2 Formulas for Riemann Zeta at natural number
Formulas for Zeta at natural 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.
2.1 Formulas of coth x family
Formulas for Zeta at natural number are obtained from coth x family. These are simple and the convergence
speed is fast.
Formula 2.1.1
When
B0=1, B2=1/6, B4=-1/30, B6=1/42,  are Bernoulli numbers and n
number, the following expression holds for
x n-1
 (n) =
(n -1)!
Especially when

1
x
n -1 2n
0 < x  2
 n -1
 +ΣΣ
r=1 s=0
is two or more natural
.
s
-n

1
(rx)
-Σ
n
rx
s! r e
r=1 2r -1


B2r x n-1+2r
2r(n -1+2r)!
x = 1, 2 ,
s
-n
 n -1 r

1
n +1
+ΣΣ
 (n)=
Σ
2 n!(n -1) r=1 s=0 s ! r ne r r=1 2r -1


B2r
2r(n -1+2r)!
s
-n
 n -1 (2r)

1
2n-1
+ΣΣ
 (n) =
Σ
n!(n -1) r=1 s=0 s! r ne 2r r=1 2r -1


n-1+2r
B2r 2
2r(n -1+2r)!
Proof
We obtained the following Riemann Zeta in Formula 1.1.3 in " 1 Zeta Generating Functions ".
B2r
1 x 1  e -xr 
0 =
+
+
log x x 2r
1
2r(2r)!
2 1!
Σ
r=1
r
Σ
r=1
(1.1)
2r+1
1 x 2  e -xr  B2r x
x1
+
-Σ
 (2)= log x -H1 +
2
1! 
2 2! Σ
r=1 r
r =1 2r(2r +1)!
 (3)=
2r+2
1 x 3  e -xr  B2r x
x1
x2
+
+
+
 (2)
log
x
H
Σ
2 2 3! Σ
2
1!
2! 
r=1 r
r =1 2r(2r +2)!
2r+3
1 x 4  e -xr  B2r x
x2
x3
x1
+
+
2
+
 (4)= 
 (3)
(
)
log x H3
Σ
4
2 4! Σ
2!
3! 
1!
r=1 r
r=1 2r(2r +3)!
2r+4
5
3
2
1
 e -xr

B2r x
x
x
x
1 x
x4
log x -H4(5)=
(2)(3)+
(4)
+
+
+
5 Σ 2r(2r +4)!
2 5! Σ
3!
4! 
2!
1!
r=1 r
r=1

For the sake of simplicity, we make the following substitutions.
Zn =  (n)
B2r x 2r+n-1
(-1)n x n  e -rx
(-1)n-1x n-1
n-1 
+
+(-1) Σ
fn =
log x -Hn-1+
n
2 n! Σ
(n -1)! 
r=1 r
r=1 2r(2r +n -1)!
Then,
Z2 = f2
-1-
x1
Z3 = f3+
Z
1! 2
x2
x1
Z4 = f4Z +
Z
2! 2 1! 3
x3
x2
x1
Z5 = f5+
Z Z +
Z
3! 2 2! 3 1! 4

k
Substituting
for the below one by one ,
1 0
x f
0! 2
1 1
1 0
Z3 =
x f3+
x f
1! 2
0!
1 1
1
1
1 0
+
Z4 =
x f4+
x f3+ x 2f2
1!
2! 1!1!
0!
Z2 =

1

1

1

1

1
1
1
1

x 1f4+ x 2f3+
x 3f2
+
+
Z5 = 0! x 0f5+
1!
2! 1!1!
3! 2!1! 1!2! 1!1!1!

