Download 05 Termwise Higher Integral (Trigonometric

05 Termwise Higher Integral (Trigonometric, Hyperbolic)
In this chapter, for the function which second or more order integral cannot be expressed with the elementary
functions among trigonometric functions and hyperbolic functions, we integrate the series expansion of these
function termwise. Therefore, sin x, cos x, sinhx, cosh x mentioned in " 4.3 Higher Integral of Elementary
Functions " are not treated here.
5.1 Termwise Higher Integral of tan x
Since both zeros of
tan x
and the primitive function
-log cos x
are
x =0 , the latter seems to be a lineal
primitive function with a fixed lower limit. Then, we assume zeros of the second or more order primitive function
x =0 .
are also
5.1.0 Higher Integral of tan x
 tan x dx
 tan x dx
x
= -log cos x
0
x x
2
=
non-elementary function
0 0
5.1.1 Termwise Higher Integral of Taylor Series of tan x
Formula 5.1.1
When
B0=1, B2=1/6, B4=-1/30, B6=1/42, 
hold for |x|<
 /2 .
 tan x dx
=Σ
k =1
0
 tan x dx
x x
=Σ
k =1
 tan x dx
x x x
3

=Σ
k=1

  tan x dx
x
0
0
22k22k -1B2k 2k+1
x
2k (2k +1)!

2
0 0
x
22k22k -1B2k 2k
x
2k (2k)!

x
0 0 0
are Bernoulli Numbers, the following expressions
n

=Σ
k =1
22k22k -1B2k 2k+2
x
2k (2k +2)!
22k22k -1B2k 2k+ n-1
x
2k (2k + n -1)!
Proof
tan x
can be expanded to Taylor series as follows.

tan x = Σ
k=1
22k22k -1B2k 2k-1
x
2k (2k -1)!
|x| <

2
Integrating both sides of this with respect to x from 0 to x repeatedly, we obtain the desired expression.
(1) Termwise Higher Definite Integral of Taylor Series of tan x
Substituting
x = /2
for Formula 5.1.1 , we obtain the following expressions.

  tan x dx
2 x
0
0
2

=Σ
k=1
22k22k -1B2k
2k (2k +1)!
-1-

 2
2k+1
(1.2')

  tan x dx
2 x x
0

=Σ
k=1
0 0
  tan x dx
2
0

3
x
x
0
4
22k22k -1B2k
2k (2k +2)!

=Σ
k =1
0

 2
2k+2
(1.3')
22k22k -1B2k
2k (2k +3)!

 2
2k+3
22k22k -1B2k
2k (2k + n -1)!

2k+ n-1
(1.4')


  tan x dx
2
0
x
x
0
n
0

=Σ
k =1
 2
(1.n')
(2) Taylor Series of the 1st order Integral of tan x
The1st order Integral of tan x is as follows.
2k
2k
x
 2 2 -1B2k

tan x dx = Σ
2k (2k)!
k=1
0
x 2k = -log cos x
From this, the following expression follows immediately.
2k
2k
 2 2 -1B 2k
2k
Σ
k =1
2k (2k)!
Substituting

Σ
k =1
= -log cos x
x
x =1 , 1/2 ,  /4

(1.t)
2
22k -1B2k
Σ
2k (2k)!
k =1
= -log cos 1

2k
2 -1B2k
Σ
2k (2k)!
k =1
2
for this one by one, we obtain the following special values.
22k22k -1B2k
2k (2k)!

|x|<

|x|<
= -log cos

 2
1
2
2k
= -log cos

This is used in 5.3.1 later.
4
5.1.2 Termwise Higher Integral of Fourier Series of tan x
Formula 5.1.2

Let
(x) =Σ
(-1)k-1
be a Dirichlet Eta Function and
kx
for |x|<  /2 .

be a ceiling function, then the following
k =1
expressions hold
1  (-1)k-1
1
2
= 0Σ
k
x
tan x dx
sin
2
0
2 k=1 k 1
x x
2
1  (-1)k-1
2
2
= 1Σ
k
x
tan x dx
sin
2
0 0
2 k=1 k 2
x x x
1  (-1)k-1
3
3
tan x dx = 2 Σ
sin 2k x 3
2
0 0 0
2 k=1 k


x


 
x
0
0


x
4
1  (-1)k-1
tan x dx = 3 Σ
sin 2k x 4
2
2 k=1 k
4

-2-
+


(1) x 0
20
(1)
+ 0
2
(1)
+ 0
2
+
0!
x1
1!
x 2 (3) x 0
- 2
2!
0!
2
(1) x 3
20
3!
-
(3) x 1
22
1!

  tan x dx
x
0
x
n
=
0
1

Σ
(-1)k-1
2n-1 k=1
k
n

sin 2k x -
n
2
n/2

k-1
+ Σ (-1)
(2k -1)
22k-2
k =1
n+1-2k
x
(n +1-2k)!
Proof
tan x can be expanded to Fourier series in a broad meaning.as follows.
1 e ix - e -ix
1- e 2ix
=i
= i1- e 2ix1 tan x =
-ix
2ix
i ix
e +e
1+ e
e 2ix + e 4ix - e 6ix +-
= i - i2e 2ix - 2e 4ix + 2e 6ix - 2e 8ix +-
= 2(sin 2x - sin 4x + sin 6x - sin 8x +-)
-2i(cos 2x - cos 4x + cos 6x - cos 8x +-) + i
(2.0)
y
40
30
20
10
-1.4 -1.2 -1.0 -0.8 -0.6 -0.4 -0.2
-10
0.2
0.4
0.6
0.8
1.0
1.2
1.4
x
-20
-30
-40
tan(x)
2*sin(2*x) - 2*sin(4*x) + 2*sin(6*x) - 2*sin(8*x)
tan x is the locus of the median (central line) of this real number part. (See the figure) However calculus of the
right side of (2.0) may be carried out, the real number does not turn into an imaginary number and the contrary
does not exist either. Therefore, limiting the range of tan x and the higher order integral to the real number, we
calculate only the real number part of the right side of (2.0) . Although the Riemann zeta
(2n) is obtained
from calculation of the imaginary number part, since it is long, it is omitted in this section.
Now, integrate the real number part of both sides of (2.0) with respect to x from 0 to x , then

x
 2(sin 2x - sin 4x + sin 6x - sin 8x +-)dx
x
tan x dx =
0
0
==-



cos 2x cos 4x cos 6x cos 8x
+
+-
2
3
4
1

cos 2x cos 4x cos 6x cos 8x
+
+-
1
2
3
4
i.e.

