IOSR Journal of Mathematics (IOSR-JM)
e-ISSN: 2278-5728,p-ISSN: 2319-765X, Volume 8, Issue 3 (Sep. - Oct. 2013), PP 33-35
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A Redefined Riemann Zeta Function Represented via Functional
Equations Using the Osborne’s Rule
Adeniji A.A. Ekakitie E.A Mohammed M.A Amusa I.S
Department of Mathematics, University of Lagos, Akoka, Nigeria.
Department of Mathematics Yaba College of Technology, Lagos, Nigeria.
Department of Mathematics Yaba College of Technology, Lagos, Nigeria.
Department of Mathematics, University of Lagos, Akoka, Nigeria.
Abstract: Intrigues most researchers about the Riemann zeta hypothesis is the ability to employ cum different
approaches with instinctive mindset to obtain some very interesting results. Motivated by their style of
reasoning, the result obtained in this work of redefining or re-representation of Riemann zeta function in
different forms by employing different techniques on two functional equations made the results better, simpler
and concise new representations of Riemann zeta function.
Keywords: Analytic Continuation, Osborne’s rule, Riemann Zeta Function
𝐓𝐇𝐄 𝐎𝐒𝐁𝐎𝐑𝐍𝐄′ 𝐒 𝐑𝐔𝐋𝐄 𝐀𝐏𝐏𝐑𝐎𝐀𝐂𝐇
Osborne′s rule gave a more comprehensive method for hyperbolic functions interms of
the Euler method.
𝑒 𝑥 +𝑒 −𝑥
𝑟𝑒𝑐𝑎𝑙𝑙 𝑓𝑟𝑜𝑚 𝑐𝑜𝑠𝑕𝑥 = 2
and the fact that 𝑒 𝜃𝑖 = 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
𝐼𝑚𝑝𝑙𝑦𝑖𝑛𝑔 𝑡𝑕𝑎𝑡 ℜ 𝑒 𝑥 = 𝑐𝑜𝑠𝑕𝑥
(1.0)
𝑠
𝑠
ℜ 𝑒 2𝑙𝑛𝑥 = cosh⁡
( 𝑙𝑛𝑥)
2
From the Riemann zeta function
𝜁 𝑠 = 𝑠 𝑠−1
2
4
∞
1
𝑠
𝑠
−
𝜋 2Γ
2
(1.1)
5
∞
−𝑛𝑛𝜋𝑥
𝑛=1 𝑒
1
3
𝑠
𝑛𝑛 2 𝜋 2 𝑥 4 − 2 𝑛𝑛𝜋𝑥 4 cosh
2
𝑙𝑛𝑥 𝑑𝑥
(1.2)
Substituting 1.0 , into equation 1.1 , then the equation 1.2 becomes
𝜁 𝑠 = 𝑠 𝑠−1
2
4
∞
1
𝑠
𝑠
𝜋 2Γ
2
−
5
1
3
𝑠
𝑛𝑛 2 𝜋 2 𝑥 4 − 2 𝑛𝑛𝜋𝑥 4 𝑒 2𝑙𝑛𝑥 𝑑𝑥
∞
−𝑛𝑛𝜋𝑥
𝑛=1 𝑒
(1.3)
considering the right arm of the r. h. s side of the equation(1.3)
∞
1
=
5
∞
1
5
4
3
𝑛𝑛 2 𝜋 2 𝑥 − 2 𝑛𝑛𝜋𝑥
∞
𝑛=1
1
3
𝑠
𝑛𝑛 2 𝜋 2 𝑥 4 − 2 𝑛𝑛𝜋𝑥 4 𝑒 2𝑙𝑛𝑥 𝑑𝑥
∞
−𝑛𝑛𝜋𝑥
𝑛=1 𝑒
5
4
∞
1
4
𝑒
𝑠
−𝑛𝑛𝜋𝑥 + 𝑙𝑛𝑥
2
𝑠
−𝑛𝑛𝜋𝑥 + 𝑙𝑛𝑥
2
(1.4)
𝑑𝑥
(1.5)
1
4
∞3
𝑠
−𝑛𝑛𝜋𝑥 + 𝑙𝑛𝑥
2
= ∞𝑛=1 1 ( 𝑛𝑛 2 𝜋 2 𝑥 𝑒
𝑑𝑥 − 1 2 𝑛𝑛𝜋𝑥 𝑒
considering the left hand side of the equation 1.6 i. e
∞
𝑛=1
𝑑𝑥
(1.6)
∞ 5 −𝑛𝑛𝜋𝑥 +𝑠 𝑙𝑛𝑥
2
𝑥4 𝑒
𝑑𝑥.