n -2
Zn = ΣCs x s fn-s
s=0
Where,
Cs
are rational numberes as follows.
1
1
+
2! 1!1!
1
1
1
1
C3 =
+
3! 2!1! 1!2! 1!1!1!
1
1
1
1
1
1
1
1
C4 = +
+
+
+
+
+
4! 3!1! 1!3! 2!2!
1!1!1!1!
2!1!1! 1!2!1! 1!1!2!
1
1
1
1
1
C5 =
+
+
+
5!
4!1! 1!4! 3!2! 2!3!
1
1
1
1
1
1
+
+
+
+
+
+
3!1!1! 1!3!1! 1!1!3! 2!2!1! 2!1!2! 1!2!2!
1
1
1
1
1
+
+
+
+
1!1!1!1!1!
2!1!1!1! 1!2!1!1! 1!1!2!1! 1!1!1!2!
C2 = -









Although these are combinations of going mad, when these are actually calculated , it is as follows.
C1 =
1
1
1
1
, C2 =
, C3 =
,  , Cs =
1!
2!
3!
s!
Then,
n -2
Zn = Σ
s=0
xs
f
s ! n-s
Returning to the original symbol ,
n -2
 (n) = Σ
s=0
xs
s!

(-1)n-s x n-s
(-1)n-1-sx n-1-s
log x -Hn-1-s+
2
(n -1-s)! 
(n -s)!
-2-
e -xr

+Σ
r
r=1
n-s
B2r x 2r+n-1-s
2r(2r +n -1-s)!
n-1-s 
Σ
r=1
+(-1)

Here, the n -1 th term must be as follows.
n-1
x
(n -1)!


B x
+Σ
2r(2r)! 
B2r x 2r
(-1)0x 0
(-1)1 x 1  e -xr
0 
+
+(-1) Σ
log x -H0+
1
0! 
2 1! Σ
r=1 r
r=1 2r(2r)!

n-1
-xr
 e
1 x1
x
=
+
log x 1
2 1! Σ
(n -1)!
r=1 r

2r
2r
r=1
However, since the content in the { } of the right side is 0 from (1.1) , the n -1 th term have to be 0 .

can be n -1 . Thus,

(-1)n-s x n-s
(-1)n-1-sx n-1-s
log x -Hn-1-s+
2
(n -1-s)! 
(n -s)!
That is, the upper limit of
n -1
 (n) = Σ
s=0
s
x
s!

+Σ
e -xr
n-s
n-1
= (-1)

+Σ
r=1
B2r x 2r+n-1-s
2r(2r +n -1-s)!
 s  log x -H 
n
(xr) 1
x
+ΣΣ
(-1)
Σ
s 
n!
s! r e
n -1
x
(-1)s
Σ
(n -1)! s=0
(-1)n
+
2
Σ
r=1
+(-1)
r
n -1
r=1
n-1
n-1-s 
n n -1
n-s-1
s
 n -1
s
s=0
n xr
r=1 s=0
n -1+2r
(-1)n-1 n -1
(-1)s
Σ
(n -1+2r)! s=0
s


B2r x 2r+n-1
2r
Here, according to " 岩波 数学公式Ⅱ " p11,
m
Σ(-1)s
s=0
s 
n
= (-1)m
m
n -1
m  n -1
Applying this,
s 
  = (-1)
n -1+2r
n -1+2r -1
Σ(-1)  s  = (-1)  n -1 
n
n -1
Σ(-1)s
= (-1)n-1
s=0
n -1
n -1
n -1
n-1
n-1
s
s=0
Furthermore,

p -1+ r
p -1

= (-1)r
r
-p
Applying this,

n -1+2r -1
n -1

= (-1)2r-1

-n
2r -1

Then,
n -1
(-1)s
Σ
s=0

n -1+2r
s

= (-1)n-1+2r-1

-n
2r -1
Substituting these (blue) for the above,
-3-

= (-1)n

-n
2r -1


n -1
1
(-x)n-1 n -1
s
-1
 (n)=
(
)
log x -Hn-1-s Σ

2
(n -1)! s=0
s
s
-n
 n -1 (rx )

1
+ΣΣ
Σ
2r -1
r=1 s=0 s !
r=1
r ne rx
 

Next, when
xn
n!