0
x
k-1
 (-1)
1 
k-1 cos 2k x
+Σ
tan x dx = - 0 Σ(-1)
k=1
2 k=1
k1
k1
-3-
+
x
0
1 1 1 1
- + - +-
1 2 3 4
Here, let

0

-cos x = sin x x

2


(-1)k-1
k=1
kx
, (x)=Σ
1  (-1)k-1

tan x dx = 0 Σ
sin 2k x 1
2
2 k=1 k

, then we obtain
(1) x 0
+
This is consistent with the real number part of the Fourier series of
20
0!
-log cos x
.
Next, integrate both sides of this with respect to x from 0 to x , then
1  (-1)k-1

sin 2kx 0Σ
1
2
2 k =1
k
 tan x dx =  
x x
0 0
0
=

1  (-1)k-1
Σ
21 k=1
k1
1  (-1)
Σ
21 k =1
k1
-
20
0!
x
1
0
0
k-1
=
(1) x 0
 dx
2
(1) x
sin 2kx 2 
1! 
2

x
2

sin 2kx -
2
2

+
(1) x
20
1
1!
Next, integrate both sides of this with respect to x from 0 to x , then
 tan x dx =  2 Σ
x x x

1
3
2
0 0 0
=
=
Σ
22 k=1

1
22
k=1
k-1

1
(-1)k-1
Σ
k=1
(-1)
k
1
(-1)k-1
k
1
3
sin 2k x 2

+

3
2

+
(1) x

3
2

+
k
1
sin 2k x sin 2k x -
20
20
2
2!
(1) x 2
20
(1) x 2
2!
-
2!
1

x
0

(-1)k-1
2Σ
2 k=1
cos 2k0
k
3
(3) x 0
22
0!
Hereafter, in a similar way we obtain the desired expressions.
(1) Termwise Higher Definite Integral of Fourier Series of tan x
x = /2
Substituting
for Formula 5.1.2 , we obtain the following expressions.

  tan x dx
2 x
0
2
=
0

 
2 x x
0
tan x dx
3
=
0 0

  tan x dx
2
0
x
0
x
4
=
0


2
0
x

0
x
tan x dx =
0



2
0
5
x

0
0
x
(1)

201!
 2
1
(1)

202!
 2
2
(1)

-
203!
 2
3
(1)

4
204!
n /2
tan x dx = Σ
6
(2.2')
k =1
 2
-
-
(3)

0
220!
 2
(3)

221!
 2
1
(3)

2
222!
(-1)k-1 (2k -1)
22k-2(n +1-2k)!
-4-
-
 2

 2
 (3)
(2.3')
22
(2.4')
+
(5)
240!
n+1-2k
+
+
 (5)
 (n)
2n-1
24
sin
n
2
(2.5')
(2.n')
(2) Fourier Series of the1st order Integral of tan x
The1st order Integral of tan x is as follows.
x
1 
k-1
tan x dx = - 0 (-1)
0
2 k =1

Σ
cos 2k x
log 2 0
+
x = -log cos x
0!
k1
From this, the following expression follows immediately.

cos 2k x
= log(2cos x)
k1
k-1
(-1)
Σ
k =1
Substituting
x =1 , 1/2


(2.f)
2
for this one by one, we obtain the following special values.
cos 2k
= log(2cos 1)
k1
cosk
= log2cos(1/2)
k1
k-1
(-1)
Σ
k =1

|x|<
k-1
(-1)
Σ
k =1
5.1.3 Dirichlet Odd Eta
Comparing Taylor series of tan x and Fourier series of tan x, we obtain Dirichlet Odd Eta
Formula 5.1.3
When
B2k , (x),  (x) denote
Bernoulli Numbers, Dirichlet Eta Function and Riemann Zeta Function
respectively, the following expressions hold.
2k
 2 -1B 2k
(3) = -Σ

k=1 2k (2k +2)!
2k+2
+
2
2!
(1) -  (3)
22k -1B2k 2k+3
3
(3) = 
(1)
3!
 Σ
k=1 2k (2k +3)!
1



22k -1B2k 2k+4  4
2
(5) = Σ

(1) +
(3) -  (5)
4!
2!
k=1 2k (2k +4)!

(5) =
22k -1B2k 2k+ 5
5
3

(1)+
(3)
3!
5!
 Σ
k=1 2k (2k +5)!
1



2k
2 -1B2k 2k+ 6  6
4
2
+
(7) = -Σ

(1)(3)+
(5)-  (7)
6!
4!
2!
k=1 2k (2k +6)!

22k -1B2k 2k+7  7
5
3
(7) = 
(1)+
(3)(5)
5!
3!
7!
 Σ
k=1 2k (2k +7)!

2k
 2 -1B 2k 2k+2n
n
(2n +1) = (-1) Σ

k=1 2k (2k +2n )!
1



n
- (-1)
(2n +1) =
(-1)n

n -1
(-1)
Σ
k =0

Σ
k=1
k
 2n-2k
(2n -2k)!
(2k +1) -  (2n +1)
22k -1B2k 2k+2n +1

2k (2k +2n +1)!
-5-
(-1)n
-
n -1

(-1)
Σ
k=0
3
(1)
 2n+1-2k
k
(2n +1-2k)!
(2k +1)
Proof

   tan x dx
2 x
0
x
=
202!
0 0

   tan x dx
2 x
0
x

=Σ
3
k=1
0 0

 2
(3)
2
-
220!

 2
22k22k -1B2k
2k (2k +2)!

 (3)
0
 2
-
(2.3')
22
2k+2
(1.3')
from these
(1)

2!

 (3)
0
 2  - 2 0!  2  -
202!
(1)
(3)
2
2
 -(3) -
=Σ
k =1
22k22k -1B2k
2k (2k +2)!

 2
2k+2
22k -1B2k 2k+2
=Σ

k =1 2k (2k +2)!
 (3)
2
22


22
22k -1B2k 2k+2  2
+
(3) = -Σ

(1) -  (3)
2!
k=1 2k (2k +2)!

i.e.
Next

  tan x dx
2
0
x
x
0
4
=
0
(1)
203!


2
0
x

0
0
x

tan x dx = Σ
4
k =1

 2
3
-
(3)
221!
22k22k -1B2k
2k (2k +3)!

1

2k+3
 2
 2
(2.4')
(1.4')
from these
(1)
203!
(1)
3!

 2
-
221!

 2
1

=Σ
k=1
22k22k -1B2k
2k (2k +3)!

 2
2k+3
22k -1B2k 2k+3
 =Σ

1!
k =1 2k (2k +3)!
(3)
 3
1

22k -1B2k 2k+3
3
(3) = 
(1)
3!
 Σ
k=1 2k (2k +3)!
1
i.e.
(3)
3



Hereafter, in a similar way we obtain the desired expressions.
5.1.4 Riemann Odd Zeta
Applying
(n) =
2n-1
2n-1 - 1
(n) for Formula 5.1.3 , we obtain Riemann Zeta Function.
Formula 5.1.4
22
 (3) = - 3
2 -1
22k -1B2k 2k+2  2

log 2
Σ
2!
k =1 2k (2k +2)!



-6-
22k -1B2k 2k+3
3

log 2
Σ
3!
k=1 2k (2k +3)!

22 1
 (3) = - 2
2 -1 
24
 (5) = 5
2 -1




22k -1B2k 2k+4  4
 2 22-1

 (3)
log 2 +
Σ
4!
2! 22
k =1 2k (2k +4)!

24 1
 (5) = 4
2 -1 


2k
2 -1B2k 2k+ 5  5
 3 22-1

 (3)
log 2 +
Σ
5!
3! 22
k=1 2k (2k +5)!

22k -1B2k 2k+ 6  6
+

log 2
Σ
6!
k =1 2k (2k +6)!
 4 22-1
 2 24-1
26
+ 7
 (3)+
 (5)
4! 22
2! 24
2 -1
26
 (7) = - 7
2 -1






log 2 
7!
7
22k -1B2k 2k+7

Σ
k=1 2k (2k +7)!
 5 22-1
 3 24-1
26 1
+ 6
 (3) +
 (5)
3! 24
5! 22
2 -1 
26 1
 (7) = - 6
2 -1 





2k
 2 -1B 2k 2k+2n
 2n
22n
 (2n +1) = (-1) 2n+1

log 2
Σ
(2n)!
2
-1 k=1 2k (2k +2n)!
 2n-2k 22k-1-1
22n n -1
n
k
- (-1) 2n+1 Σ(-1)
 (2k +1)
(2n -2k)! 22k-1
2
-1 k=1

n
 (2n +1) =
-
(-1)n

22n
22n-1
(-1)n

Σ

k=1

2k
2 -1B2k
 2n+1
log 2
 2k+2n+1 2k (2k +2n +1) !
(2n +1) !


