × sin  Γ(1 − s)Γ(1 − k + ks)Γ(s + r)ζ(1 − k + ks)ζ(s + r) πk(1 − s) 2  sin  π(1 − s − r) 2  (2π)k+1 x −s ds. 26 ANUSHREE GUPTA AND BIBEKANANDA MAJI Since k and r both odd, so we must use     k−1 πk(1 − s) kπs 2 sin = (−1) , cos 2 2    πs  r+1 π(1 − s − r) . = (−1) 2 sin sin 2 2 Use these two trigonometric identities to see k+r Z (−1) 2 (2π)k+1−r 1 Vk,r (x) = Γ(1 − s)Γ(1 − k + ks)Γ(s + r)ζ(1 − k + ks)ζ(s + r) π2x 2πi (1+λ)    πs   (2π)k+1 −s kπs ds. sin × cos 2 2 x
Here we multiply and divide by cos πs and then use Euler’s reflection formula to deduce 2 that k+r Z Γ(1 − k + ks)Γ(s + r) (−1) 2 (2π)k−r 1 ζ(1 − k + ks)ζ(s + r) Vk,r (x) = x 2πi (1+λ) Γ(s)
 −s cos kπs (2π)k+1 2
× ds. x cos πs 2 Next, we shall use (4.9), that is,
cos kπs k−1 2
= (−1) 2 πs cos 2 (k−1) X j=−(k−1) ′′  iπjs i exp − 2 j  . This is one of the important changes in this proof. Plugging this expression in the above integral, we obtain 2k+r−1 Z (k−1) (−1) 2 (2π)k−r X ′′ j 1 Γ(1 − k + ks)Γ(s + r) Vk,r (x) = i ζ(1 − k + ks) x 2πi (1+λ) Γ(s) j=−(k−1) ×ζ(s + r) e iπj 2 (2π)k+1 x !−s ds. Henceforth the simplification of the vertical integral Vk,r (x) is exactly same as in Theorem 3.1. If we continue the calculations in the same spirit, then the final expression of Vk,r (x) will be ! k+1−2r 2k+r−1 (k−1) ∞ X ′′ X {} (−1) 2 (2π) 2 k,0 j Sk,r (n) G0,k i X(−j) . Vk,r (x) = 2k−1 r, − k1 , · · · , − (k−1) xk 2 k n=1 j=−(k−1) (5.53) An important point is to note that the argument of the Meijer G-function is X(−j). Finally, collecting all the residual terms (5.51), (5.52), and together with (5.53), one can complete the proof of (3.5).  ON RAMANUJAN’S FORMULA FOR ζ(1/2) AND ζ(2m + 1) 27 Proof of Corollary 3.7. Setting r = −(2m + 1), with m ≥ 1, in Theorem 3.7, and using the identity ζ ′ (−2km) = (−1)km (2km)!ζ(2km + 1) , 2 (2π)2km we complete the proof.  5.4. Derivation of Ramanujan’s formula for odd zeta values. Proof of Corollary 3.9. Plugging k = 1 in Corollary 3.8, we deduce that ∞ X n=1  x 2m 1 ζ (2m + 2) (−1)m D1,−(2m+1) (n)e−nx = − ζ(2m + 1) + + ζ(2m + 1) 2 x 2 2π m (−1)m (2π)2m+1 X (−1)i+1 B2i+2 B2m−2i  x 2i+1 + 2 (2i + 2)!(2m − 2i)! 2π + i=0 ∞ m 2m+2 X (−1) (2π) x 1,0 S1,−(2m+1) (n)G0,1 n=1 ! (2π)2 n {} . x −(2m + 1) (5.54) From (2.2), we know D1,−(2m+1) (n) = σ−(2m+1) (n), and S1,−(2m+1) (n) = σ2m+1 (n). (5.55) The equation (4.7) yields that 1,0 G0,1 (2π)2 n {} x −(2m + 1) ! =  x 2m+1 4π2 n e− x . 4π 2 n (5.56) Now substituting (5.55) and (5.56) in (5.54), one can derive that ∞ X  x 2m 1 ζ (2m + 2) (−1)m σ−(2m+1) (n)e−nx + ζ(2m + 1) = + ζ(2m + 1) 2 x 2 2π n=1 + m (−1)m (2π)2m+1 X (−1)i+1 B2i+2 B2m−2i  x 2i+1 2 (2i + 2)!(2m − 2i)! 2π i=0 ∞  x 2m X σ2m+1 (n) − 4π2 n m + (−1) e x . 2π n2m+1 n=1 Finally, invoking Euler’s formula (1.1) for ζ(2m + 2) and using the fact that σ−(2m+1) (n) and upon simplification, one can derive (3.6). σ2m+1 (n) n2m+1 =  28 ANUSHREE GUPTA AND BIBEKANANDA MAJI Proof of Corollary 3.10. Letting k = 1 and r = 2m + 1 with m ≥ 1 in Theorem 3.7, and 2m+2 with the fact that ζ(−(2m + 2)) = 0 and ζ(−(2m + 1)) = − B2m+2 , we find that ∞ X D1,2m+1 (n)e−nx = n=1 ζ(2m + 2) B2m+2 + (2m + 1)! 4m + 4 x2m+2 ∞  x −(2m+1) 4π2 n X 1 S (n) e− x 1,2m+1 x(2π)2m n=1 4π 2 n  2m+2 ∞ X 2π B2m+2 B2m+2 + (−1)m σ2m+1 (n)e−nx = ⇒ 4m + 4 x 4m +4 n=1  2m+2 X ∞ 4π 2 n m+1 2π + (−1) (5.57) σ2m+1 (n)e− x . x n=1 + (−1)m+1 Here we have used (2.2) to write D1,2m+1 (n) and S1,2m+1 (n)n2m+1 in terms of σ2m+1 (n). Finally, to obtain (3.7), replace x = 2α and αβ = π 2 in (5.57).  