B2r x n-1+2r+
2r (n -1+2r)!
n >1 ,
n -1
Σ(-1)s
s=0
 s  log x = 0
n -1
Furthermore, according to my work " 04 Higher Integral " 4.6.2 ,
n-1
n -1
(-1)s
Σ
s=0
s 
n
Hn-s =
(-1)
n
Hr =(1+ r)+ 
Usint this,
n -1
(-1)
Σ
s=0
s
 s H
n -1
n-1-s
(-1)n-2
=
n -1
(-1)n-1
=n -1
Substituting these (blue) for the above, we obtain
B2r x n-1+2r
2r(n -1+2r)!
s
-n
 n -1 (xr )

1
xn
x n-1
 (n) =
+ΣΣ
(n -1)!(n -1) 2 n! r=1 s=0 s! r ne xr Σ
r=1 2r -1


i.e.
x n-1
 (n) =
(n -1)!

1
x
n -1 2n
 n -1
 +ΣΣ
r=1 s=0
-n

(xr) s 1
Σ
s! r ne xr r=1 2r -1


B2r x n-1+2r
2r(n -1+2r)!
Example 1

re
1
6
r
r
r
r
(5)=
+Σ 1+
+
+
+
1! 2! 3! 4!  r e
25!4
1

(xr)
(xr)
x3
x2
(3)=
+Σ 1+
+
1!
2!
2!2 23! r=1
2
2+2r
 2r -1
-5
B
-Σ
 2r -1 2r(4+2r)!
-5
B 2
1
2
2r (2r) (2r) (2r)
+
+
+
(5)=
+Σ1+
Σ  2r -1 2r(4+2r)!
1!
2!
3!
4!  r e
5!4

1
2
3
1
3 xr
4
5 r
r=1
4

2
3

-Σ
r=1
B2r x
0<x 2
2r(2+2r)!

2r
r=1
4

5 2r
r=1
-3
4+2r
2r
r=1
Example 2 (7)
According to the formula at x=2 this is calculated. As the result of calculating the series to the 8 th term,
the significant 10 digits were obtained.
Note
In this formula, the convergence is the fastst around x=2.
Formula 2.1.1'
When
B0=1, B2=1/6, B4=-1/30, B6=1/42,  are Bernoulli numbers and n
-4-
is two or more natural
number, the following expression holds for
 (n)=
n-1

x
(n -1)!
0 < x  2
1
(n -1)x
+
- log x
2n
n -1

1-n

(xr)s 1
Σ
s ! r ne xr r=1 2r
 
 n -2
+ ΣΣ
r=1 s=0
Especially when
.
n-1+2r
B2r x
2r(n -1+2r)!
x =1 ,
s
1-n
 n -2 r

1
n 2+1
+ ΣΣ
 (n) =
Σ
r
2 n!(n -1)
2r
r=1 s=0 s ! r n e
r=1
 
B2r
2r(n -1+2r)!
Proof
We start with the middle of the previous proof .
n-1-s n-1-s
s
n -2
 (n) = Σ
s=0
x
s!

(-1)
x
(n -1-s)!
(-1)n-s x n-s
log x -Hn-1-s+
2
(n -s)!

+Σ
r=1
n-1
= (-1)
(-1)n
+
2

+Σ
r=1
n-1-s 
n n -2
B2r x 2r+n-1-s
2r(2r +n -1-s)!
+(-1)
Σ
n-s
r=1
r
n -1
log x -Hn-1-s
s
 
n
(xr )
x
+ΣΣ
-1
(
)
Σ
s 
n!
s!
n -2
x
(-1)s
(n -1)! Σ
s=0
n-1
e -xr
s
 n -2
s
s=0
r=1 s=0
n -1+2r
(-1)n-1 n -2
s
-1
(
)
(n -1+2r)! Σ
s=0
s