22k -1
22n n -1
k
Σ(-1) (2n +1-2k) ! 22k (2k +1)
22n-1 k=1
2n+1-2k
-7-
5.2 Termwise Higher Integral of tanh x
Since both zeros of
tanh x
and the primitive function
log cosh x
are
x =0 , the latter seems to be a lineal
primitive function with a fixed lower limit. Then, we assume zeros of the second or more order primitive function
x =0 .
are also
5.2.0 Higher Integral of tanh x
 tanh x dx
 tanh x dx
x
= log cosh x
0
x x
2
=
non-elementary function
0 0
5.2.1 Termwise Higher Integral of Taylor Series of tanh x
Formula 5.2.1
When
B0=1, B2=1/6, B4=-1/30, B6=1/42, 
hold for |x|<
are Bernoulli Numbers, the following expressions
 /2 .
 tanh x dx
x
22k22k -1B2k 2k
x
2k (2k)!

=Σ
k=1
0
 tanh x dx
x x
=Σ
k=1
0 0
 tanh x dx
x x x
22k22k -1B2k 2k+1
x
2k (2k +1)!

2
3
0 0 0
22k22k -1B2k 2k+2
x
2k (2k +2)!

=Σ
k=1

  tanh x dx
x
x
n
0
0

=Σ
k=1
22k22k -1B2k 2k+ n-1
x
2k (2k + n -1)!
Proof
tanh x
can be expanded to Taylor series as follows.

tanh x = Σ
k =1
22k22k -1B2k 2k-1
x
2k (2k -1)!
|x| <

2
Integrating both sides of this with respect to x from 0 to x repeatedly, we obtain the desired expression.
(1) Termwise Higher Definite Integral of Taylor Series of tanh x
Substituting

x = 1/2
for Formula 5.2.1 , we obtain the following expressions.
1
2

=Σ
tanh x dx
k=1
0
1
2
  tanh x dx
0
x
0
2
22k22k -1B2k
2k (2k)!

=Σ
k=1
 
22k22k -1B2k
2k (2k +1)!
-8-
1
2
2k
 
1
2
(1.1')
2k+1
(1.2')
1
2 x
   tanh x dx
0
x

=Σ
3
k =1
0 0
22k22k -1B2k
2k (2k +2)!
 
1
2
2k+2
(1.3')

1
2
  tanh x dx
0
x
x
0

=Σ
n
k=1
0
22k22k -1B2k
2k (2k + n -1)!
 
1
2
2k+ n-1
(1.n')
(2) Taylor Series of the 1st order Integral of tanh x
The 1st order Integral of tanh x is as follows.
2k
2k
x
 2 2 -1 B 2k

tanh xdx = Σ
2k (2k)!
k=1
0
x 2k = log cosh x
From this, the following expression follows immediately.
2k
2k
 2 2 -1 B 2k
2k
Σ
k =1
2k (2k)!
Substituting

Σ
k =1
= log cosh x
x
x =1 , 1/2 ,  /4
2

(1.t)
2
= log cosh 1
22k -1B2k
Σ
2k (2k)!
k =1

2k
2 -1B2k
Σ
2k (2k)!
k =1

for this one by one, we obtain the following special values.
22k22k -1B2k
2k (2k)!

|x|<
|x|<
= log cosh

 2
2k
= log cosh
1
2

4
This is used in 5.4.1 later.
5.2.2 Termwise Higher Integral of Fourier Series of tanh x
Formula 5.2.2
Let
(x) be a Dirichlet Eta Function,
 tanh x dx
x
=
0

x x
0 0
then the following expressions hold for
x >0 .
-2k x
1 
(1) x 0
x1
k-1 e
+
Σ(-1)
1!
20 k=1
20 0!
k1
-2k x
1 
(1) x 1 (2) x 0
x2
k-1 e
+ 1
+
tanh x dx = - 1 Σ(-1)
2!
0!
2 k=1
2
20 1!
k2

2
-2k x
1 
x3
k-1 e
+
tanh x dx = 2 Σ(-1)
3!
0 0 0
2 k=1
k3
(1) x 2 (2) x 1 (3) x 0
+ 1
1! 220! 0!
20 2!
2

-2k x
x
x
(-1)n-1 
xn
k-1 e
 tanh x dx n =
+
Σ(-1)
n!
0
0
2n-1 k=1
kn
x x x
3
 
-9-
n -1
- Σ(-1)
k
(k +1)
2k
k =0
n-1- k
x
(n -1- k)!
Proof
tanh x can be expanded to a Fourier series as follows. (See the following figure).
1- e -2x
e x - e -x
=
= 1- e -2x1 - e -2x + e -4x tanh x = x
-x
-2x
1+ e
e +e
e -6x +-
= 1 - 2e -2x - 2e -4x + 2e -6x - 2e -8x +-
= 1 - 2(cos 2ix - cos 4ix + cos 6ix - cos 8ix +- )
- 2i(sin 2ix - sin 4ix + sin 6ix - sin 8ix +- )
y
(2.0)
1.0
0.8
0.6
0.4
0.2
0.0
0
1
2
3
4
5
x
tanh(x)
1 - 2*(exp(-2*x) - exp(-20002*x))/(exp(-2*x) + 1)
Integrate both sides of (2.0) with respect to x from 0 to x , then

x
 1 - 2e
x
tanh xdx =
0
-2x
- e -4x + e -6x - e -8x +- dx
0
 
e
e
=x +
1
2
= x+

e -2x e -4x e -6x e -8x
+
+-
1
2
3
4
-2x
-4x
+
e
-6x
-
3
e
 
-8x
4
+-
-
i.e.

x
k =1
0
Here, let

k-1
tanh x dx = Σ(-1)

(-1)k-1
k=1
kx
(x)=Σ
0

k =1
x

(-1)k-1
k=1
k1
+x -Σ
, then we obtain
 tanh x dx = Σ(-1)
x
e -2k
k1
k-1
e -2k
k1
x
+ x - (1)
log cosh x .
Next, integrate both sides of this with respect to x from 0 to x , then
Of course, this is also the Fourier series expansion of
- 10 -
x
0

1 1 1 1
- + - +-
1 2 3 4
 tanh x dx
x x

x
2
=
0 0
0

-2k x
1 
k-1 e
+ x - (1) dx
Σ(-1)
20 k =1
k1

-2k x
1 
x2
x1
k-1 e
= - 1 Σ(-1)
- (1)
+
2!
1!
2 k=1
k2
-2k x
=-

+
0
0
(1) x 1 (2) x
x
+ 1
2!
0!
20 1!
2
2
1 
k-1 e
-1
(
)
Σ
21 k =1
k2
x
Hereafter, in a similar way we obtein the desired expressions.
(1) Termwise Higher Definite Integral of Fourier Series of tan x
x = 1/2 > 0
Substituting
1
2

for Formula 5.2.2 , we obtain the following expressions.

k-1
= Σ(-1)
tanh x dx
k =1
0
1
2

0
x
0
(1)
e -k 1 1
+
- 0
1
1! 2
2 1!
k
 
1
2
-k
1 2
1
1 
k-1 e
+
tanh x dx = - 1 Σ(-1)
2 k=1
k 2 2! 2
(1) 1 1 (2)
- 0
+ 1
2 1! 2
2 0!
 
