3 Formulas for Riemann Zeta at odd number
Formulas for Zeta at odd 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.
3.1 Formulas of cot x family
Formula 3.1.1
When
B0=1, B2=1/6, B4=-1/30, B6=1/42, 
Harmonic numbers, the following expressions hold for
s
are Bernoulli numbers and
0 < x < 2
Hs =Σ1/t
are
t=1
.
2s
22s-2B2s (rx) sin rx
1  n
s 
(2n +1) = - ΣΣ(-1)
x r=1 s=0
(2s)!
r 2n+2

2s
2s
 n
2 -2B2s H2n+1-2s
2 -2B2s
B2rx
+ΣΣ
Σ
2r
s=0 (2s )!(2n +1-2s ) !
r=1 s=0 (2s )!(2n +1+2r -2s ) !
n 2n
+(-1) x
x =
(2n +1) =
Especially when
(-1) 
n
2r

,
2s
2s
2 -2B2s H2n+1-2s  n
2 -2B2s
B2r
+ΣΣ
Σ
2r
s=0 (2s )!(2n +1-2s ) !
r=1 s=0 (2s )!(2n +1+2r -2s ) !

2n
n
n
2r

Proof
We obtained the following Riemann Zeta in Formula1.4.3 in 1.4 .
2r+2
 B 2r x
1  sin rx x 2
 (3) =
log x -H3+
3! 
x r=1 r 4
r =1 2r(2r +3)!
Σ
Σ

1  sin rx x 4
 (5) = Σ 6 +
log x -H5 -Σ

5!
x r=1 r
r =1
B2rx
2r+4
2r(2r +5)!
+
x2
 (3)
3!
2r+6
 B2r x
1  sin rx x 6
x2
x4
+
 (7) = Σ 8  (5) (3)
log x -H7 +Σ
7! 
3!
5!
x r=1 r
r =1 2r(2r +7)!
2r+8
 B2rx
1  sin rx x 8
x4
x6
x2
 (9) = Σ 10 + log x -H9 -Σ
+
(7)(5)+
(3)
x r=1 r
5!
7!
9!
3!
r =1 2r(2r +9) !

Substituting
(k)
for the below one by one, we obtain
1  n -1
sin rx
(-1)sCs x 2s 2n+2-2s
Σ
Σ
x r=1 s=0
r
n -1
Cs
- (-1)nx 2nΣ
log x -H2n+1-2s
s=0 (2n +1-2s)!
2r
Cs
 n -1
B2rx
n 2n
+ (-1) x ΣΣ
2r
r =1 s=0 (2n +1+2r -2s)!
 (2n +1) = -
Where,
Cs
are rational numberes as follows.
C0 = -1 , C1 =
1
1
1
1
, C2 =
, C3 =
3!
5! 3!3!
7!
-1-


1
1
1
,
+
+
3!3!3!
3!5! 5!3!
C4 =

1
9!
 
1
1
1
+
+
+
3!7! 5!5! 7!3!

1
1
1
1
+
+
3!3!5! 3!5!3! 5!3!3!
3!3!3!3!

In fact, these are the coefficients of Taylor series of csc x, and are given by
number as follows.
22s-2
Cs =
B
(2s)! 2s
22s , factorial and Bernoulli
s =0, 1, 2, 
Substituting this for the above,
2s
22s-2B2s (rx ) sin rx
1  n -1
2n
s 
(2n +1) = - ΣΣ(-1)
- (-1)nx 
2n+2
x r=1 s=0
(2s)!
r

2s

1  sin rx
2
xΣ
r=1 r

2s
2s
 n -1
2 -2B2s
2 -2B2s
B2sx
log x -H2n+1-2s-ΣΣ
Σ
2r
s =0 (2s )!(2n +1-2s ) !
r =1 s =0 (2s )!(2n +1+2r -2s ) !
n -1
The inside of the parenthesis of the right side can be transformed as follows.
2s
2s
 n -1
2 -2B2s
2 -2B2s
B2sx
log x -H2n+1-2s -ΣΣ
Σ
2r
s =0 (2s )!(2n +1-2s ) !
r =1 s =0 (2s )!(2n +1+2r -2s ) !
n -1
n -1
=Σ
s=0
2s
2s
2s
n -1 2 -2B2s H2n+1-2s
2 -2B2s log x
(2s)!(2n +1-2s) ! Σ
s=0 (2s )!(2n +1-2s ) !
2s
2 -2B2s
B2rx
- ΣΣ
2r
r =1 s =0 (2s )!(2n +1+2r -2s ) !
 n -1
2r
2r
2n
2 -2B2n  B2rx
+
(2n )! Σ
r =1 2r(1+2r) !
Here, from Lemma 3.4.1 and Lemma 3.4.2c ,
n -1
Σ
s=0
2s
2n
2 -2B2s
2 -2B2n
=(2n)!
(2s)!(2n +1-2s) !
B2rx