1
n xr
r e
n-1-2r
B2r x
2r
Here, according to " 岩波 数学公式Ⅱ " p11,
m
(-1)s
Σ
s=0
s 
n
= (-1)m
m
n -1
m  n -1
Applying this,
s 
 
n -1+2r
-1
(
)
Σ
 s  = (-1) 
n
n -2
(-1)s
Σ
s=0
n -2
= (-1)n-2
n -1
n -2
n-2
s
s=0
 

n -1
= (-1)n(n -1)
1
n -1+2r -1
n -1+2r -1
= (-1)n
n -2
n -2
= (-1)n

Furthermore.

p -1+ r
p -1

= (-1)r
r
-p
Applying this,

n -1+2r -1
n -2
 
=
n -1-1+2r
n -1-1
= (-1)2r


n -1+2r -1
n -2

-(n -1)
2r
  
=
1-n
2r
Then,
n -2
Σ(-1)s
s=0

n -1+2r
s

= (-1)n

-5-
= (-1)n
 
1-n
2r


Substituting these (blue) for the above,
n-1
n -2
n-1 x
 (n) = (-1)
(n -1)!
 n -2
+ΣΣ
r=1 s=0
Next, when
Σ(-1)s
s=0
s
n -1
log x -Hn-1-s +
1-n

(xr )s 1
(-1)n-1
n
-1
+
(
)
s ! r ne xr Σ
2r
r=1 (n -1+2r)!
 
(n -1) x n
2
n!
n-1-2r
B2r x
2r
n >1 ,
n -2
Σ(-1)s
s=0
 s  = (-1)
n -1
n
Then,
n -2
Σ(-1)s
s=0
 s  log x = (-1) log x
n -1
n
Furthermore, according to my work " 04 Higher Integral " 4.6.2 ,
n-1
n -1
(-1)s
Σ
s=0
s 
n
Hn-s =
(-1)
n
Hr =(1+ r)+ 
Usint this,
n -2
(-1)
Σ
s=0
s
 s H
n -1
n-1-s
(-1)n-2
(-1)n
=
=
n -1
n -1
Then,
n-1
(-1)
n-1
n -2
n -1
x
s
-1
(
)
log x -Hn-1-s
(n -1)! Σ
s=0
s
 
n-1
(-1)n
x
n
(-1) log x (n -1)!
n -1

n-1
= (-1)
n-1
x
=
(n -1)!

1
- log x
n -1


Thus,
n-1
x
 (n) =
(n -1)!

1
- log x
n -1

 n -2
+ ΣΣ
r=1 s=0
(n -1)x n
+
2n!
1-n

(xr)s 1
s ! r ne xr Σ
2r
r=1
 
n-1+2r
B2r x
2r(n -1+2r)!
i.e.
n-1
x
 (n) =
(n -1)!

1
(n -1)x
+
- log x
2n
n -1
 n -2
+ ΣΣ
r=1 s=0

1-n

(xr)s 1
s ! r ne xr Σ
2r
r=1
 
n-1+2r
B2r x
2r(n -1+2r)!
Example 1
2
x
 (3)= 2!

3

xr
1
2x
- log x +
+Σ 1+
1!
2
23! r=1


r e
-6-
1
3 xr

-Σ
r=1
 2r 
-2
B2r x 2+2r
2r(2+2r)!
0<x 2

52+1 
r1 r2 r3
+
1+
+
+
 (5)=
1! 2! 3!
25!4 Σ
r=1
re
1
5 r

-Σ
r=1
 2r 
-4
B2r
2r(4+2r)!
Example 2 (7)
When x=2, this is calculated according to the formula. As the result of calculating the series to the 7 th term,
the significant 10 digits were obtained.
-7-
2.2 Formula of tanh x family
Formula for Zeta at natural number is obtained from tanh x family. This is simple and the convergence speed
is fast.
Formula 2.2.1
When
B0=1, B2=1/6, B4=-1/30, B6=1/42,  are Bernoulli numbers and n
number, the following expression holds for
(n) =
2n-1
2n-1 - 1
Especially when x = 1 ,
 (n) =

2n-1
2n-1 -1
0< x 
s
 n -1 (xr )
xn
-ΣΣ
2n! r=1 s=0 s!

is two or more natural
.
(-1)r
r ne xr

+Σ
r=1
 2r -1
-n

n-1+2r

22r-1B2r
2r(n -1+2r)!

