4 Formulas for Riemann Zeta at even number
Formulas for Zeta at even number obtained in " 1 Zeta Generating Functions " were automorphisms which were
expressed by the lower zetas. In this chapter, removing the lower zetas, we obtain the explicit formulas.
4.1 Formulas of cot x family
Formula 4.1.1
When
B0=1, B2=1/6, B4=-1/30, B6=1/42,  are Bernoulli numbers, the following expression
holds for
0 < x < 2
.
2s
2s
s
1  n -1 (-1) B2s 2 -2(rx) sin rx
 (2n) = - ΣΣ
(2s)!
x r=1 s=0
r 2n+1
Especially when
x =
B2nx
2n
2(2n)!
2
2n
-2-
 22n+1-2
x

,
B2n(2)
2n
 (2n) =
2(2n)!
Proof
We obtained the following Riemann Zeta in Formula1.4.3 in 1.4 .
2
1
1  sin rx
xΣ
r=1
r3
1  sin rx
 (4) = Σ 5 +
x r=1 r
1x
 x
+
2 3! 2 2!
1  sin rx
 (6) = Σ 7 x r=1 r
1  sin rx
 (8) = Σ 9 +
x r=1 r

1 x6  x5 x2
x4
+
+
 (4) (2)
2 7! 2 6! 3!
5!
 (2) =
Substituting
(k)
1 x4  x3 x2
+
 (2)
2 5! 2 4! 3!
1 x8  x7 x2
x4
x6
+
 (6) (4)+
 (2)
5!
7!
2 9! 2 8! 3!
for the below one by one, we obtain
1  n -1
(-1)n n -1
2s sin rx
s
Cs
(2n) = - ΣΣ(-1) Cs x 2n+1-2s 2 Σ
x r=1 s=0
s=0
r
Where,
Cs
1
9!
1
1
1
1
, C2 =
, C3 =
3!
5! 3!3!
7!

 3!5! + 5!3!  + 3!3!3! ,
1
1
1
 3!7! + 5!5! + 7!3!  +  3!3!5! + 3!5!3! + 5!3!3!  - 3!3!3!3!
1
1
1
1

And, these are calculable by the following expression.
2s
Cs =
2n-1
2n
are rational numberes as follows.
C0 = -1 , C1 =
C4 =

x
x
(2n +1-2s) ! (2n -2s) !
2 -2
B
(2s)! 2s
s =0, 1, 2, 
-1-
1
1
1
Substituting this for the above,
2s
1  n -1
sin rx
s 2 -2
 (2n) = - ΣΣ(-1)
B2sx 2s 2n+1-2s
(2s)!
x r=1 s=0
r

22s-2B2s
22s-2B2s
n -1
(-1)n 2n n -1
2n-1
- x
x Σ
Σ
2
s=0 (2s)!(2n +1-2s)!
s=0 (2s)!(2n -2s)!

Here, following expressions hold.
n -1
22s-2B2s
Σ
s=0
n -1
Σ
s=0
22n-2B2n
=(2s)!(2n +1-2s)!
(2n)!1!
22s-2B2s
(2s)!(2n -2s)!
22n+1-2B2n
=(2n)!0!
Then, substituting these for the above,
2s
2s
s
1  n -1 (-1) B2s2 -2(rx) sin rx
 (2n) = - ΣΣ
(2s)!
x r=1 s=0
r 2n+1
(-1)nx 2n
+
2

22n-2B2n  22n+1-2B2n
(2n)!
(2n)!
x

Thus, we obtain the desired expression.
Formula 4.1.2
Let Bernoulli numbers and Euler numbers are as follows.
B0=1, B2=1/6, B4=-1/30, B6=1/42, B8=-1/30, 
E6=-61, E8=1385, 
E0=1, E2=-1, E4=5,
Then, the following expression holds for
 n -1
 (2n)=ΣΣ
2s
E2s(rx)
(2s)!
r=1 s=0
Especially when
x =
 (2n) = -
0 < x < 2
cosrx
r 2n
.
E2nx
-
2n
2(2n)!
+
2n
2n
2n-1
 2 2 -1B2nx
(2n)!
2
,
1
22n-2
 ΣΣ
 n -1
E2s( r)
r=1 s=0
2s
(2s)!
(-1)r
r 2n
-
E2n
2n
2(2n)!