0
x
0 0
-k
1
1
1 
k-1 e
+
tanh x dx = 2 Σ(-1)
3
3! 2
2 k=1
k
(1) 1 2 (2) 1
- 0
+ 1
2 2! 2
2 1! 2

 
3
 
1
2

0
x

0
(2.1')
 
2
 
1
2 x
0
(-1)n-1
 
1
2
n
tanh x dx =
0
(2.2')
3
1
(3)
  - 2 0!  
2
1 1
e -k
+
n! 2
2n-1 k=1
kn
n -1 (-1)k
1 n-1- k
(k +1)
-Σ
k =0
2k (n -1- k)! 2
x
0
 

k-1
Σ(-1)
1
2
0
(2.3')
n
 
(2.n')
(2) Taylor Series of the 1st order Integral of tanh x
The 1st order Integral of tanh x is as follows.
-2k x
x

k-1 e
(-1)
tanh x dx =
k =1
0
k1

Σ
+ x - log 2 = log cosh x
From this, the following expression follows immediately.
-2k x

k-1 e
= log(2cosh x) - x
(-1)
1
Σ
k =1
Substituting

k
x =1 , 1/2 ,  /2
k-1
(-1)
Σ
k =1
e -2k
k1
x >0
x >0
for this one by one, we obtain the following special values.
= log(2cosh 1) - 1
- 11 -
(2.f)

k-1
(-1)
Σ
k =1

k-1
Σ(-1)
k =1
e -k
k1
e -k
k1

1
2

-



-
= log 2cosh

= log 2cosh
2
1
2

2
5.2.3 Dirichlet Eta Function
Comparing Taylor series of tanh x and Fourier series of tanh x, we obtain Dirichlet Eta Function.
Formula 5.2.3
When
B2k , (x) denote

Bernoulli Numbers and Dirichlet Eta Function, the following expressions hold.
= Σ(-1)
0
22k -1B2k
1 1
+Σ
2 0!
k=1 2k (2k -1)!

k-1 -k
e
k=1
k-1
2k
 2 -1 B 2k
1 1
e -k
-Σ
+
1
2 1!
2k (2k)!
k=1
k
k-1
2k
 2 -1 B 2k
1 1
1
e -k
+Σ
+
(1)
2
2 2!
1!
k =1 2k (2k +1)!
k

(1) = Σ(-1)
k=1

(2) = Σ(-1)
k=1

(3) = Σ(-1)
k=1
k-1

(4) = Σ(-1)k-1
k=1


(n) = Σ(-1)
k=1
k-1
2k
 2 -1 B 2k
1 1
1
e -k
+
(1)+
Σ
2 3!
2!
k =1 2k (2k +2)!
k3
2k
 2 -1 B 2k
1 1
e -k
+Σ
4
2 4!
k =1 2k (2k +3)!
k
1
1
+
(1) (2) +
3!
2!
1
(2)
1!
1
(3)
1!
2k
2
-1B2k


(-1)n 1
e -k
n
+ (-1) Σ
2 n!
k=1 2k (2k + n -1)!
kn
n -1 (-1)k
-Σ
(n -k)
k!
k=1
Proof
tanh
1
= 1 - 2e -1 - e -2 + e -3 - e -4 +- 
2

1
tanh
2

=Σ
k=1
 0 = Σ(-1)
k-1 -k
k =1

0
1
2
e
 
1
2
2k-1
(1.0')
22k -1B2k
1
+Σ
2
k =1 2k (2k -1)!


tanh x dx
22k22k -1B2k
2k (2k -1)!
(2.0')
k-1
= Σ(-1)
k =1
1
e -k
+
- (1)
2
k1
- 12 -
(2.1')

1
2

=Σ
tanh x dx
k =1
0

 (1) = Σ(-1)
k-1
k=1
1
2

0
x
0
22k22k -1B2k
2k (2k)!
 
1
2
2k
(1.1')
2k
 2 -1 B 2k
1
e -k
+
Σ
2k (2k)!
2
k=1
k1
-k
1
1 
1 2
k-1 e
+
tanh x dx = - 1 Σ(-1)
2 k=1
k 2 2! 2
(1) 1 1 (2)
- 0
+ 1
2 1! 2
2 0!
 
2
 
1
2
  tanh x dx
0
x

=Σ
2
k=1
0

 (2) = Σ(-1)
k-1
k=1
22k22k -1B2k
2k (2k +1)!
 
1
2
 
1
2
0
(2.2')
2k+1
(1.2')
2k
 2 -1 B 2k
1 1
1
e -k
+
(1)+Σ
2 2!
1!
k =1 2k (2k +1)!
k2
Hereafter, in a similar way we obtain the desired expressions..
5.2.4 Riemann Zeta Function
Applying
(n) =
2n-1
2n-1 - 1
(n) for Formula 5.2.3 , we obtain Riemann Zeta Function.
Formula 5.2.4
21
 (2) = 1
2 -1
Σ(-1)
22
 (3) = 2
2 -1


k-1
k=1

k-1
(-1)
Σ
k=1
2k
 2 -1 B 2k
1
e -k
log 2
+
+
Σ
1!
22!
k=1 2k (2k +1)!
k2
2k
 2 -1 B 2k
1
e -k
log 2
+
Σ
2!
23!
k=1 2k (2k +2)!
k3
1
2 -1  (2)
+
1!
21
2k
 2 -1 B 2k
1
e -k
log 2
+Σ
+
4
3!
24!
k=1 2k (2k +3)!
k


23
 (4) = 3
2 -1


k-1
(-1)
Σ
k=1
22-1  (3)
21-1  (2)
+
1!
2!
22
21

 (n) =
Σ(-1)
(-2)
Σ
2 -1 
2n-1

2n-1-1
k=1
n-1
n-1
k-1

k =1
n-1- j
n -2
-1  (n -j)
e -k
j-1 2
-1
+
(
)
Σ
j!
j=1
2n-1- j
kn
22k -1B2k
1
log 2
+
2k (2k + n -1)! 2n !
(n -1)!
- 13 -



5.3 Termwise Higher Integral of cot x
Since both zeros of
cot x
and the primitive function
log sin x
are
x = /2 , the latter seems to be a lineal
primitive function with a fixed lower limit. Then, we assume zeros of the second or more order primitive function
are also
x = /2 .
5.3.0 Higher Integral of cot x
 cot x dx
x
= log sin x

2
 cotx dx
x x
2
 
=
non-elementary function
2 2
5.3.1 Termwise Higher Integral of Taylor Series of cot x
Formula 5.3.1
When
(x) on 0< x < 
<n>
B2k denote Bernoulli Numbers and functions f
n 1
xn
(x)=
log x -Σ
n!
k=1 k

<n>
f
 -Σ

k=1
are as follows.
22kB2k 2k+n
x
2k (2k +n)!
n =1, 2, 3, 
the following expressions hold .
 cot x dx
x
22kB2k 2k
x
2k (2k)!

= f<0>(x) = log x -Σ

k=1
2
  cot x dx
x
x


2
2

x
x
x
  
1

(x) x2
0!