2r
2 -1B2r x
2r(n -1+2r)!
-n
 n -1 r s (-1)r

1
- ΣΣ
+
Σ
2n!
2r -1
r=1 s=0 s!
r=1
r ne r

Proof
We obtained the following Dirichlet Eeta in Formula1.2.3 in 1.2 .
2r
2r
-rx
 2 -1B 2r x
1 x1 
r e
(1)=
- (-1) 1 2 1!
2r(2r)!
Σ
r=1
r
Σ
r=1
2r
2r+1
-rx
 2 -1B 2r x
1 x2 
x1
r e
(2)= (1)
- (-1) 2 +Σ
+
1!
2 2! Σ
2r(2r +1)!
r=1
r=1
r
(3)=
2r
2r+2
-rx
 2 -1B 2r x
1 x3 
x2
x1
r e
- (-1) 3 -Σ
(1)+
(2)
2!
2 3! Σ
1!
2r(2r +2)!
r=1
r=1
r
2r+3
2r
-rx
 2 -1B2r x
1 x4 
x2
x1
x3
r e
(4)= - (-1) 4 +Σ
+
(1)(2)+
(3)
2!
1!
2 4! Σ
3!
2r(2r + 3) !
r=1
r=1
r

2r+n-1
2r
-rx
n -1 (-1)sx s

 2 -1B2r x
(-1)n x n
r e
n
(n)= - (-1)
+(-1) Σ
-Σ
(n -s)
2 n! Σ
s!
2r(2r + n -1) !
r=1
r=1
s=1
rn
Substituting (k ) for the below one by one in a similar way as the proof of Formula 2.1.1 , we obtain
2r+n-1
2r
 n -1
Cn x n  n -1 x s (-1)r
2 -1B2r x
(-1)s
n
(n)=
-ΣΣ
+(-1) ΣΣ
n-s
2
2r
r=1 s=0 s! r
r=1 s=0 s!(n -1+2r -s )!
e rx
Where,
Cs is the same as the coefficient in the Formula 2.1.1 and is given by Cs =
1
. Then ,
s!
2r
 n -1
 n -1
2 -1B2r x
(-1)s
x s (-1)r
xn
n
(n)=
-ΣΣ
+(
-1
)
Σ
Σ
2n ! r=1 s=0 s! r n-s e rx
2r
r=1 s=0 s!(n -1+2r -s )!

n -1+2r
 n -1
 n -1
x s (-1)r
xn
n
s
-1
=
-ΣΣ
+(
-1
(
)
)
Σ
Σ
2n ! r=1 s=0 s! r n-s e rx
r=1 s=0
s
Here, as seen in the proof of Formula 2.1.1 ,
n -1
Σ(-1)s
s=0

n -1+2r
s

= (-1)n

-n
2r -1

-8-

2r+n-1
2r+n-1
2r
2 -1B2r x
2r(n -1+2r)!
Applying this,

s
-n
 n -1 (xr )

(-1)r
xn
(n) =
- ΣΣ
+
Σ
s! r n e xr
2n !
2r -1
r=1 s=0
r=1
And using
(n) =
2n-1
2n-1 - 1
(n)

2r
2 -1B2r x
n-1+2r
2r(n -1+2r)!
, we obtain the desired exression.
Example 1

 r e +Σ  2r -1


2 -1 B
r
r
r
r
1
2
(-1)
(5)=
-Σ 1+
+
+
+
+
Σ
  2r(4+2r)! 
1! 2! 3! 4!  r e
2 - 1  25!
(3)=
22
22-1
4
4

xr x 2r 2
x3
-Σ 1+
+
1!
2!
23! r=1

1
(-1)r
2

3 xr
3
2+2r
2r
2 -1B2r x
2r(2+2r)!
-3
r=1
4
r
5 r
r =1

-5
r=1
2r -1

2r
Example 2 (7)
When x=3/2, this is calculated according to the formula. As the result of calculating the series to
the 10 th term, the significant 10 digits were obtained.
Note
In this formula, the convergence is the fastst around x =3/2 .
-9-