2r
Σ
r=1 2r(2r +1) !
= log x - 1 +
1  sin rx
xΣ
r=1
r2
Using these,
2s
2s
 n -1
2 -2B2s
2 -2B2s
B2sx
log x -H2n+1-2s -ΣΣ
Σ
2r
s =0 (2s )!(2n +1-2s ) !
r =1 s =0 (2s )!(2n +1+2r -2s ) !
n -1
2s
2n
2s
n -1 2 -2B2s H2n+1-2s
2 -2B2n
log x - Σ
=(2n )!
s=0 (2s )!(2n +1-2s ) !

n
- ΣΣ
r =1 s=0
2s
2 -2B2s
B2rx
(2s)!(2n +1+2r -2s) ! 2r
2r
2n
2 -2B2n
+
(2n)!

log x -1+
i.e.
2s
2s
 n -1
2 -2B2s
2 -2B2s
B2sx
log
x
-H

2n+1-2s ΣΣ
Σ
2r
s =0 (2s )!(2n +1-2s ) !
r =1 s =0 (2s )!(2n +1+2r -2s ) !
n -1
n
= -Σ
s=0
2s
2s
 n
2 -2B2s H2n+1-2s
2 -2B2s
B2rx
- ΣΣ
2r
(2s)!(2n +1-2s) !
r =1 s=0 (2s)!(2n +1+2r -2s) !
2n
1 2 -2B2n  sin rx
+
x
(2n )! Σ
r=1
r2
Substituting this in the above parenthesis
-2-
2r
2s
2s
22s-2B2s (rx )
1  n -1
s 
(2n +1) = - ΣΣ(-1)
x r=1 s=0
(2s)!
-Σ
n
s=0
sin rx
r
2n+2
- (-1)nx
2n

2s
2s
n

2 -2B2s H2n+1-2s
2 -2B2s
B2rx
- ΣΣ
2r
(2s)!(2n +1-2s) !
r=1 s=0 (2s )!(2n +1+2r -2s ) !
2r
2n
1 2 -2B2n  sin rx
+
x
(2n)! Σ
r=1
r2

2s
2n
2n
 sin rx
22s-2B2s (rx) sin rx
1  n -1
(-1)n 2 -2B2n x
s 
= - ΣΣ(-1)
Σ
x
x r=1 s=0
(2s)!
(2n )!
r=1
r 2n+2
r2

n 2n
+(-1) x
2s
2s
2 -2B2s H2n+1-2s  n
2 -2B2s
B2rx
+
Σ
Σ
Σ
2r
s=0 (2s )!(2n +1-2s ) !
r=1 s=0 (2s )!(2n +1+2r -2s ) !
n
2r

2r

Here,
2n
2 -2B2n x
(-1)
Σ
(2n)!
r=1

2n
n
sin rx
r
2
2n
2n
2 -2B2n (rx )
= Σ(-1)
(2n)!
r=1

n
sin rx
r
2n+2
Thus,
2s
22s-2B2s (rx)
1  n
s 
(2n +1) = - ΣΣ(-1)
x r=1 s=0
(2s)!

n 2n
+(-1) x
sin rx
r 2n+2
2s
2s
 n
2 -2B2s H2n+1-2s
2 -2B2s
B2rx
+ΣΣ
Σ
2r
s=0 (2s )!(2n +1-2s ) !
r=1 s=0 (2s )!(2n +1+2r -2s ) !
n
Note
When
x = ,
the convergence speed of this formula is the fastest in the Trigonometric functions family.
Example 1
(5) = 

4

1
269
1
7
+Σ +
21600 r=1 (2r +5)! 6(2r +3)! 360(2r +1)!


B2r
2r
2r

Example 2 (7)
According to the formula at
x =
this is calculated. As the result of calculating the series to the 12 th term,
the significant 10 digits were obtained.
Formula 3.1.2
s
Let
Hs =Σ1/t
be Harmonic number, and let Bernoulli numbers and Euler numbers are as follows.
t=1
B0=1, B2=1/6, B4=-1/30, B6=1/42, B8=-1/30, 
-3-
E0=1, E2=-1,
E4=5,
Then, the following expression holds for

n
E2s(rx)
Σ
n
s=0
E8=1385, 
.
cos rx
r 2n+1
(2s)!
r=1 s=0
-(-1) x
0 < x < 2
2s
(2n +1) = ΣΣ
n 2n
E6=-61,
2r
n

E2s H2n-2s
E2s
B2rx
+ Σ
2r
(2s)!(2n -2s) ! Σ
r=1 s=0 (2s )!(2n +2r -2s ) !