Substituting k = r = 1 in Theorem 3.7 and setting x = 2α and αβ = π 2 , one can obtain the following identity of Schlömilch [29], and rediscovered by Ramanujan [26, Ch. 14, Sec. 8, Cor. (i)], [27, p. 318]. Corollary 5.2. For α, β > 0 with αβ = π 2 , ∞ X ∞ X n n α+β 1 α +β = − 2nα − 1 2nβ − 1 e e 24 4α n=1 n=1 In particular, if we consider α = β = π, then ∞ X n=1 n 1 1 = − . −1 24 8π e2nπ Proof of Theorem 3.11. The proof of this theorem travels in the same direction as of Theorem 3.7 since we are dealing k odd and r = −1. In this case, the integrand function is F (s) = Γ(s)ζ(ks)ζ(s + 1)x−s . One can easily check that the integrand function has poles are at s = 0, k1 and −1. At s = 0, we have a pole order 2, but the other two points are simple pole. In view of (5.47), we know   x 1 R0 = log . 2 (2π)k One can find that R−1 = ζ(−k)x . 2 ON RAMANUJAN’S FORMULA FOR ζ(1/2) AND ζ(2m + 1) 29 Now along with the residue at k1 , Theorem 3.7, with r = −1, leads to       k+3 ∞ X x 1 ζ(−k) (2π) 2 1 1 1 − k1 −nx +1 x + + Dk,−1 (n)e = log + Γ ζ 2k−1 2 (2π)k k k k 2 xk 2 n=1 ! (k−1) ∞ X ′′ X {} k,0 × Sk,−1 (n)G0,k ij (5.58) (k−1) X(−j) . 1 −1, − , · · · , − k k n=1 j=−(k−1) Calculations of the above right hand side sum goes along the same line of (5.48). For completeness, we mention it briefly. First, employ Lemma 4.6 with b = −1 to see   1 ! k−1 k exp −kX(−j) {} (2π) 2 k,0 √ G0,k . (k−1) X(−j) = 1 X(−j) −1, − k , · · · , − k k Substituting it in (5.58), together with definition (3.1) of X(j), the right side infinite sum takes the shape of  (k−1) ∞ n1  X X ′′ iπj 1 Sk,−1 (n) k 1+ j − iπj exp −e 2k (2π) k i e 2 n x n=1 j=−(k−1) (k−1) ′′ X = j=−(k−1) ∞ X ∞ X 1 m=1 d=1 (k−1) =− X j=−(k−1) ′′ ∞ X m=1 d  exp −e  iπj 2k  1+ k1 (2π) log 1 − exp −e iπj 2k d m1  k x 1+ k1 (2π)  m  1  k x . Now substituting (5.59) in (5.58), one can finish the proof of (3.8). (5.59)  Proof of Corollary 3.12. Plugging k = 1 in (3.8), we arrive at    ∞ ∞  x  π2 X X 4π 2 m x 1 −nx + log 1 − exp − − − D1,−1 (n)e = log 2 2π 6x 24 x m=1 n=1 ⇒ ∞ X X n=1 d|n ∞ X 1 −nx e d ∞ X ∞  x  π2 X 1 x 1 − 4π2 md x + = log − + e 2 2π 6x 24 d m=1 d=1 ∞  x  π2 X 1 1 1 x  4π2 n . ⇒ + = log − + nx n(e − 1) 2 2π 6x 24 x n e − 1 n=1 n=1 Finally, replacing x = 2α and β = π 2 /α and simplifying we obtain (3.9).  6. Concluding Remarks On page 332, Ramanujan recorded an intriguing formula for ζ(1/2) and also mentioned the following interesting infinite series X nr , N x) − 1 exp(n n=1 30 ANUSHREE GUPTA AND BIBEKANANDA MAJI where N ∈ N, r ∈ Z and N − r is any even integer. This infinite series can be written as Z c+i∞ 1 Γ(s)ζ(s)ζ(N s − r)x−s ds, 2πi c−i∞
for c > max 1, 1+r N . Recently, Dixit and the second author obtained a beautiful generalization of Ramanujan’s formula for odd zeta values while studying the above integral. In the current paper, we have studied a variant of the above integral, Z c+i∞ ∞ X 1 −nx Γ(s)ζ(ks)ζ(s − r)x−s ds, Dk,r (n)e = 2πi c−i∞ n=1 where for k ∈ N, r ∈ Z, and c > max (1/k, 1 + r). The analysis of this integral can be divided into the following four cases. Case 1: When k and r are both are even integers, Theorem 3.1 gives us