2r
2.3 Formulas of csch x family
Formulas for Zeta at natural number are obtained from csch x family. These are simple and the convergence
speed is the fastest.
Formula 2.3.1
When
B0=1, B2=1/6, B4=-1/30, B6=1/42,  are Bernoulli numbers and n
number, the following expression holds for
2n
 (n) = n
2 -1

0< x 
.
n-1
-x(2r-1)
s
 n -1 x(2r -1) e
x
+ Σ
2(n -1)!(n -1) Σ
s!
r=1 s=0
(2r -1)n

-n
1 
+ Σ
2 r=1 2r -1
Especially when
is two or more natural

n-1+2r
22r-2B2r x
2r(n -1+2r)!

x =1 ,
2n
 (n) = n
2 -1

s
 n -1 (2r -1)
1
e -(2r-1)
+ Σ
2(n -1)!(n -1) Σ
s!
r=1 s=0
(2r -1)n

-n
1 
+ Σ
2 r=1 2r -1

22r-2B2r
2r(n -1+2r)!

Proof
We obtained the following Dirichlet Eeta in Formula1.3.3 in 1.3 .
0
2r
2r
-(2r -1)x
 e
1 x0
1  2 -2B2r x
x
- Σ
log -H0 +Σ
1
2
2r(2r)!
2 0!
2 r=1
r=1 (2r -1)

=

(1.1)
2r
2r+1
-(2r-1)x
 e
1 x1
1  2 -2B2r x
x
(2)= + Σ
log -H1 +Σ
2
2 1!
2 r =1
2r(2r +1)!
2
r=1 (2r -1)


2r+2
2r
)
 e -(2r -1 x
x1
1 x2
1  2 -2B2r x
x
log -H2 +Σ
- Σ
+
(2)
3
2 2!
2 r =1 2r(2r +2) !
1!
2
r=1 (2r -1)

(3)=

2r+3
2r
)
 e -(2r -1 x
x1
x2
1 x3
1  2 -2B2r x
x
(4)= log -H3 +Σ
+ Σ
(2)+ (3)
4
1!
2 3!
2 r =1 2r(2r +3) !
2!
2
r=1 (2r -1)



Substituting
(k )
for the below one by one in a similar way as the proof of Formula 2.1.1 , we obtain
n-1-s n-1-s
-(2r-1)x
s
n -2
x
(n) = Σ
s=0 s !


e
x
log 2 -Hn-1-s +Σ
n-s
r=1 (2r -1)
(-1)
x
2(n -1-s)!


2r
2r+n-1-s
(-1)n-1-s  2 -2B2r x
Σ
2
2r(2r +n -1-s)!
r=1

Here, the n -1 th term must be as follows.
n-1
x
(n -1)!

2r
2r
-(2r-1)x
 e
(-1)0  2 -2B2r x
(-1)0x 0
x
log 2 -H0 +Σ
1
2 Σ
2r(2r)!
20!
r=1 (2r -1)
r=1


However, since the content in the { } of the right side is 0 from (1.1) , the n -1 th term have to be 0 .
That is, the upper limit of

can be n -1. Thus,
- 10 -

n -1
(n) = Σ
s=0

xs
s!
x

(-1)n-1-sx n-1-s
e -(2r-1)
x
log 2 -Hn-1-s +Σ
n-s
2(n -1-s)!
r=1 (2r -1)