Proof
We obtained the following Riemann Zeta in Formula1.4.3 in 1.4 .

 (3) = Σ
r=1

 (5) = Σ
r=1

 (7) = Σ
r=1

 (9) = Σ
r=1
2r+2
 B2r x
cosrx x 2
log x -H2 +Σ
2! 
r =1 2r(2r +2)!
r3
2r+4
 B2r x
cosrx x 4
x2
+
+
 (3)
log x H4 Σ
4! 
2!
r =1 2r(2r +4)!
r5

cosrx x 6
+
log
x
H
Σ


6
6!
r =1
r7

cosrx x 8
+
log
x
H
Σ


8
8!
r =1
r9
B2rx
2r+6
2r(2r +6)!
+
x2
x4
 (5) (3)
2!
4!
2r+8
x2
x4
x6
+
(7)(5)+
(3)
4!
6!
2r(2r +8) ! 2!
B2rx

Substituting
(k)
 (2n +1)
for the below one by one, we obtain
2s
n -1
 n -1 C s x
n 2n
-1
=
+
(
)
cosrx
x
2n+1-2s
r=1 s=0
s=0
ΣΣ
Σ
r
n 2n
- (-1) x
Where,
Cs
(-1)sCs
log x -H2n-2s
(2n -2s)! 
 n -1
Σ
Σ
r =1 s=0
2r
(-1)sCs
B2rx
2r
(2n +2r -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 = +
+
+
+
+
+
6!2! 4!4! 2!6!
4!2!2! 2!4!2! 2!2!4!
8!
2!2!2!2!
C0 =



According to Mr. Sugimoto , these are given by the following expression.
Cs = (-1)s
E2s
E2s
=
(2s)!
(2s)!
s =0, 1, 2, 
Substituting this for the above,
 n -1
(2n +1) = ΣΣ
r=1 s=0
Σ
n -1
s=0
2s
E2s(rx)
(2s)!
cos rx
r
2n+1
+ (-1)nx
2n

2r
 n -1
E2s log x -H2n-2s
E2s
B2rx
-ΣΣ
2r
(2s)!(2n -2s) !
r=1 s=0 (2s )!(2n +2r -2s ) !
The inside of the parenthesis of the right side can be transformed as follows.
-4-

n -1
Σ
s=0
 n -1
E2s log x -H2n-2s
E2s
B2rx
-ΣΣ
2r
(2s)!(2n -2s) !
r=1 s=0 (2s )!(2n +2r -2s ) !
n -1
n -1
E2s log x
E2s H2n-2s
=Σ
-Σ
s=0 (2s )!(2n -2s ) !
s=0 (2s )!(2n -2s ) !

E2s
B2rx
2r
(2s)!(2n +2r -2s) !
n
- ΣΣ
r =1 s=0
2r
2r
2r
E2n  B2rx
+
(2n)! Σ
r =1 2r(2r) !
Here, from Lemma 3.4.1 and Lemma 3.4.2c ,
n -1
Σ
s=0

E2s
E2n
=(2s)!(2n -2s) !
(2n )!
B2rx
2r

Σ
r=1 2r(2r ) !
= log x + Σ
r=1
cos rx
r
Using these,
n -1
Σ
s=0
 n -1
E2s log x -H2n-2s
E2s
B2rx
-ΣΣ
2r
(2s)!(2n -2s) !
r=1 s=0 (2s )!(2n +2r -2s ) !
-1
n
E2n
E2s H2n-2s
log x - Σ
=(2n)!
s=0 (2s )!(2n -2s ) !

2r
 cos rx
E2s
E2n
B2rx
log x + Σ
+
r
2r
(2s)!(2n +2r -2s) !
(2n)!
r=1

n
- ΣΣ
r =1 s=0
Since
n -1
Σ
s=0
2r

H2n-2n = 0 ,
 n -1
E2s log x -H2n-2s
E2s
B2rx
-ΣΣ
2r
(2s)!(2n -2s) !
r=1 s=0 (2s )!(2n +2r -2s ) !
2r
2r
 n
E2n  cos rx
E2s H2n-2s
E2s
B2rx
= -Σ
-ΣΣ
+
r
2r
(2n )!Σ
s=0 (2s )!(2n -2s ) !
r=1
r =1 s=0 (2s )!(2n +2r -2s) !
n
Substituting this in the above parenthesis
2s
E2s(rx)
 n -1
(2n +1) = ΣΣ
(2s)!
r=1 s=0
n 2n
- (-1) x
Σ
n
s=0
cos rx
r 2n+1
2n
 cos rx
(-1)nE2n x
+
Σ
r
(2n)!
r=1
2r
n

E2s H2n-2s
E2s
B2rx
+ Σ
2r
(2s)!(2n -2s) ! Σ
r=1 s=0 (2s )!(2n +2r -2s ) !