Proof
We obtained the following Riemann Zeta in Formula1.4.3 in 1.4 .
2
1
cosrx 1 x
 x
+
2 2! 2 1!
r=1
r2
 cosrx
1 x4  x3 x2
+
 (4) = Σ 4 +
 (2)
2 4! 2 3! 2!
r=1
r

 (2) = Σ

 (6) = Σ
r=1

 (8) = Σ
r=1
cosrx 1 x 6 
+
2 6! 2
r6
cosrx 1 x 8 
+
2 8! 2
r8
x4
x5 x2
+
 (4) (2)
4!
5! 2!
x7 x2
x4
x6
+
 (6) (4)+
 (2)
7! 2!
4!
6!
-2-

Substituting
(k)
for the below one by one, we obtain
 n -1
(2n) =ΣΣCs x 2s
r=1 s=0
Where,
Cs
cos rx
r
2n-2s
+
s
s
 x 2n-1 n -1 (-1) Cs
(-1)n 2n n -1 (-1) Cs
x Σ
-(-1)n
2
2 Σ
s=0 (2n -2s ) !
s=0 (2n -1-2s )!
are rational numberes as follows.



1
1
1
1
1
1
1
1
, C1 =
, C2 = , C3 =
,
+
+
+
0!
2!2!2!
2!
4! 2!2!
6!
4!2! 2!4!
1
1
1
1
1
1
1
1
C4 = +
+
+
+
+
+
8!
2!2!2!2!
6!2! 4!4! 2!6!
4!2!2! 2!4!2! 2!2!4!
C0 =



And, these are calculable by the following expression.
Cs = (-1)s
E2s
E2s
=
(2s)!
(2s)!
Substituting this for the above,
 n -1
 (2n) = ΣΣ
E2sx
2s
(2s)!
r=1 s=0
cosrx
r 2n-2s
E2s
(-1)n 2n n -1
+
x Σ
2
s=0 (2s)!(2n -2s)!
n
- (-1)
 x 2n-1
2
E2s
n -1
Σ
s=0 (2s)!(2n -1-2s)!
Here, following expressions hold.
E2s
n -1
Σ
s=0 (2s)!(2n -2s)!
n -1
Σ
s=0
E2n
(2n)!
=-
22n22n-1B2n
E2s
=
(2s)!(2n -1-2s)!
(2n)!
Then, substituting these for the above,
 n -1
 (2n)=ΣΣ
2s
E2s(rx)
(2s)!
r=1 s=0
Especially when
x =
cosrx
r
2n
E2nx
-
2n
+
2(2n)!
2n
2n
2n-1
 2 2 -1B2nx
(2n)!
2
,
 n -1
 (2n) = ΣΣ
E2s( r)
2s
(2s)!
r=1 s=0
i.e.
 n -1
 (2n) = ΣΣ
r 2n
E2s( r)
2s
r=1 s=0
cos  r
cos  r
2n
(2) B2n 2n
+
2 -1
2(2n)!
2(2n)!
-
(2s)!
r
 ΣΣ
E2s( r)
2n
E2n
2n
E2n
2n
+ 22n-1 (2n)
2(2n)!
From this,
 (2n) = -
 n -1
1
22n-2
r=1 s=0
2s
(2s)!
(-1)r
r 2n
-
E2n
2n
2(2n)!