1

x2
0!
0
<1>
2
=f
cot x dx 3 = f <2>(x) -


0
f<1>
f

 2

 2
<2>
-
1

x1!
2


1
f

 2
<1>
2 2 2
 
x

x
1
4



cot x dx = f <3>(x) - 0! x - 2

2
2

0
f

 2
<3>
1

x1!
2

-
1

x2!
2

<2>

<1>

1
 f  2

2
f
 2

  cot x dx
x
x


2
2
n
=f
<n -1>
n -2
(x) -Σ
k=0
where
1

x2
k!
k
<0>
<1>
Proof
2k
22k B2k 2k
 2 B 2k
x cot x = 1+Σ(-1)
x = 1+Σ
x 2k
(2k)!
(2k)!
k =1
k=1


 
 2  n =2, 3, 4, 



f   = 0 , f   = - log 2
2
2
2
n -1-k >
f<
k
- 14 -
From this
2k
1  2 B2k 2k-1
cot x = -Σ
x
x k =1 2k (2k -1)!
Integrate the both sides with respect to x from  /2
 cot x dx = log x -Σ
x


k=1
x,
to
then

2k
22kB2k 2k
 2 B 2k

x - log
2k (2k)!
2 Σ
k=1 2k (2k)!

 2
2k

2
Here
log


-Σ
2
k=1
22kB2k
2k (2k)!

 2
2k
=0
Because, the following equation is known ( 岩波数学公式Ⅱ p151)
2k
 2 B 2k
2k
log x -Σ
k=1
Then, substituting
x = /2
0< x <
= log sin x
x
2k (2k)!
for this, we obtain the above.
Thus,
 cot x dx = log x -Σ
x


k=1
22kB2k 2k
x = f<0>(x)
2k (2k)!
2
Next, integrate the both sides with respect to

x
x


2
2
f
x
cot x dx 2
=
x
to
x,
<0>
(x)dx = f<1>(x) - f<1>

then

 2
2
Furthermore, integrate the both sides with respect to

x
 /2
from
x
x
  
cot x dx 3 =
2 2 2
 f
x

<1>
x
(x) - f<1>
from
 /2
to
x , then

 2  dx
2
<2>
=f
<2>
(x) - f

 2
1

x1!
2

1

 f  2
<1>
Hereafter, in a similar way we obtein the desired expressions.
Although Formula 5.3.1 is a general expression anyway, if it is expand actually, it is very complicated.
Then, this was devised and was made the same easy expression as tan x. It is as follows.
Formula 5.3.1'
When
B2k denote Bernoulli Numbers, the following expressions hold for 0< x <  .
2k
2k
x
x
 2 2 -1B 2k
 2k+ n-1
n
 cotx dx = -Σ
x2k (2k + n -1)!
2
k=1


 
2
Proof

2

cot x = -tan x -

2

- 15 -


tan x = Σ
k=1
22k22k -1B2k 2k-1
x
2k (2k -1)!

|x|<
2
From these

cotx = -Σ
k =1
22k22k -1B2k

x2k (2k -1)!
2


2k-1
0< x < 
Integratingboth sides of this with respect to x from  /2 to x repeatedly, we obtain the desired expressions.
(1) Termwise Higher Definite Integral of Taylor Series of cot x
When
n 1
xn
(x)=
log x -Σ
n!
k=1 k

<n>
f
x =0
substituting

0

cot x dx
2 2
0 x
 
1
cot x dx = 0!


2
2 2
 
0

2
x


3
x

4
cot x dx = -

0

0
 2
1
= 0!
2
x

k=1
22kB2k 2k+n
x
2k (2k +n)!
 2
1
0!

 2
f<1>
<2>
f

 2

 2
0
<3>
f


   cot x dx
0
x
x



2
2
2
n
n -2
=Σ
k=0
(-1)k+1
k!

(1.2')
1
+
1!
 2
2
2
n =1, 2, 3, 
for Formula 5.3.1 , we obtain the following expressions.
x

 -Σ

1
1!
1
2!
<n -1-k >
1

 2  2
+
k
 2 f


f<1>
1

 2  2


 2 f  2
f<2>
(1.3')
2
<1>

 2
(1.4')
(1.n')
(2) Taylor Series of the 1st order Integral of cot x
The 1st order Integral of cot x is as follows.
2k
x
 2 B 2k
 cot x dx

= log x -Σ
k =1
2k (2k)!
x 2k = log sin x
2
From this, the following expression follows immediately.
2k
 2 B 2k
2k
x
Σ
k=1 2k (2k)!
Substituting

Σ
k=1

Σ
k=1

Σ
k=1
0< x < 
= -log sin x + log x
x =1 , 1/2 ,  /4
for this one by one, we obtain the following special values.
2k
2 B2k
2k (2k)!
B2k
2k (2k)!
B2k
2k (2k)!
= -log sin 1
= -log sin

 2
0< x < 
2k
= -log sin
- 16 -
1
- log 2
2

4
+ log

4
(1.t)
Moreover, from the last expressions of this and 5.1.1 (2) , the next expression follows
2k
 2 -2B 2k
 2k

= -log
This is used in 5.5.1 later.
Σ
k=1
 2
2k (2k)!
4
5.3.2 Termwise Higher Integral of Fourier Series of cot x
Formula 5.3.2
(x) be Dirichlet Eta Function and 
0< x <  .
Let
for
be the ceiling function, then the following expressions hold
1
1  1
= 0 Σ 1 sin 2kx 2
2 k=1 k
 cot x dx

x

(1)

 - 2 0! x - 2 
0
0
2
  cot x dx
x
x


2
2
 cot x dx
x
x
x



2
2
2

x

x


2
2
1  1
2
= 1 Σ 2 sin 2kx 2
2 k=1 k

2
3
=
4
cot x dx =

22
Σ
k =1
3
sin 2kx 3
2
k
1

1
1
23
1
Σ
k =1 k


sin 2kx 4
4
2
(1)

 - 2 1! x - 2 
1
0
(1)

2
(3)

3
(3)

0

1
 - 202! x - 2  + 220! x - 2 

-
(1)
203!

x-
2

+
221!

x-
2


  cot x dx
x
x


2
2
n
1
=

n-1 Σ
k =1
2
n/2
+Σ
k =1
1
k

2kx n sin
n
2

(-1)k (2k -1)

x2k-2 (n +1-2k)!
2
2


n+1-2k
Proof
cot x can be expanded to Fourier series in a broad meaning.as follows.
1+ e 2ix
e ix + e -ix
= -i
= -i1 + e 2ix1
cotx = i ix
-ix
2ix
1- e
e -e
+ e 2ix + e 4ix + e 6ix +
= -i - i2e 2ix + 2e 4ix + 2e 6ix + 2e 8ix +
= 2(sin 2x + sin 4x + sin 6x + sin 8x +)
-2i(cos 2x + cos 4x + cos 6x + cos 8x +) - i
(2.0)
cot x is the locus of the median (central line) of this real number part. (See the figure of the following page.)
However calculus of the right side of (2.0) may be carried out, the real number does not turn into an imaginary
number and the contrary does not exist either. Therefore, limiting the range of cot x and the higher order
integral to the real number, we calculate only the real number part of the right side of (2.0).
Although the Riemann zeta
(2n) is obtained from the calculation of the imaginary number part,
long, it is omitted in this section.
- 17 -
since it is
y
40
30
20
10
0
0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0
-10
x
-20
-30
-40
cot(x)
2*sin(2*x) + 2*sin(4*x) + 2*sin(6*x) + 2*sin(8*x)
 /2
Now, integrate the real number part of both sides of (2.0) with respect to x from
to x , then
 cot x dx =  2(sin 2x + sin 4x + sin 6x * sin 8x +)dx
x
x


2
2


cos 2x cos 4x cos 6x cos 8x
+
+
+
+
=2
3
4
1
x

2

 
cos 2x cos 4x cos 6x cos 8x
=+
+
+
+
1
2
3
4
-

1 1 1 1
- + - +-
1 2 3 4
i.e.
k-1
 (-1)
1  cos 2k x
-Σ
cot x dx = log sin x = - 0 Σ
k=1
2 k=1 k 1

k1

x
2

-cos x = sin x -
Here, let

2


(-1)k-1
k=1
kx
, (x)=Σ
1
1  1
cot x dx = 0 Σ 1 sin 2kx 2
2 k=1 k



x
, then we obtain
(1)