Here,
(-1)nE2n x
(2n )!
2n

Σ
r=1
2n
 cos rx
cos rx
E2n(rx )
=
Σ
2n+1
r
(2n)!
r=1 r
Thus,

2s
n
(2n +1) = ΣΣ
E2s(rx)
r=1 s=0
n 2n
-(-1) x
Σ
n
s=0
(2s)!
cos rx
r
2n+1
n

E2s H2n-2s
E2s
B2rx
+ΣΣ
2r
(2s)!(2n -2s) ! r=1 s=0 (2s)!(2n +2r -2s) !
2r

Example (7)
When x =1/64, this is calculated according to the formula. As the result of calculating the series to the 24 th
term, the significant 10 digits were obtained.
-5-
-6-
3.2 Formulas of tan x family
Formula 3.2.1
Let Bernoulli numbers and Tangent numbers are as follows.
B0=1, B2=1/6, B4=-1/30, B6=1/42, B8=-1/30, 
T1=1, T3=2,
Then, the following expression holds for
22n
 (2n +1) =
22n-1

- (-1) x
x =
 (2n +1) =
0< x 
.
2s
2s
s
1  n (-1) B2s 2 -2(rx) (-1)rsin rx
Σ
(2s)!
xΣ
r=1 s=0
r 2n+2

n 2n
Especially when
T7=272 , T9=7936 , 
T5=16,
n
Σ
Σ
r =1 s=0
22s-2B2s
22r-1B2rx 2r
2r
(2s)!(2n +1+2r -2s)!

,
2n
(-1)n-1(2)
22n-1


n
Σ
Σ
r =1 s =0
22s-2B2s
(2s)!(2n +1+2r -2s)!
T

2r-1
 2
2r
Proof
We obtained the following Riemann Zeta in Formula1.5.3 in 1.5 .
2r
 2 -1B 2r
1 
r-1 sin rx
+
(1) =
(-1)
2
2r(2r +1)!
x
Σ
r=1
r
Σ
r=1
x 2r
2r
 2 -1B 2r
1 
x2
r-1 sin rx
2r+2
-Σ
+
(3) = Σ(-1)
(1)
x
2r(2r +3)!
3!
x r=1
r=1
r4
2r
 2 -1B 2r
1 
x2
x4
r-1 sin rx
2r+4
+Σ
+
(5) = Σ(-1)
(3)(1)
x
2r(2r +5)!
3!
5!
x r=1
r=1
r6
(7) =
2r
2
4
6
 2 -1B2r
1 
x
x
x
2r+6
r-1 sin rx
x
(5)(3)+
(1)
-Σ
+
(-1)
8
xΣ
3!
5!
7!
r=1
r=1 2r(2r +7) !
r

Substituting
(k)
for the below one by one, we obtain
r-1
1  n
sin rx
2s (-1)
s
(2n +1) = - ΣΣ(-1) Cs x
x r=1 s=0
r 2n+2-2s
n 2n
- (-1) x

n
Σ
Σ
r =1 s=0
22r-1B2rx 2r
Cs
2r
(2n +1+2r -2s)!
Where, Cs is the same as the coefficient in Formula 3.1.1 and is given by
Cs =
22s-2
B . Then ,
(2s)! 2s
2s
2s
s
1  n (-1) B2s 2 -2(rx) (-1)rsin rx
(2n +1) = ΣΣ
(2s)!
x r=1 s=0
r 2n+2
-7-

n 2n
Σ
Σ
r =1 s=0
- (-1) x
And using
n
22r22r-1B2r
22s-2B2s
2r
(2s)!(2n +1+2r -2s)!
22r22r-1B2r
22n
(2n +1) = 2n
(2n +1) , T2r-1 =
2r
2 -1

 2
2r
, we obtain the desired
expression.
Example1
16
(5) = 15
4 
Σ
r=1 
1
1
7
+
(2r +5)! 6(2r +3)! 360(2r +1)!