2r
2r+n-1-s
(-1)n-1-s  2 -2B2r x
Σ
2r(2r +n -1-s)!
2
r=1
n-1
n -1
n -1
x
x
s
-1
(
)
log 2 -Hn-1-s
Σ
(n -1)! s=0
s
(-1)n-1
=
2
 
 n -1
s
x(2r -1)
r=1 s=0
s!
+ ΣΣ


e -x(2r-1)
(2r -1)n
n -1+2r
1  (-1)n-1 n -1
s
- Σ
-1
(
)
2 r=1 (n -1+2r)! Σ
s=0
s


n-1+2r
22r-2B2r x
2r
Here, as seen in the proof of Formula 2.1.1 ,
s
n -1
(-1)
=Σ(-1)  s  H
n -1
-n
n -1+2r
=
-1
-1
(
)
(
)
Σ
 s 
2r -1
n -1
(-1)s
Σ
s=0
n -1
x
log 2 = 0
n-1
n -1
s
n-1-s
s=0
n -1
s
n
s=0
Substituting these for the above,
n-1
-x(2r-1)
s
 n -1 x(2r -1) e
x
+ Σ
(n) =
2(n -1)!(n -1) Σ
s!
r=1 s=0
(2r -1)n

-n
1 
+ Σ
2 r=1 2r -1
And using
(n) =
2n
(n )
2n - 1

n-1+2r
22r-2B2r x
2r(n -1+2r)!
, we obtain the desired exression.
Note
In this formula, the convergence is the fastst around x=1. We had better not think other than this.
Example 1

2r -1 (2r -1)
8 1 
(3) = 7
+Σ 1+
+
1!
2!
8 r=1

2

-(2r -1)

-3
1 
+
Σ
3
2 r=1 2r -1
(2r -1)
e

2r
2 -2B2r
2r(2+2r)!

Example 2 (7)
According to the formula at x=1 this is calculated. As the result of calculating the series to the 6 th term,
the significant 10 digits were obtained.
- 11 -
Formula 2.3.1'
When
B0=1, B2=1/6, B4=-1/30, B6=1/42,  are Bernoulli numbers and n
0 < x 
number, the following expression holds for

2n
2n-1
 (n) =
n-1
x
2(n -1)!

.
1
x
-log
2
n -1

 n -2
s
x(2r -1)
r=1 s=0
s!
+ΣΣ
e -x(2r-1)
(2r -1)n
n-1+2r
22r-2B2r x
2r(n -1+2r)!
 
1  1-n
+ Σ
2 r=1 2r
Especially when
is two or more natural

x = 1, 2 ,
2n
 (n) = n
2 -1

1
2(n -1)!

 +ΣΣ
 n -2
1
+log 2
n -1
r=1 s=0
(2r -1)s e -(2r-1)
s!
(2r -1)n
 
1  1-n
+ Σ
2 r=1 2r
 (n) =

2n
2n-1
22r-2B2r
2r(n -1+2r)!
s
 n -2 (4r -2)
2n-2
e -(4r-2)
+ΣΣ
(n -1)!(n -1) r=1 s=0
s!
(2r -1)n
 
1  1-n
+ Σ
2 r=1 2r
n-1+2r
22r-2B2r 2
2r(n -1+2r)!

Proof
We start with the middle of the previous proof .
n-1-s n-1-s
s
n -2
(n) = Σ
s=0
x
s!

(-1)
x
2(n -1-s)!


2r
2r+n-1-s
(-1)n-1-s  2 -2B2r x
Σ
2r(2r +n -1-s)!
2
r=1
n-1
n -2
n -1
x
x
(-1)s
log 2 -Hn-1-s
Σ
(n -1)! s=0
s
(-1)n-1
=
2
 
 n -2
s
x(2r -1)
r=1 s=0
s!
+ ΣΣ

Here, as seen in the proof of Formula 2.1.1' ,
Σ(-1)s
s=0

n -1+2r
s

= (-1)n


e -x(2r-1)
(2r -1)n
n -1+2r
1  (-1)n-1 n -2
- Σ
(-1)s
Σ
2 r=1 (n -1+2r)! s=0
s
n -2
x

e -(2r-1)
x
+
log
Hn-1-s Σ
n-s
2
r=1 (2r -1)
 
1-n
2r
- 12 -

n-1+2r
22r-2B2r x
2r

s
n -1
Σ(-1)  s  H
n -2
(-1)s
Σ
s=0
n -2
n -1
log
x
x
= (-1)n log
2
2
s
s=0
n-1-s
Substituting these for the above,
n-1
x
(n) =
2(n -1)!