1
5
𝑠
−𝑛𝑛𝜋𝑥 + 𝑙𝑛𝑥
𝑛𝑛 2 𝜋 2
𝐿𝑒𝑡 I =
∞
𝑥4𝑒
1
14
𝑑𝑥 and introducing the integration by part:
2
9
4
4
(1.7)
𝑠
4
𝑡𝑕𝑒𝑛 , 𝐼 = (18+4𝑠 ) 9 𝑥 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 − 9 (𝑛𝑛)3 𝜋 3
From equation (1.8)
∞
𝑆𝑢𝑝𝑝𝑜𝑠𝑒 𝑀1 =
4
1
4
4
9
(1.8)
9
𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +𝑙𝑛𝑥 𝑑𝑥, 𝑏𝑦 𝑓𝑢𝑟𝑡𝑕𝑒𝑟𝑖𝑛𝑔 𝑡𝑕𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛 𝑏𝑦 𝑝𝑎𝑟𝑡 𝑓𝑟𝑜𝑚
13
𝑠
4
𝑀1 = (𝑛𝑛)3 𝜋 3
𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 + 𝑛𝑛𝜋
9
13
13
𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 2.0 𝑖𝑛𝑡𝑜 1.8 , 𝑤𝑒 𝑜𝑏𝑡𝑎𝑖𝑛:
18
∞ 9 −𝑛𝑛𝜋𝑥 +𝑠 𝑙𝑛𝑥
2
𝑥4𝑒
𝑑𝑥
1
𝑠
4
∞ 13 −𝑛𝑛𝜋𝑥 + 𝑠 𝑙𝑛𝑥
2
𝑥4𝑒
𝑑𝑥
1
4𝑠
4
13
−
4𝑠
26
𝑠
(1.9),
𝑀1
(2.0)
4
∞
13
𝑠
𝐼 = (18+4𝑠 )[9 𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 − 9 𝑛𝑛 3 𝜋 3 𝑀 + 26 𝑀 = 13 𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 + 13 𝑛𝑛𝜋 1 𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 𝑑𝑥
(2.1)
∞ 13
𝑠
𝑠
26
4 13
4
𝐹𝑟𝑜𝑚 2.0 , 𝑤𝑒 𝑕𝑎𝑣𝑒 𝑀1 =
[ 𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 + 𝑛𝑛𝜋
𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 𝑑𝑥 ,
26 + 4𝑠 13
13
1
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33 | Page
A Redefined Riemann Zeta Function Represented Via Functional Equations Using The Osborne’s
𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 2.0 𝑖𝑛𝑡𝑜 1.8
𝐼=
18(𝑛𝑛 )2 𝜋 2
4
18+4𝑠
9
18(𝑛𝑛 )2 𝜋 2
9
9
4
4
𝑠
𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 −
𝑠
2
4(𝑛𝑛 )2 𝜋 2
18
26
9
18+4𝑠
26+4𝑠
468 𝑛𝑛 2 𝜋 2
−𝑛𝑛𝜋𝑥 + 𝑙𝑛𝑥
13
4
4
13
4
𝑠
[ 𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 +
13
𝑠
2
−𝑛𝑛𝜋𝑥 + 𝑙𝑛𝑥
= 18+4𝑠 9 𝑥 𝑒
− 1053 +396𝑠+36𝑠 2 (13 𝑥 𝑒
(2.2)
𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑖𝑛𝑔 𝑡𝑕𝑒 𝑟𝑖𝑔𝑕𝑡 𝑕𝑎𝑛𝑑 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑡𝑕𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1.6 𝑖. 𝑒
4
13
𝑛𝑛𝜋
144
∞ 13 −𝑛𝑛𝜋𝑥 +𝑠 𝑙𝑛𝑥
2
𝑥4𝑒
𝑑𝑥
1
(2.1)
4
) − 1053 +234 𝑠 (𝑛𝑛) 𝜋