Example (6)
According to the formula at
x =
this is calculated. As the result of calculating the series to the 18400 th
term, the significant 10 digits were obtained.
-3-
By-products
n -1
Σ
s=0
n -1
Σ
s=0
22s-2B2s
(2s)!(2n -2s)!
22n+1-2B2n
=(2n)!0!
22n22n-1B2n
E2s
=
(2s)!(2n -1-2s)!
(2n)!
-4-
4.2 Formulas of tan x family
Formula 4.2.1
When
B0=1, B2=1/6, B4=-1/30, B6=1/42,  are Bernoulli numbers, the following expression
holds for
0< x 
22n
 (2n) =
Especially when
.
(-1)sB2s 22s-2(rx)2s-1 (-1)rsin rx B2n(2x)2n
+
2(2n)!
(2s)!
r 2n
 n -1
ΣΣ
22n-2 r=1 s=0
x =
,
B2n(2)
2n
 (2n) =
2(2n)!
Proof
We obtained the following Dirichlet Eta in Formula1.5.3 in 1.5 .
1 x2
1 
r-1 sin rx
+
(2) =
(-1)
3
2 3!
x
Σ
r=1
r
1 x4 x2
1 
r-1 sin rx
+
(4) = Σ(-1)
(2)
2 5! 3!
x r=1
r5
1 x6 x2
1 
x4
r-1 sin rx
+
+
(6) = Σ(-1)
(4)(2)
2 7! 3!
5!
x r=1
r7
1 x8 x2
1 
x4
x6
r-1 sin rx
+
(8) = Σ(-1)
(6)(4)+
(2)
2 9! 3!
5!
7!
x r=1
r9

Substituting
(k)
for the below one by one , we obtain
Cs
1  n -1
(-1)nx 2n n -1
r-1 sin rx
2s
s
+
(2n)=- ΣΣ(-1) Cs (rx) (-1)
Σ
2
x r=1 s=0
s=0 (2n +1-2s)!
r 2n+1
Where,
Cs
is the same as the coefficient in Formula 4.1.1 and is given by
 n -1
(2n) = ΣΣ
r=1 s=0
22s-2
Cs =
B . Then ,
(2s)! 2s
(-1)s22s-2B2s(rx)2s-1 (-1)rsin rx
(2s)!
r 2n
22s-2B2s
(-1)n x 2n n -1
+
Σ
2
s=0 (2s)!(2n +1-2s)!
Here,
n -1
Σ
s=0
22s-2B2s
22n-2B2n
=(2n)!1!
(2s)!(2n +1-2s)!
Substituting this for the above,
 n -1
(2n) = ΣΣ
r=1 s=0
(-1)sB2s 22s-2(rx)2s-1 (-1)rsin rx 22n-2B2nx 2n
+
2(2n)!
(2s)!
r 2n
-5-
applying
(2n) =
 (2n) =
22n-1
22n-1-1
22n
(2n) =
 n -1
22n-2
Σ
Σ
r=1 s=0
22n
22n-2
(2n)
to this,
(-1)sB2s 22s-2(rx)2s-1 (-1)rsin rx
(2s)!
r 2n
2n
2n
22n 2 -2B2nx
+ 2n
2(2n)!
2 -2
i.e.
 (2n) =
22n
 n -1
ΣΣ
22n-2 r=1 s=0
(-1)sB2s 22s-2(rx)2s-1 (-1)rsin rx B2n(2x)2n
+
2(2n)!
(2s)!
r 2n
Example (6)
When x=1/64, this is calculated according to the formula. As the result of calculating the series to the 32 th
term , the significant 10 digits were obtained.
Formula 4.2.2
Let Bernoulli numbers and Euler numbers are as follows.
B0=1, B2=1/6, B4=-1/30, B6=1/42, B8=-1/30, 
E0=1, E2=-1, E4=5,
E6=-61, E8=1385, 
Then, the following expression holds for
 (2n) = Especially when
22n
22n-2
x =
 (2n) = -
0 < x  .
 ΣΣ
E2s(rx)
 ΣΣ
E2s( r)
 n -1
r=1 s=0
2s
(-1)rcosrx
(2s)!
r 2n
,
1
22n-2
 n -1
r=1 s=0
2s
(-1)r
(2s)!
r 2n
Proof
We obtained the following Dirichlet Eta in Formula1.5.3 in 1.5 .