(-1)n
=
n -1
1
x
-log
2
n -1
 n -2
s
x(2r -1)
r=1 s=0
s!
 +ΣΣ
 
1  1-n
+ Σ
2 r=1 2r
And using
(n) =
2n
(n ) ,
2n - 1
e -x(2r-1)
(2r -1)n
n-1+2r
22r-2B2r x
2r(n -1+2r)!
we obtain the desired exression.
Note
When x=1, the convergence speed of this formula is the fastest in this chapter.
Example 1
(3)=
8
7
8
(3)=
7
 
1
4


2r -1
1
+log 2 +Σ 1+
1!
2
r=1


1 
4r -2
+Σ 1+
2 r=1
1!


-(2r -1)
 (2r -1)
)
e -(4r -2
e
+
3
 
1  -2
+
3
2Σ
2r
r=1
(2r -1)
 
1  -2
2Σ
2r
r=1
2r
2 -2B2r
2r(2+2r)!
2r
2+2r
2 -2B2r 2
2r(2+2r)!


Example 2 (7)
According to the formula at x=1 this is calculated. As the result of calculating the series to the 6 th term,
the significant 10 digits were obtained.
- 13 -
2.4 Formulas of trigonometric function family
The following formulas for zeta at natural number are obtained also from each family of cot x, tan and csc x.
(1) For a complex number
x = u + v i s.t. 0 < |x|  2 , v 0
(-ix)n-1
 (n) =
(n -1)!

1
ix
+
n -1 2n
(-irx)s e i rx
s!
rn
 + ΣΣ
 n -1
r=1 s=0
n-1 
Σ
r =1
+ (-ix)
(-ix)n-1
 (n)=
(n -1)!

1
(n -1)ix
i
-log x +
2
2n
n -1

(2) For a complex number
 (n) =
 n -2
r=1 s=0
2
2n-1 -1
B2rx
2r
22r-1B2rx 2r
2r(n -1+2r)!

  2r(n -1+2r)!
1-n
2r

(-ix)n  n -1 (-irx)s (-1)re i rx
- Σ
2 n! Σ
s!
r=1 s=0
rn
n-1 
-(-ix)


Σ
r=1

-n
2r -1
(
)
 n -1 -(2r -1) i xs e 2r -1 ix
(-ix)n-1
+ Σ
n
2(n -1)!(n -1) Σ
s!
r=1 s=0
(2r -1)
-n
(-ix)n-1 
Σ
2
r =1 2r -1

2n
 (n) = n
2 -1

(-ix)n-1
2(n -1)!
 n -2
s
-(2r -1) i x
r=1 s=0
s!
+ΣΣ
(-irx)s e i rx
s!
rn
x = u + v i s.t. 0 < |x|   , v 0
n-1
2n
 (n) = n
2 -1
Σ
r =1
2r
 2r(n -1+2r)!
 +ΣΣ
n-1 
+ (-ix)
B2rx
-n
2r -1

1
x i
-log +
2 2
n -1

 
(2r -1)ix


(-ix)n-1  1-n
n
2 Σ
2r
r =1
(2r -1)
e
22r-2B2rx 2r
2r(n -1+2r)!
2r
2r
2 -2B2rx
2r(n -1+2r)!

These were drawn from Formula 1.4.4, 1.5.4 and 1.6.4 in " 1 Zeta Generating Functions " in a way similar
to previous 3 sections.
However, these are the same as the formulas in previous 3 sections. That is, By replacing
reduce to the previous 3 formulas. Conversely, by replacing
x
with
-ix
x
with
ix
, these
, the formulas in previous 3 sections
reduce to these. Therefore, these were described only as reference.
2012.04.07
K. Kono
Alien's Mathematics
- 14 -