2r
2 -1B2r
2r
2r
Example2 (7)
According to the formula at
x =
this is calculated. As the result of calculating the 3000 terms,
the significant 4 digits were obtained.
As seen in Example2, the convergence of the formula at

Σ
r=1
T2r-1
(2r +1)!
If the formula at
x =

 2
2r
x =
22r-1B2r
=Σ
2r(2r +1)!
r=1

is vary slow. Then, from Lemma 3.4.2t ,
2r
= log 2
is transformed as follows using this, the convergence is faster.
Formula 3.2.1'

2 
(2n +1) =(-1)n-1 2n
2n
2 -1
 n-1
ΣΣ
r =1s =0
2s
2r
2 -2B2s
2 -1B2r
2r
(2s)!(2n +1+2r -2s)!
2r

2n
2 -2B2n
log 2
+
(2n )!
Example2' (7)
The same calculation as Example2 was performed by this formula . As the result of calculating the 55 terms,
the significant 6 digits were obtained.
-8-
Formula 3.2.2
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, 
0< x 
Then, the following expression holds for
 (2n +1) = -
22n
 ΣΣ

22n-1
n
2s
E2s(rx)
- (-1) x
2n
(-1)rcosrx
(2s)!
r=1 s=0
n
.

n
Σ
Σ
r =1 s=0
r 2n+1
22r-1B2rx 2r
E2s
2r
(2s)!(2n +2r -2s)!

Proof
We obtained the following Riemann Zeta in Formula1.5.3 in 1.5 .
2r

 2 -1B 2r
r-1 cosrx
+
(1) = (-1)
x 2r
1
2r(2r)!
Σ
r=1
Σ
r=1
r

(3) = Σ(-1)
r=1
r-1

(5) = Σ(-1)r-1
r=1

(7) = Σ(-1)
r=1
r-1
2r
 2 -1B 2r
x2
cosrx
2r+2
-Σ
+
(1)
x
2r(2r +2)!
2!
r=1
r3
2r
 2 -1B 2r
x2
cosrx
x4
2r+4
+Σ
+
(3)(1)
x
2r(2r +4)!
2!
4!
r=1
r5
cos rx
r
7

-Σ
r=1
2r
4
6
2 -1B2r 2r+6 x 2
x
x
x
(5)(3)+
(1)
+
4!
6!
2!
2r(2r +6) !

Substituting
(k)
for the below one by one, we obtain
r-1
cosrx
2s (-1)
(2n +1) =
Cs x
2n+1-2s
r=1 s=0

n
ΣΣ
Where,
Cs
r
22r-1B2rx 2r
+(-1) x ΣΣ(-1) Cs
2r(2r +2n -2s)!
r =1 s=0
 n
n 2n
s
is the same as the coefficient in the Formula 3.1.2 and is given by
Cs = (-1)s
E2s
E2s
=
(2s)!
(2s)!
Then,

2s
n
(2n +1) = -ΣΣ
r=1 s=0
E2s(rx)
n 2n
+ (-1) x
And using
(2n +1) =
(2s)!

n
Σ
Σ
r =1 s=0
(-1)rcosrx
r 2n+1
22r-1B2rx 2r
E2s
2r
(2s)!(2n +2r -2s)!
22n
(2n +1) ,
22n - 1
we obtain the desired expression.
-9-
3.3 Formulas of csc x family
Formula 3.3.1
When
B0=1, B2=1/6, B4=-1/30, B6=1/42, 
Harmonic numbers, the following expressions hold for
s
Hs =Σ1/t
are Bernoulli numbers and
0< x 
are
t=1
.
2s
2s
s
1  n (-1) 2 -2B2s (2r -1) x sin(2r -1) x
(2n +1) = - 2n+1
ΣΣ
2n+2
(2s)!
2
-1 x r=1 s=0
(2r -1)
22n+1
+
(-1)n(2x)
22n+1-1
x =
(2n +1) =
Especially when
2s
2s
2r
2 -2B2s H2n+1-2s  n 2 -2B2s 2 -2B2r x 2r
-ΣΣ
Σ
s=0 (2s )!(2n +1-2s ) !
r=1 s=0 (2s )!(2n +1+2r -2s ) ! 2r

2s
2s
2r
2 -2B2s H2n+1-2s  n 2 -2B2s 2 -2B2r  2r
-ΣΣ
Σ
s=0 (2s )!(2n +1-2s ) !
r=1 s=0 (2s )!(2n +1+2r -2s ) ! 2r


2n
n
,
(-1)n(2 )
2n
22n+1 - 1

n
Proof
Originally, this should be derived from Formula1.6.3 in 1.6 . However, since it is complicated, we derive this
from the previous two sections. From Formula3.1.1 and Formula3.2.1 ,
2s
22s-2B2s (rx)
1  n
s 
(2n +1) = - ΣΣ(-1)
x r=1 s=0
(2s)!