1 x2
r-1 cosrx
+
(2) = (-1)
2
2 2!
Σ
r=1

r
(4) = Σ(-1)
r=1
2n
E2nx
2(2n)!
r-1
1 x4 x2
cosrx
+
(2)
2 4! 2!
r4
-6-
-
E2n
2n
2(2n)!



(6) = Σ(-1)
r-1
r=1

(8) = Σ(-1)
r-1
r=1
1
cosrx
+
2
r6
1
cosrx
8
2
r
x6 x2
x4
+
(4)(2)
6! 2!
4!
x8 x2
x4
x6
+
(6)(4)+
(2)
8! 2!
4!
6!

(k)
Substituting
for the below one by one , we obtain
 n -1
(2n) = ΣΣCs x (-1)
r=1 s=0
Where,
Cs
2s
r-1
cosrx
r 2n-2s
s
(-1)n x 2n n -1 (-1) Cs
Σ
2
s=0 (2n -2s)!
is the same as the coefficient in the Formula 4.1.2 and is given by
Cs = (-1)s
E2s
E2s
=
(2s)!
(2s)!
Then ,
 n -1
(2n) = -ΣΣ
r=1 s=0
2s
E2s(rx)
(-1)rcosrx
(2s)!
r 2n
-
E2s
(-1)n x 2n n -1
Σ
2
s=0 (2s)!(2n -2s)!
Here,
n -1
Σ
s=0
E2n
E2s
=(2n)!
(2s)!(2n -2s)!
Substituting this for the above,
 n -1
(2n) = -ΣΣ
r=1 s=0
And applying
(2n) =
2s
E2s(rx)
22n
22n-2
(-1)rcosrx
(2s)!
(2n) to this,
r 2n
2n
E2nx
+
2(2n)!
we obtain the desired expression.
-7-
4.3 Formulas of csc x family
Formula 4.3.1
B0=1, B2=1/6, B4=-1/30, B6=1/42, 
holds for 0 < x   .
When
 (2n) = -
22n
 n -1
22n-1
Σ
Σ
r=1 s=0
are Bernoulli numbers , the following expression
(-1)sB2s 22s-2{(2r -1)x }2s-1 sin(2r -1)x
(2s)!
(2r -1)2n
+
Especially when
x =
2n
2n-1
 2 B2nx
2
(2n)!
,
B2n(2)
2n
 (2n) =
2(2n )!
Proof
We obtained the following Dirichlet Lambda in Formula1.6.3 in 1.6 .
1
1  sin {(2r-1)x } x
+
2!
xΣ
r=1
(2r -1)3
1  sin {(2r-1)x } x 3
 (4) = Σ
4!
x r=1 (2r -1)5

1  sin {(2r-1)x } x 5
+
6!
xΣ
r=1
(2r -1)7
1  sin {(2r-1)x } x 7
 (8) = Σ
8!
x r=1 (2r -1)9


 (2) =
 (6) =
Substituting
4

x2
+
 (2)
4 3!
4
+
x2
x4
 (4)(2)
3!
5!