a transformation formula, which allows us to derive Ramanujan’s formula for ζ(1/2) and Wigert’s formula for ζ(1/k). Case 2: When k is even and r is odd, i.e., k − r is an odd integer. In this case, Theorem 3.4 allows us to obtain a new identity for ζ(−1/2) analogous to Ramanujan’s formula for ζ(1/2). Case 3: When k and r are both odd, i.e., k − r is an even integer. This is the most interesting case. Theorem 3.7 allows us to derive Ramanujan’s formula for odd zeta values. Corollary 3.8 can be conceived as a one-variable generalization of Ramanujan’s formula for odd zeta values, which connects two odd zeta values ζ(2m + 1), and ζ(2km + 1) and a zeta
value at a rational argument ζ k1 + 2m + 1 . Case 4: When k is odd and r is even, i.e., k − r is an odd integer. In this case, the integrand function will have infinitely many poles, namely, at all odd negative integers. In the current paper, we have not discussed this case. It would be interesting to study this case too. Ramanujan’s formula (1.2) for ζ(2m + 1) has been generalized by many mathematicians in various directions. Ramanujan himself gave a huge generalization [10, Entry 20, p. 430]. Berndt [7] derived Euler’s formula for ζ(2m) as well as Ramanujan’s formula for ζ(2m + 1) from a single transformation formula for a generalized Eisenstein series. The Dirichlet Lfunction L(s, χ) and the Hurwitz zeta function ζ(s, a) are natural generalizations of the Riemann zeta function. A character analogue of Ramanujan’s formula was obtained by Merrill [22] and more generally, for any periodic function was obtained by Bradley [14]. Dixit, Kumar, Gupta and the second author [17] found a generalization for Hurwitz zeta function, which allowed them to connect many odd zeta values from a single formula. Dixit and Gupta [16] established an interesting analogue of Ramanujan’s formula which connects square of odd zeta values. Very recently, Banerjee and Kumar [6] found an analogue of Ramanujan’s formula (1.2) over an imaginary quadratic fields. For more references on the generalizations of Ramanujan’s formula, readers are encouraged to see [13]. ON RAMANUJAN’S FORMULA FOR ζ(1/2) AND ζ(2m + 1) 31 Acknowledgements. The authors want to thank IIT Indore for the conductive research environment. The second author is grateful to SERB for the Start-Up Research Grant SRG/2020/000144. References [1] M. Abramowitz and I. Stegun, Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. Number 55 in National Bureau of Standards Applied Mathematics Series. 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Titchmarsh, The Theory of the Riemann Zeta-function, Clarendon Press, Oxford, 1986. [31] S. Wigert, Sur une extension de la série de Lambert, Arkiv Mat. Astron. Fys. 19 (1925), 13 pp. [32] W. W. Zudilin, One of the numbers ζ(5), ζ(7), ζ(9) and ζ(11) is irrational (Russian), Uspekhi Mat. Nauk 56 No. 4 (2001), 149–150; translation in Russian Math. Surveys 56 No. 4 (2001), 774–776. Anushree Gupta, Department of Mathematics, Indian Institute of Technology Indore, Simrol, Indore, Madhya Pradesh 453552, India. Email address: [email protected], [email protected] Bibekananda Maji, Department of Mathematics, Indian Institute of Technology Indore, Simrol, Indore, Madhya Pradesh 453552, India. Email address: [email protected], [email protected]