n 2n
+(-1) x
sin rx
r 2n+2
2s
2s
 n
2 -2B2s H2n+1-2s
2 -2B2s
B2rx
+ΣΣ
Σ
2r
s=0 (2s )!(2n +1-2s ) !
r=1 s=0 (2s )!(2n +1+2r -2s ) !
n
2s
2s
s
1  n (-1) B2s 2 -2(rx)
(2n +1) = ΣΣ
x r=1 s=0
(2s)!

n 2n
n
Σ
Σ
r =1 s=0
- (-1) x
2r
(-1)rsin rx
r
2n+2
2s
2r
2 -2B2s
2 -1B2rx
2r
(2s)!(2n +1+2r -2s) !
2r
Adding both,
2s
2s
s
1  n (-1) 2 -2B2s (rx) sin rx -(-1)rsin rx
(2n +1) +(2n +1) = - ΣΣ
x r=1 s=0
(2s)!
r 2n+2
+ (-1) x
n
Σ
s=0
n 2n
2s
2 -2B2s H2n+1-2s
(2s)!(2n +1-2s) !

n 2n
- (-1) x
n
Σ
Σ
r =1 s=0
2s
2r
2 -2B2s
2 -2B2rx
2r
(2s)!(2n +1+2r -2s) !
2r
Here,


22n -1
22n+1 - 1
2n
+1
=

(2n +1)
(
)
22n
22n
2s
(-1)s 22s-2B2s (rx ) sin rx -(-1)rsin rx
(2s)!
r 2n+2
(2n +1) +(2n +1) = 1+

n
Σ
Σ
r=1 s=0

n
= 2ΣΣ
r=1 s=0
(-1)s 22s-2B2s (2r -1) x
(2s)!
- 10 -
2s
sin(2r -1) x
(2r -1)
2n+2

Thus,
2s
2s
s
1  n (-1) 2 -2B2s (2r -1) x sin(2r -1) x
(2n +1) = - 2n+1
ΣΣ
2n+2
(2s)!
-1 x r=1 s=0
2
(2r -1)
2s
2s
2r
2n
n 2 -2B2s H2n+1-2s
 n 2 -2B2s 2 -2B2r x 2r
(-1)n(2x)
+
-ΣΣ
Σ
s=0 (2s )!(2n +1-2s ) !
r=1 s=0 (2s )!(2n +1+2r -2s ) ! 2r
22n+1-1
22n+1


Note
As seen in the proof, these are the weighted averages of Formula3.1.1 and Formula3.2.1 and are not original
formulas. These are only complicated, and there is not much value.
Formula 3.3.2
s
Let
Hs =Σ1/t
be Harmonic number, and let Bernoulli numbers and Euler numbers are as follows.
t=1
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+1
(2n +1) =

0< x 
E2s(2r -1)x
n
ΣΣ
22n+1-1 r=1 s=0
(-1)n (2x )2n
22n+1-1
-
.
(2s)!

2s
cos(2r -1)x
(2r -1)
2n+1
 n
E2s H2n-2s
E2s 22r-2B2r x 2r
Σ
Σ
Σ
s =0 (2 s)!(2 n -2 s) !
r =1 s =0 (2 s)!(2 n +2 r -2 s) ! 2 r
n

x =  /2 ,
(2n +1) =
Especially when
(-1)n-1 2n
22n+1-1

 n
E2s H2n-2s
E2s 22r-2B2r 1
Σ
Σ
Σ
s =0 (2 s)!(2 n -2 s) !
r =1 s =0 (2 s)!(2 n +2 r -2 s) ! 2 r
n

 2 
2r
Proof
From Formula3.1.2 and Formula3.2.2 ,

2s
n
(2n +1) = ΣΣ
E2s(rx )
(2s)!
r=1 s=0
n 2n
-(-1) x
Σ
n
s=0

cos rx
n
(2n +1) = -ΣΣ
r=1 s=0
2n+1
2r
n

E2s H2n-2s
E2s
B2rx
+ Σ
2r
(2s)!(2n -2s) ! Σ
r=1 s=0 (2s )!(2n +2r -2s ) !
2s
E2s(rx)
(2s)!

n 2n
+ (-1) x
r
n
Σ
Σ
r =1 s=0

(-1)rcos rx
r 2n+1
2r
2r
E2s
2 -1B2rx
2r
(2s)!(2n +2r -2s) !
Adding both,

n
(2n +1) +(2n +1) = ΣΣ
r=1 s=0
n 2n
-(-1) x

2s
E2s(rx)
(2s)!
cos rx -(-1)rcos rx
r
2n+1
2r
2r
 n
E2s H2n-2s
E2s
2 -2B2rx
-ΣΣ
Σ
2r
s=0 (2s )!(2n -2s ) !
r=1 s=0 (2s )!(2n +2r -2s ) !
n
- 11 -