x4
x6
x2
+
 (6)(4)+
(2)
5!
7!
4 3!
(k)
for the below one by one , we obtain
2s-1
s
 n -1 (-1) C s x
(2n) =sin(2r -1)x +(-1)n
2n+1-2s
r=1 s=0
ΣΣ
Where,
Cs
(2r -1)
is the same as the coefficient in Formula 4.1.1 and is given by
 n -1
(2n) = -ΣΣ
r=1 s=0
Σ
s=0
Cs =
4
Σ
s=0
Cs
(2n -2s)!
22s-2
B . Then ,
(2s)! 2s
(-1)s22s-2B2s{(2r -1)x }2s sin(2r -1)x
(2s)!
(2r -1)2n
2s
2n-1 n -1
2 -2B2s
n x
+ (-1)
4 Σ
s=0 (2s)!(2n -2s)!
Here,
n -1
 x 2n-1 n -1
22s-2B2s
22n+1-2B2n
=(2n)!0!
(2s)!(2n -2s)!
Substituting this for the above,
-8-
2s
2s
s
1  n -1 (-1) B2s 2 -2{(2r -1)x } sin(2r -1)x
(2n) = - ΣΣ
(2s)!
x r=1 s=0
(2r -1)2n
2n+1
-2B2nx 2n-1
 2
+
4
(2n )!
applying
(2n) =
 (2n) = -
22n
22n-1
(2n) =
22n
 n -1
22n-1
Σ
Σ
r=1 s=0
22n+1
22n+1-2
(2n )
to this,
(-1)sB2s 22s-2{(2r -1)x }2s-1 sin(2r -1)x
(2s)!
(2r -1)2n
2n
2n-1
 2 B2nx
+
2
(2n)!
Formula 4.3.2
Let Bernoulli numbers and Euler numbers are as follows.
B0=1, B2=1/6, B4=-1/30, B6=1/42, B8=-1/30, 
E0=1, E2=-1, E4=5,
E6=-61, E8=1385, 
Then, the following expression holds for
22n
 (2n) =
Especially when
2s
E2s{(2r -1)x }
 n -1
ΣΣ
(2s)!
22n-1 r=1 s=0
x =  /2
0 < x  .
4n
2n-1
cos(2r -1)x  2 B2nx
+
4
(2n)!
(2r -1)2n
,
B2n(2)
2n
 (2n) =
2(2n)!
Proof
We obtained the following Dirichlet Lambda in Formula1.6.3 in 1.6 .
1

(2) = Σ
r=1

(4) = Σ
r=1

(6) = Σ
r=1

(8) = Σ
r=1

Substituting
cos {(2r-1)x } x
+
1!
(2r -1)2
cos {(2r-1)x } x 3
3!
(2r -1)4
cos {(2r-1)x } x 5
+
5!
(2r -1)6
Cs
4

x2
+
(2)
4 2!

x2
x4
+
(4)(2)
4!
4 2!
cos {(2r-1)x } x 7  x 2
x4
x6
+
(6)(4)+
(2)
7! 4 2!
4!
6!
(2r -1)8
(k)
for the below one by one , we obtain
Cs x 2s
 n -1
cos(2r -1)x-(-1)n
2n-2s
r=1 s=0
(2n) = ΣΣ
Where,

(2r -1)
 x 2n-1
4
is the same as the coefficient in the Formula4.1.2 and is given by
-9-
(-1)s
Cs
Σ
(2n -1-2s)!
s=0
n -1
Cs = (-1)s
E2s
E2s
=
(2s)!
(2s)!
Then ,
E2sx
 n -1
(2n) = ΣΣ
r=1 s=0
2s
(2s)!(2r -1)2n-2s
cos(2r -1)x
n
- (-1)
 x 2n-1
4
E2s
n -1
Σ
s=0 (2s)!(2n -1-2s)!
Here,
n -1
Σ
s=0
22n22n-1B2n
E2s
=
(2n)!
(2s)!(2n -1-2s)!
Substituting this for the above,
E2sx
 n -1
(2n) = ΣΣ
r=1 s=0
2s
(2s)!(2r -1)2n-2s
cos(2r -1)x
n
- (-1)
And applying
(2n) =
22n
22n-1
(2n) to this,
2n
2n
2n-1
 2 2 -1B2n x
4
(2n)!
we obtain the desired expression.
Example (6)
When x=1/64, this is calculated according to the formula. As the result of calculating the series to the 25 th
term, the significant 10 digits were obtained.
2012.04.10
K. Kono
Alien's Mathematics
- 10 -