 - 2 0! x - 2 
0
0
2
log sin x
 /2 to x , then
This is consistent with the real number part of the Fourier series of
Next, integrate both sides of this with respect to x from
  cot x dx =  
x
x
x
2



2
2
2
=

1  1
1
2kx 0 Σ 1 sin
2
2 k=1 k

2
1  1
2kx 1 Σ 2 sin
2
2 k=1 k

1  1
2
= 1 Σ 2 sin 2kx 2
2 k =1 k

- 18 -
.
(1)

 - 2 0! x - 2  dx
0
0
(1)

x
 - 2 1! x - 2  
1
0
(1)

 - 2 1! x - 2 
0
1

2
 /2
Next, integrate both sides of this with respectto x from
 cot x dx =   2
x
x
x
x
3




2
2
2
2
2 Σ
=
2
1
=
2
Σ
k=1
1

2
=
22
Σ
k =1


1
k =1

1
1
3
sin 2kx 3
2
k

1
2
sin 2kx 2
2
k

1

-
3
2

-


1
Σ
k=1 k
sin 2kx 3
(1)
-
3
sin 2kx 3
2
k
1
to x , then
0
2 2!
(1)
202!
-
201!

x - 2 
202!
(1)
(1)




xx-
2
2
2

x - 2 


2
+
dx
x


2
2
1

1
2
2
Σ
k=1
(3)
222!
cos 2k
k

x-
3

2

0
Hereafter, in a similar way we obtain the desired expressions.
(1) Termwise Higher Definite Integral of Fourier Series of cot x
Substituting
x =0
for Formula 5.3.2 , we obtain the following expressions.
  cot x dx
x
0


2
2
2
=
   cot x dx
0
x
x

 
2
2 2
3



2
2
2
   cot x dx
0
x
x



2
2
2

 2
(1)
0
2 2!
x
x
201!
=-
   cot x dx
0
(1)
4
5
=
(1)
203!
=-
1
(2.2')

 2

 2
(1)

2 4!
+
0
 2
(3)

221!
 2
1
(3)

2
2
4

2 0!
+
3
 2
0
(3)
2
2
2 2!
 2
+
 (3)
(2.3')
22
(2.4')
-
(5)
4
-
 (5)
2 0!
(2.5')
24

   cot x dx
0
x
x



2
2
2
n
n
= (-1)
n /2
Σ
k=1
(-1)k-1 (2k -1)

(n +1-2k) !  2 
2k-2
2
n+1-2k
-
 (n)
n-1
2
sin
n
2
(2.n')
(2) Fourier Series of the 1st order Integral of cot x
The 1st order Integral of cot x is as follows.
x
1  cos 2k x
 cotx dx = - 2
0
0
Σ
k=1
k1
- log 2 = log sin x
0< x < 
From this, the following expression follows immediately.

Σ
k=1
cos 2k x
= -log(2sin x)
k1
Substituting

Σ
k=1
x =1 , 1/2
0< x < 
for this one by one, we obtain the following special values.
cos 2k
= -log(2sin 1)
k1
- 19 -
(2.f)

Σ
k =1
cosk
k1
= -log2sin(1/2)
5.3.3 Dirichlet Odd Eta & Riemann Odd Zeta
Comparing Taylor series of cot x and Fourier series of cot x, we obtain Dirichlet Odd Eta and Riemann
Odd Zeta.
Formula 5.3.3
When
B2k , (x),  (x) denote
Bernoulli Numbers, Dirichlet Eta Function and Riemann Zeta Function
respectively, the following expressions hold.
B2k

log =
1 +Σ
 2k
2k (2k +1)!
2 1

2
B2k
+Σ
(3) =  2k+2 -  (3)
log -Σ
2!
k =1 k
k=1 2k (2k +2)!
3
3 1

1 
B2k
-Σ
 (3) =  2k+3
log -Σ
2
2
+3
!
(
)
 3!
k
k
k
k=1
k =1
4
4 1


2
B2k
2k+4
-Σ
+
(5) =

 (3) -  (5)
log -Σ
2!
4!
k =1 k
k=1 2k (2k +4)!
5 1

1 5
3
B2k
2k+5
-Σ
+
 (5) =

 (3)
log -Σ
3!
 5!
k=1 k
k =1 2k (2k +5)!
k=1


 


 
(7)= (7)= -
6
6!
1





B2k

log -Σ  +Σ 2k (2k +6)! 
1
k
6
k=1
7
2k+6
4
-
4!
k=1
 7! log -Σ  -Σ 2k (2k +7)! 
k =1
B2k

1
k
7
2k+7
+
k =1
(3)+
5
5!
2
2!
(5) - (7)
(3)-
3
3!

(5)

(2n +1) = (-1)
n
 2n
 (2n)! 
2n
log -Σ
k =1
n -1
 2n-2k
k=1
(2n -2k) !
+ (-1)nΣ(-1)k-1
(2n +1) = (-1)
n
n
+ (-1)
1
k
 2n+1
  (2n +1) ! 
1
1

n -1
(-1)
Σ
k=1


B2k
k =1
2k (2k +2n ) !
-Σ
(2k +1) - (2n +1)
2n +1
log  - Σ
k=1
1
k
 2n-2k+1
k-1

 2k+2n
(2n -2k +1) !


B2k
k=1
2k (2k +2n +1) !
-Σ

 2k+2n+1
(2k +1)
Proof
From Fourier series (2.2') and Taylor series (1.2')
1
log 2  1
 0 <1>
=
f
0
0! 2
2
2 1!
 
 
1
=1!


 2
1

(2.2")
 2  log 2 -1 -Σ

k =1
- 20 -
22kB2k
2k (2k +1)!

 2
2k+1
Hence we obtain

B2k
k =1
2k (2k +1)!
log = 1 + Σ
 2k
Next, substitute (2.2") for Taylor series (1.3') , then
   cot x dx
0
x
x

 
2
2 2
3
<2>
= -f
<2>
= -f

 2
1
+
1!

log 2
1!
 2

 2  2
f<1>

 2
From this and Fourier series (2.3') , we obtain
2
1
log 2
 3 2 ( )
2!
2
i.e.

1
2
1 (3)

= -f<2>
+ 2
2
2 0!

 2
 
1
1 (3)
<2> 
=
3
+

(
)
f
2
22 0!
22
 
log 2
2!

 2
log 2
1!

 2
2
2
(a)
From (a), it is as follows.
(3) = -2 f
2 <2>
2
= -2
2

 2 -
2!
  2  log 2 -Σ  -Σ

1
2!
2
2!

2

=-
log 2 -  (3)

2
log -Σ
k =1

1
k
2
k=1
k=1
22kB2k
2k (2k +2)!



2
2
2
<3>
= -f

log 2 -  (3)
2!

1
B2k
+Σ
 2k+2 -  (3)
2
2
+2
!
(
)
k
k k
k=1

 2 +
= -f
 2
2k+2
2
Next, substitute (2.2") and (a) for Taylor series (1.4') one by one, then
0 x
x
1
 1 <2> 
4
<3> 
   cot x dx


f  
1!  2 
2

2

 2 f  2
<1>
1
1 (3)
3
+

(
)
22
22 0!
1
 2  2 
-

1
2!