4
∞ 13 −𝑛𝑛𝜋𝑥 +𝑠 𝑙𝑛𝑥
2
𝑥4𝑒
𝑑𝑥
1
1
𝑠
𝑠
∞3
∞ 1
3
−𝑛𝑛𝜋𝑥 + 𝑙𝑛𝑥
∞
2
𝑑𝑥 = 𝑛𝑛𝜋 1 𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 𝑑𝑥
𝑛=1 1 2 𝑛𝑛𝜋𝑥 4 𝑒
2
𝑠
∞ 1
𝑆𝑢𝑝𝑝𝑜𝑠𝑒 𝑀2 = 1 𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 𝑑𝑥
=
𝑎𝑝𝑝𝑙𝑦𝑖𝑛𝑔 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛 𝑏𝑦 𝑝𝑎𝑟𝑡 𝑜𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (2.4)
5
5
4
𝑠
9
16𝑛𝑛𝜋
𝑠
(2.3)
(2.4)
∞ 9 −𝑛𝑛𝜋𝑥 +𝑠 𝑙𝑛𝑥
2
𝑥4 𝑒
𝑑𝑥
1
16
[5 𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 + 45 𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 + 45 (𝑛𝑛)2 𝜋 2
16𝑛𝑛𝜋𝑠901∞𝑥54𝑒−𝑛𝑛𝜋𝑥+𝑠2𝑙𝑛𝑥𝑑𝑥]
𝑀2 =
5+2𝑠
(2.5)
𝑟𝑒𝑐𝑎𝑙𝑙 𝑀2 =
∞
1
10
1
2
3𝑛𝑛𝜋
𝑠
𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 𝑑𝑥 𝑡𝑕 𝑒𝑛 𝑓𝑟𝑜𝑚 (1.6) 𝑀2 =
𝑀2 = 𝑛𝑛𝜋 (15+6𝑠)
4
5
5
4
𝑥 𝑒
𝑠
2
−𝑛𝑛𝜋𝑥 + 𝑙𝑛𝑥
+
16𝑛𝑛𝜋
16𝑛𝑛𝜋𝑠901∞𝑥54𝑒−𝑛𝑛𝜋𝑥+𝑠2𝑙𝑛𝑥𝑑𝑥,
45
9
4
𝑥 𝑒
𝑠
2
−𝑛𝑛𝜋𝑥 + 𝑙𝑛𝑥
∞
1
𝑠
𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 𝑑𝑥, 𝑤𝑒 𝑜𝑏𝑡𝑎𝑖𝑛
1
∞ 9 −𝑛𝑛𝜋𝑥 +𝑠 𝑙𝑛𝑥
2
𝑥4𝑒
𝑑𝑥
1
16
+ 45 (𝑛𝑛)2 𝜋 2
(2.6)
𝑓𝑟𝑜𝑚 𝑡𝑕𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1.6 , 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝐼 𝑎𝑛𝑑 𝑀2 , 𝑡𝑜 𝑜𝑏𝑡𝑎𝑖𝑛
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1.6 , 𝑏𝑒𝑐𝑜𝑚𝑒𝑠
=
18(𝑛𝑛 )2 𝜋 2
4
18+4𝑠
9
9
𝑠
468 𝑛𝑛 2 𝜋 2
13
4
𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 − 1053 +396𝑠+36𝑠 2
13
𝑠
−
−
∞ 13 −𝑛𝑛𝜋𝑥 +𝑠 𝑙𝑛𝑥
2
𝑥4𝑒
𝑑𝑥
1
144
𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 − 1053 +234 𝑠 𝑛𝑛 4 𝜋 4
−
10𝑛𝑛𝜋15+6𝑠45𝑥54𝑒−𝑛𝑛𝜋𝑥+𝑠2𝑙𝑛𝑥+16𝑛𝑛𝜋45𝑥94𝑒−𝑛𝑛𝜋𝑥+𝑠2𝑙𝑛𝑥+1645𝑛𝑛2𝜋21∞𝑥94𝑒−𝑛𝑛𝜋𝑥+𝑠2𝑙𝑛𝑥𝑑𝑥
−16𝑛𝑛𝜋𝑠901∞𝑥54𝑒−𝑛𝑛𝜋𝑥+𝑠2𝑙𝑛𝑥𝑑𝑥
(2.7)
∞
𝑟𝑒𝑐𝑎𝑙𝑙 𝑀2 =
=
18 𝑛𝑛 2 𝜋 2 4
18+4𝑠
9
9
1
∞
𝑠
𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 𝑑𝑥 , 𝐼 =
1
1
𝑠
5
𝑠
𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 𝑑𝑥 , 𝑀1 =
468 𝑛𝑛 2 𝜋 2
𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 −
1053 +396𝑠+36𝑠 2
−
13
𝑠
4
−𝑛𝑛𝜋𝑥 + 𝑙𝑛𝑥
2
𝑥4𝑒
13
144
∞
𝑠
1
𝑛𝑛 4 𝜋 4
1053 +234 𝑠
9
𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 𝑑𝑥
∞ 13 −𝑛𝑛𝜋𝑥 +𝑠 𝑙𝑛𝑥
2
𝑥4𝑒
𝑑𝑥
1
−
40𝑥54𝑒−𝑛𝑛𝜋𝑥+𝑠2𝑙𝑛𝑥𝑛𝑛𝜋75+30𝑠+160𝑛𝑛𝜋𝑥94𝑒−𝑛𝑛𝜋𝑥+𝑠2𝑙𝑛𝑥𝑛𝑛𝜋675+3270𝑠−16𝑛𝑛2𝜋2𝑛𝑛𝜋675+270𝑠
𝑀1+16𝑛𝑛𝜋𝑠𝑛𝑛𝜋1375+540𝑠𝐼
(2.8)
𝑐𝑜𝑙𝑙𝑒𝑐𝑡𝑖𝑛𝑔 𝑙𝑖𝑘𝑒 𝑡𝑒𝑟𝑚𝑠, 𝑤𝑒 𝑕𝑎𝑣𝑒:
=
18(𝑛𝑛 )2 𝜋 2 4
18+4𝑠
144
1053 +234𝑠
𝑠
468 𝑛𝑛
2𝜋 2
4
13
𝑠
𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 −
1053 +396𝑠+36𝑠 2 13
13
𝑠
16 𝑛𝑛 2 𝜋 2
4 4 ∞ 4 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥
𝜋 1 𝑥 𝑒
𝑑𝑥 −
𝑛𝑛𝜋 675 +270𝑠
9
𝑛𝑛
9
𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 −
𝑀1 +
5
40𝑥 4 𝑒 −𝑛𝑛𝜋𝑥
𝑠
+ 𝑙𝑛𝑥
2
𝑛𝑛𝜋 75+30𝑠
16𝑛𝑛𝜋𝑠
𝑛𝑛𝜋 1375 +540𝑠
18(𝑛𝑛 )2 𝜋 2
18+4𝑠
4
9
𝑠
468 𝑛𝑛 2 𝜋 2
13
4
160 𝑛𝑛𝜋 𝑥 4 𝑒 −𝑛𝑛𝜋𝑥
𝑠
+ 𝑙𝑛𝑥
2
𝑛𝑛𝜋 675 +3270𝑠
−
𝐼
(2.9)
𝑟𝑒𝑐𝑎𝑙𝑙 𝑓𝑟𝑜𝑚 1.6 , 𝑏𝑦 𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑦𝑖𝑛𝑔 𝑤𝑒 𝑜𝑏𝑡𝑎𝑖𝑛
=
∞
𝑛=1
9
+
5
𝑠
𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 − 1053 +396𝑠+36𝑠 2 (13 𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 ) −
9
40𝑥 4 𝑒
𝑠
−𝑛𝑛𝜋𝑥 + 𝑙𝑛𝑥
2
𝑛𝑛𝜋 (75+30𝑠)
+
160𝑛𝑛𝜋𝑥94𝑒−𝑛𝑛𝜋𝑥+𝑠2𝑙𝑛𝑥𝑛𝑛𝜋(675+3270𝑠)−1441053+234𝑠(𝑛𝑛)4𝜋41∞𝑥134𝑒−𝑛𝑛𝜋𝑥+𝑠2𝑙𝑛𝑥𝑑𝑥−16(𝑛𝑛
)2𝜋2𝑛𝑛𝜋(675+270𝑠)𝑀1+16𝑛𝑛𝜋𝑠𝑛𝑛𝜋(1375+540𝑠)𝐼
(3.0)
=
∞
−𝑛𝑛𝜋𝑥
𝑛=1 𝑒
18(𝑛𝑛 )2 𝜋 2 4
18+4𝑠
9
9
𝑠
𝑥 4 𝑒 2𝑙𝑛𝑥 −
468 𝑛𝑛
2𝜋 2
4
13
𝑠
( 𝑥 4 𝑒 2𝑙𝑛𝑥 ) −
1053 +396𝑠+36𝑠 2 13
5 𝑠
40𝑥 4 𝑒 2𝑙𝑛𝑥
𝑛𝑛𝜋 (75+30𝑠)
9 𝑠
+
160 𝑛𝑛𝜋 𝑥 4 𝑒 2𝑙𝑛𝑥
𝑛𝑛𝜋 (675 +3270 𝑠)
−
1441053+234𝑠(𝑛𝑛)4𝜋41∞𝑥134𝑒−𝑛𝑛𝜋𝑥+𝑠2𝑙𝑛𝑥𝑑𝑥−16𝑛𝑛2𝜋2𝑛𝑛𝜋(675+270𝑠)𝑀1+16𝑛𝑛𝜋𝑠𝑛𝑛𝜋(1375+540𝑠)𝐼
(3.1)
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A Redefined Riemann Zeta Function Represented Via Functional Equations Using The Osborne’s
∞
𝑠
𝑠
𝑒 −𝑛𝑛𝜋𝑥 = 𝜓 𝑥 𝑎𝑛𝑑 𝑒 𝑙𝑛𝑥 2 = 𝑥 2
𝐿𝑒𝑡
𝑛 =1
Then equations 3.1 becomes
= 𝜓(𝑥)𝑥
𝑠
2
18(𝑛𝑛 )2 𝜋 2 4
18+4𝑠
144
(𝑛𝑛)4 𝜋 4
1053 +234𝑠
9
9
4
468 𝑛𝑛 2 𝜋 2
4
5
13
4
2
4
𝑠
𝑠
−
𝜋 2Γ
2
∞ 13 −𝑛𝑛𝜋𝑥 +𝑠 𝑙𝑛𝑥
2
𝑥4𝑒
𝑑𝑥
1
𝑠
𝜓(𝑥)𝑥 2
160𝑛𝑛𝜋 𝑥 4
𝑥 − 1053 +396𝑠+36𝑠 2 (13 𝑥 ) − 𝑛𝑛𝜋 (75+30𝑠) + 𝑛𝑛𝜋 (675+3270 𝑠) −
16(𝑛𝑛 )2 𝜋 2
16𝑛𝑛𝜋𝑠
− 𝑛𝑛𝜋 (675+270𝑠) 𝑀1 + 𝑛𝑛𝜋 (1375 +540𝑠) 𝐼