Here,
22n+1 - 1
(2n +1)
22n
(2n +1) +(2n +1) =

n
Σ
Σ
r=1 s=0
2s
E2s(rx)
cos rx -(-1)rcos rx
(2s)!
r
2n+1

n
= 2ΣΣ
r=1 s=0
2s
E2s{(2r -1)x}
cos {(2r -1)x}
(2s)!
(2r -1)
2n+1
Thus,
(2n +1) =
22n+1

n
ΣΣ
-1 r=1 s=0
E2s(2r -1)x
2n+1
2
(-1)n (2x )2n
22n+1-1

(2s)!
2s
cos(2r -1)x
(2r -1)
2n+1
 n
E2s H2n-2s
E2s 22r-2B2r x 2r
-ΣΣ
Σ
s =0 (2 s)!(2 n -2 s) !
r =1 s =0 (2 s)!(2 n +2 r -2 s) ! 2 r
n

Note
These are also the weighted averages of Formula3.1.2 and Formula3.2.2 . However, the formula at
x = /2
is original and is worthy.
Example1
(5)=
4
31


1
1
83
5
+Σ
+
288 r =1 (4+2r) ! 2(2+2r) ! 24(2r)!


2r
2 -2B2r
2r

 2 
2r
Example 2 (7)
According to the formula at
x = /2
this is calculated. As the result of calculating the series to the 13 th
term, the significant 10 digits were obtained.
- 12 -
3.4 Lemmas
Lemma 3.4.1
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 expressions hold.
22s-2B2s
n
Σ
s=0 (2s)!(2n +1-2s)!
E2s
n
Σ
s=0 (2s)!(2n -2s)!
=0
=0
Proof

x csch x = 1 -Σ
s=1
22s-2B2s 2s
x
(2s)!
|x|< 
i.e.
0
2
4
2 -2B0
2 -2B2 2 2 -2B4 4
2x
=x x -
0!
2!
4!
e x - e -x
Here, let
C2s = -22s-2B2s .
Then
C0
C 2 2 C 4 4 C6 6
2x
=
+
x +
x +
x +
0!
2!
4!
6!
e x - e -x
1
1 2 1 4 1 6 1 8
e x - e -x
=
+
x +
x +
x +
x +
1! 3!
2x
5!
7!
9!
The Cauchy product is

1 = C0 +

C2
C0
+
x2 +
0!3! 2!1!
In order to hold this for arbitrary
x


C2
C4
C0
+
+
x4 +
0!5! 2!3! 4!1!
, the followings are necessary.
C0
C0
C2
C2
C4
= 0,
= 0 ,
+
+
+
0!5! 2!3! 4!1!
0!3! 2!1!
C0 =1 ,
That is,
n
Σ
s=0
-22s-2B2s
C2s
n
=
=0
(2s)!(2n +1-2s)! Σ
s=0 (2s)!(2n +1-2s)!
Next,

sech x =Σ
s=0
E2s 2s
x
(2s)!
|x|<

2
i.e.
2
e x + e -x
E0
E1 1 E2 2 E3 3 E4 4
+
x +
x +
x +
x +
0! 1!
2!
3!
4!
=
1
1 1 1 2 1 3 1 4
e x + e -x
=
+
x +
x +
x +
x +
0! 1!
2
2!
3!
4!
The Cauchy product is
1 = E0 +


E1
E0
+
x1 +
0!1! 1!0!


E1
E2
E0
+
+
x 2 +
0!2! 1!1! 2!0!
- 13 -
In order to hold this for arbitrary
x
, the followings are necessary.
E0
E1
E0
E1
E2
= 0,
= 0 ,
+
+
+
0!2! 1!1! 2!0!
0!1! 1!0!
E0 =1 ,
That is,
n
Σ
s=0
Since
Es
=0
s !(n -s)!
E1 = E3 = E5  = 0 ,
we obtain
E2s
n
Σ
s=0 (2s)!(2n -2s)!
=0
Lemma 3.4.2c
When
for
B0=1, B2=1/6, B4=-1/30, B6=1/42, 
0 < x < 2
.
B2rx