From this and Fourier series (2.4') , we obtain
log 2
3!
i.e.
1
22
Substitute
1
22


 2

 2
f<3>
1
3
1 (3)
- 2
2 1!
 (3) = -f<3>

 2

 2

 2
= -f
-


1
 2 - 2  2
<3>
1
2
log 2
3!

 2
(3)+
1 (3)
22 0!

3
(b)
for (b), then
1

 2   (3) = -f  2 
<3>
1
=3!
1

3
 2 
log

2
log 2
3!
3
-Σ
k =1
1
k


 2

+Σ
k =1
- 21 -
3
22kB2k
2k (2k +3) !

 2
2k+3
-
log 2
3!

 2
3
1
=3!

3
 2  log -Σ  +Σ

1
k
3
k =1
k=1
22kB2k
2k (2k +3)!

 2
2k+3
From this, we obtain
 2
1
3
 (3) = -2
=-

2

-1
  2 

1
3!
 3! 
3
log -Σ
k=1
3
1
k

22kB2k

1
+Σ
log -Σ
k=1 k
k=1 2k (2k +3)!

B2k
-Σ
 2k+3
2
+3
!
2
(
)
k k
k=1

3

 2

2k+3

Hereafter, in a similar way we obtein the desired expressions.
5.3.4 Another set of Riemann Odd Zeta
Substituting
22n - 1
(2n +1) = 2n  (2n +1)
2
for Formula 5.3.3 , we obtain another set of Riemann
Odd Zeta.
Formula 5.3.4
2
 2! 
 2k (2k +2)!  
1


B 
+


 (3)
log
Σ
Σ
 4! 
2!
2k (2k +4)!
k
1

B 


log
Σ
Σ
 6! 

2k (2k +6)!
k


3

 (5)
(
)
 4!
2!
22
3
=
( )
23-1
24
 (5) =
25-1
26
 (7)= - 7
2 -1
26
- 7
2 -1

(2n +1) = (-1)
k =1
4
6
1
k

k =1
k =1
6

22n+1 -1
2k+4
2k
2k+6
k =1
 2n
-1  (2n) ! 
22n
2
2k
2
22n
2
2k+2
k=1
k=1
2n+1
B2k

-Σ
4
4
n
+ (-1)n
2
log -Σ
2n
log -Σ
k=1
1
k
n -1
 2n-2k
k=1
(2n -2k) !
Σ(-1)k-1
- 22 -


B2k
k=1
2k (2k +2n ) !
-Σ
(2k +1)

 2k+2n
n =2, 3, 4, 
5.4 Termwise Higher Integral of coth x
coth x is x = sinh -11 = 0.8813 ,
we assume zeros of the second or more order integral of coth x are also x =0.8813 .
Where, since the zero of coth x is not 0.8813 , these seem to be a collateral higher integral.
Since the zero of the first order integral
log sinh x
of
5.4.0 Collateral Higher Integral of coth x
When
 = 2sinh -11 = 1.762747174
 coth x dx
x
= log sinh x

2
x x
 coth x dx
2
 
=
non-elementary function
2 2
5.4.1 Termwise Higher Integral of Taylor Series of coth x
Formula 5.4.1
Let
 = 2sinh -11 (= 1.762747174)
and
B0=1, B2=1/6, B4=-1/30, B6=1/42, 
<n>
(x) on x >0 be as follows.
22k B2k
n 1

xn
<n>
+Σ
f (x)=
log x -Σ
x 2k+n
2
+
!
2
(
)
k
n
k
n!
k
k=1
k=1
Then the following expressions hold for x >0 .
f
are Bernoulli Numbers and functions


 coth x dx
x
(x) = log x +Σ
=f

22k B2k 2k
x
2k (2k)!

<0>
n =1, 2, 3, 
k=1
2
  coth x dx
x
x


2
2

x
x
x
  
1

(x) x2
0!

<1>
2
=f
3
coth x dx = f <2>(x) -
1

x0!
2



0
f
0
f<1>

 2

 2
<2>
-
1

x1!
2


1

 2
<1>
f
2 2 2
  coth x dx
x
x


2
2
4
=f
1

(x) x2
0!

<3>
0

 f  2
<3>
1

x1!
2

-

  coth x dx
x
x


2
2
n
=f
<n -1>
n -2
(x) -Σ
k=0
1

x2
k!

Proof

x coth x = 1+Σ
k=1
22k B2k 2k
x
(2k)!
- 23 -
k
f
 f  2
1

x2!
2

<n -1-k >

 2

1
<2>

2
f

 2
<1>
From this
22k B2k
1 
coth x = +Σ
x 2k-1
x k=1 2k (2k -1)!
Integrate the both sides with respect to x from  /2



x , then
22k B2k 2k
22k B2k


+
x - log
2k (2k)!
2 Σ
k=1 2k (2k)!
 coth x dx = log x +Σ
x
to
k=1

 2
2k

2
Here
log


+Σ
2
k =1
22k B2k
2k (2k)!

 2
2k
=0
Because, the following equation is known ( 岩波数学公式Ⅱ p150)
2k-1
B2k 2k
 2
log x +Σ
k=1
Then, substituting
k (2k)!
x = /2
= log sinh x
x
x >0
for this, we obtain the above.
Thus,
22k B2k 2k
x = f<0>(x)
2k (2k)!
 coth x dx = log x +Σ
x


k=1
2
Next, integrate the both sides with respect to x from

x
x


2
2
coth x dx 2
=

x

 /2
to x , then
f<0>(x)dx = f<1>(x) - f<1>

x
 2
2
Furthermore, integrate the both sides with respect to x from
x

x
  
coth x dx 3 =
 f
x
<1>
(x) - f<1>

2 2 2
 /2
to x , then

 2  dx
2
<2>
=f
<2>
(x) - f

 2
1

x1!
2


1
 f  2
<1>
Hereafter, in a similar way we obtein the desired expressions.
(1) Termwise Higher Definite Integral of Taylor Series of coth x
When
22k B2k
n 1

xn
+Σ
f (x)=
log x -Σ
x 2k+n
n!
k =1 k
k=1 2k (2k +n )!
substituting x =0 for Formula 5.4.1 , we obtain the following expressions.

<n>

0
x


2
2
coth x dx
 
0
x

x

 
2
2 2
1
=0!
2
coth x dx
3
= -
1
0!

 2
0

0
 2
f<1>

 2
<2>
f
n =1, 2, 3, 

 2
- 24 -
(1.2')
+
1
1!

 2
1
f<1>

 2
(1.3')
 
x
x

1
 coth x dx = 0!


2
2
2
0
4


0
 2 f  2
<3>
1
+
1!


1
 2 f  2
1
2!
<2>

 2
2
f<1>

 2
(1.4')

 
0
x



2
2
x

n -2
coth x dx = -Σ
n
k=0
(-1)k
k!