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 3.0 𝑖𝑛𝑡𝑜 1.2
𝜁 𝑠 = 𝑠 𝑠−1
9
40𝑥 4
18(𝑛𝑛 )2 𝜋 2 4
18+4𝑠
9
9
𝑥4 −
468 𝑛𝑛 2 𝜋 2
1053 +396𝑠+36𝑠
4
13
( 𝑥4 )−
2
13
5
40𝑥 4
𝑛𝑛𝜋 (75+30𝑠)
(3.2)
9
+
160 𝑛𝑛𝜋 𝑥 4
𝑛𝑛𝜋 (675 +3270 𝑠)
−
1441053+234𝑠(𝑛𝑛)4𝜋41∞ 𝑥134𝑒−𝑛𝑛𝜋𝑥+𝑠2𝑙𝑛𝑥𝑑𝑥−16(𝑛𝑛)2𝜋2𝑛𝑛𝜋(675+270𝑠)𝑀1+16𝑛𝑛𝜋𝑠𝑛𝑛𝜋(1375+540𝑠)
𝐼 .
(3.3)
from the above represenation the ζ s is valid for 𝑠 > 1
Conclusion
This representation really shows how the complex part of the equation could be eliminated by
introducing a hyperbolic function from Enoch and Adeyeye(2012) result which was the reason why employ the
technique of the Osborne’s Rule approach. Also from the result obtained, by observation one see how much
primes numbers are been played around the synthetic process. This actually obeys the rules guiding the Number
theory (every integer is a product of prime). This method helps in removing the "𝑖"in the result using the Euler’s
Method though we could not see how the integers are been expressed in the products of primes much but this
validate some of the claims of researchers like Euler (1748), Odgers (2004). Thus, the Osborne’s approach is
more preferable.
References
[1]
[2]
[3]
[4]
[5]
E. Bombieri, (2000), Problems of the millennium: The Riemann hypothesis, Clay mathematical Institute.
O. Enoch, (2012), A New Representation of the Riemann Zeta Function Via Its Functional Equation
O. Enoch, (2012), A General Representation of the Zeros of the Riemann Zeta Function via Fourier series Expansion
O.O.A. Enoch and F.J. Adeyeye, (2012), A Validation of the Real Zeros of the Riemann Zeta Function via the Continuation Formula
of the Zeta Function, Journal of Basic & Applied Sciences, 2012, 8, 1
P. Sarnak, (2004), Problems of the Millennium: The Riemann Hypothesis by Princeton University and courant Institute of
Mathematics.
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