2r

Σ
r=1 2r(2r)!
B2rx

=Σ
r=1
2r
Σ
r=1 2r(2r +1)!
Especially, when
x =
B2r

=
cosrx
+ log x
r
1  sin rx
+ log x - 1
xΣ
r=1
r2
,
2r
Σ
r=1 2r(2r)!
B2r

are Bernoulli numbers, the following expressions hold
= log

2
2r
Σ
r=1 2r(2r +1)!
= log - 1
Proof
The 1st order integrals of Taylor series and Fourier series of

x


cot x dx = log x -Σ
r=1
2
x


were as follows. (See " 5 Termwise Higher Integral ")
2 B2r 2r
x
2r(2r)!
0< x < 
1
1  1
2rx 0 Σ 1 sin
2
2 r=1 r

cot x dx =
cot x
2r
 - log 2
0< x < 
2
From these,

Σ
r=1
Replacing

22rB2r 2r
 1
x = Σ 1 cos(2rx) + log 2x
2r(2r)!
r=1 r
x
with
x /2 ,
B2rx
2r
Σ
r=1 2r(2r)!

=Σ
r=1
cosrx
+ log x
r
Next, the 2nd order integrals of Taylor series and Fourier series of cot x were as follows.
x x
22rB2r 2r+1

2
  cot x dx


2
2
= x(log x -1) -Σ
r=1
-
2

log

2
2r (2r +1)!

-1

-Σ
r=1
- 14 -
x
22rB2r
2r(2r +1)!

 2

2r+1

x
x


2
2
cot x dx 2 =
1  1
2
2rx 1 Σ 2 sin
2
2 r=1 r


 - x - 2  log 2
From these,
22rB2r
2r+1
x
2r (2r +1) !

x(log x -1) -Σ
r=1
=Giving
x =0

22rB2r
2r(2r +1) !

r=1

 2

2r+1
1  1

2
- xlog 2
2 sin( rx)
2
2Σ
r=1 r


to this,
 2 log 2 -1 -Σ

-
 2 log 2 -1 -Σ



r=1
22rB2r
2r (2r +1)!

 2
=
2r+1

2
log 2
Therefore,

x(log x -1) -Σ
r=1
22rB2r 2r+1
1  1
= - Σ 2 sin(2rx) - x log 2
x
2 r=1 r
2r(2r +1)!
From this,

Σ
r=1
Replacing
22rB2rx 2r
1  1
= log(2 x) - 1 +
2
2 sin( rx)
2r(2r +1)!
2x Σ
r=1 r
x
with
x /2 ,
B2rx

2r
Σ
r=1 2r(2r +1)!
= log x - 1 +
1  sin(rx)
xΣ
r=1
r2
Lemma 3.4.2t
When
for
B0=1, B2=1/6, B4=-1/30, B6=1/42, 
are Bernoulli numbers, the following expressions hold
x   .

Σ
r=1
22r-1B2rx 2r
=
2r(2r)!

Σ
r=1
(-1)r
r
cosrx + log 2
22r-1B2rx 2r
1  (-1)r
= Σ 2 sin rx + log 2
Σ
2r(2r +1)!
x r=1 r
r=1
Especially, when x =  ,


Σ
r=1
2r
22r-1B2r
= log 2
2r(2r +1)!
Proof
The 1st order integrals of Taylor series and Fourier series of

r=1
0
x
r-1

0
0
were as follows. (See " 5 Termwise Higher Integral ")
22r22r-1B2rx 2r
2r(2r)!
 tan x dx = Σ
1
(-1)
 tan x dx = 2 Σ r
x
tan x
r=1
1

sin 2r x -
|x|<
1
2
 + log 2
From these,

Σ
r=1
22r22r-1B2rx 2r
 (-1)r-1
= -Σ
cos(2rx) + log 2
2r(2r)!
r
r=1
- 15 -
|x|<

2

2
Replacing

Σ
r=1
x
with
2r
x /2 ,
2 -1B2rx 2r
 (-1)r
=Σ
cosrx + log 2
2r(2r)!
r
r=1
Next, the 2nd order integrals of Taylor series and Fourier series of
2r
2r
x x
 2 2 -1B 2r
2
2r+1
 tan x dx
 tan x dx
=Σ
r=1
0 0
x x
2
0 0
=
2r(2r +1)!
tan x
were as follows.
x
2
1  (-1)r-1
2
sin
r
x
Σ
2
21 r=1 r 2

 + x log 2
From these,

Σ
r=1
Replacing

Σ
r=1
22r22r-1B2rx 2r
1  (-1)r-1
=sin(2rx) + log 2
2x Σ
2r(2r +1)!
r=1
r2
x
with
2r
x /2 ,
2 -1B2rx 2r
1  (-1)r
= Σ 2 sin rx + log 2
2r(2r +1)!
x r=1 r
2012.04.08
2012.12.05 Renewal
K. Kono
Alien's Mathematics
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