 2
k
n -1-k >
f<

 2
(1.n')
2
(2) Taylor Series of the 1st order Integral of coth x
The 1st order Integral of coth x is as follows.
22k B2k
x

 coth x dx = log x +Σ 2k (2k)! x

2k
0< x < 
= log sinh x
k=1
2
From this, the following expression follows immediately.
22k B2k 2k

x
Σ
k =1 2k (2k)!
Substituting

Σ
k =1

Σ
k =1

Σ
k =1
x =1 , 1/2 ,  /4
22k B2k
2k (2k)!
B2k
2k (2k)!
B2k
2k (2k)!
0< x < 
= log sinh x - log x
(1.t)
for this one by one, we obtain the following special values.
= log sinh 1
= log sinh

 2
2k
= log sinh
1
+ log 2
2

4
- log

4
Moreover, from the last expressions of this and 5.2.1 (2) , the next expression follows
2k
 2 -2 B 2k
 2k


Σ
k =1
 2
2k (2k)!
= log coth
+ log
4
4
5.4.2 Termwise Higher Integral of Fourier Series of coth x
Formula 5.4.2
Let
 = 2sinh -11 = 1.762747174,
then the following expressions hold for x >0 .
1  e -2k x

coth x dx = log sinh x = - 0 Σ 1 + x 2
2 k=1 k



x
+

2
- log 2
2
  coth x dx
x
x


2
2
2
1
1  e -2k x

= 1Σ 2 +
x2!
2
2 k=1 k


2
+
 /2-log 2
1!

1  e -k 
- 1Σ 2
2 k=1 k
- 25 -
x-

2

1  e -2k x 1

coth x dx = - 2 Σ 3 +
x2
3!
2 k =1 k


x
x
 
x

3
3
+
 /2-log 2

x - 2 
2!
2
2 2 2
1  e -k 

- 1 Σ 2 x2
2 k=1 k

1  e -2k x 1

 coth x dx = 3 Σ 4 +
x2
4!

2 k=1 k

 
x
2
x

4

1  e -k 
+ 2Σ 3
2 k=1 k

4
+
 /2-log 2

x-
3!

2

3
2

1  e -k  1

x
Σ
2
21 k=1 k 2 2!
   coth x dx
x
x


2
2

n
=
(-1)n
2n-1

Σ
k =1

1  e -k 

x
Σ
2
22 k=1 k 3
2

+
x
1
e -2k

+
x
2
n!
kn
n -2
-Σ
j=0

+
(-1)n-j 1

x
2
2n-1- j j !

n

-
1  e -k 
Σ
23 k=1 k 4
 /2-log 2
(n -1)!
j 
e -k
Σ k
k=1

x - 2 
n-1

n-j
Proof
coth x can be expanded to a Fourier series as follows. (See the following figure).
1+ e -2x
e x + e -x
=
= 1+ e -2x1 + e -2x + e -4x +
coth x = x
-x
-2x
e -e
1- e
= 1 + 2e -2x + 2e -4x + 2e -6x + 2e -8x +-
= 1 + 2(cos 2ix + cos 4ix + cos 6ix + cos 8ix + )
+ 2i(sin 2ix + sin 4ix + sin 6ix + sin 8ix + )
e -6x +
(2.0)
y
60
50
40
30
20
10
0
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7
x
coth(x)
2*exp(-2*x) + 2*exp(-4*x) + 2*exp(-6*x) + 2*exp(-8
Now, let
from
 /2 =sinh -11 (= 0.881373587) .
 /2
Integrate the both sides of (2.0) with respect to x
to x , then
- 26 -

x

 2e
x
coth x dx =
-2x

2
+ e -4x+ e -6x+ e -8x+ + 1dx
2
=-

e -2x e -4x e -6x e -8x

+
+
+
+ - x 2
2
3
4
1


x

2
=-

e
-2x
+
1
e
-4x
+
2

e
-6x
+
3

e
-8x
4


+ - x -

2


ee -2
e -3
e -4
+
+
+
+
+
1
2
3
4
i.e.
1  e -2k x

coth x dx = - 0 Σ 1 + x 2
2 k=1 k



x
 +Σ

k=1
e -k
k1

2
Here, the following equation holds,
-k 

Σ
k =1
e

=
- log 2
2
k1
(w)
Because,

log(1-x) = -Σ
k =1
xk
k
-1 x < 1
From this,

Σ
k =1
e -2kx
= -log1-e -2x
k
x >0
On the other hand,
1 = sech x =
e x1- e -2x
e x- e -x
=
2
2
From this,
1
1- e -2x
=
1
ex
ex
 log
=
 -log1-e -2x = x - log 2
log
-2x
2
2
1- e
Then,

Σ
k =1
e -2kx
= -log1-e -2x = x - log 2
k
Substituting
x = sech -11 =

2
for this, we obtain (w).
Thus, using (w), we obtain
1  e -2k x

coth x dx = log sinh x = - 0 Σ 1 + x 2
2 k=1 k



x


+
2
2
This is consistent with the real number part of the Fourier series of
Next, integrate both sides of this with respect to x from
 /2
- 27 -
log sinh x
to x , then
.
- log 2
  coth x dx
x
x


2
2
2
 -Σ
x
=


k=1
e -2k

+ x1
2
k
x


 dx
 2
+
- log 2
2

=
1
1  e -2k x

+
xΣ
1
2
2
2!
2 k =1 k


 /2-log 2
2
+
1!

x - 2 
x

2
1
1  e -2k x

= 1Σ 2 +
x2
2!
2 k =1 k

 /2-log 2
2
+

x - 2 
1!
1  e -k 
- 1Σ 2
2 k=1 k
Hereafter, in a similar way we obtein the desired expressions.
(1) Termwise Higher Definite Integral of Fourier Series of coth x
x =0
Substituting
for Formula 5.4.2 , we obtain the following expressions.
  coth x dx
x
0


2
2

2
=
 (2)
1
2!
21

 2
2
log 2
+
1!

 2
1  e -s
- 1 Σ 2
2 0! s=1 s
(2.2')
2
 (3)
 3 log 2  2
3
coth x dx = - 2 +
2
3! 2
2!
2
 
2 2
1
1  e -s
  e -s
+ 2 Σ 3
+ 1
Σ
2 0! s=1 s
2 1! 2 s=1 s 2
4
x
x
3
 (4)

log 2  3
4
+
 coth x dx
=
2
4! 2
3!

23

 
0
2
x
 
x
 
 
 
0

2
2
 
2
1
- 1
2 2!

2
 2 Σ

s=1

1
e -s
- 2
2
2 1!
s
(2.3')
 

 2 Σ

s=1


1  e -s
e -s
Σ
s 3 230! s=1 s 4
(2.4')

   coth x dx
0
x
x



2
2
2
n
n
= (-1)
 (n)
2
n -1
n!
n-1

 2

n -2
1
j=0
2n-1- r r!
-(-1)n Σ
n
log 2
+
(n -1)!
r 
s=1
 2

n-1

e -s
 2 Σ s

n-r
(2.n')
(2) Fourier Series of the 1st order Integral of coth x
The 1st order Integral of coth x is as follows.
-2k x
x


coth x dx = -
1  e
Σ
20 k=1 k 1

+ x-

+
2

2
- log 2 = log sinh x
x >0
2
From this, the following expression follows immediately.
-2k x

Σ
k =1
e
k1
= x - log(2sinh x)
x >0
- 28 -
(2.f)
Substituting

Σ
k=1

Σ
k=1

Σ
k=1
x =1 , 1/2 ,  /2
for this one by one, we obtain the following special values.
e -2k
k1
= 1 - log(2sinh 1)
e -k
k1
=
e -k
k1

1
1
- log 2 sinh
2
2


=
2


- log 2 sinh
2


5.4.3 Riemann Zeta Function
Comparing Taylor series of coth x and Fourier series of coth x, we obtain Riemann Zeta Function.
Formula 5.4.3
When
 = 2sinh -11 = 1.762747174 and B2k are Bernoulli Numbers,
the following
expressions hold.

 (2) = Σ
k=1

 (3) = Σ
k=1
e -k
k2

-
 1! log -Σ  +Σ

k =1

2 1
2
e -k
+

log
Σ
2!
k=1 k
k3

+

e -k
k4

 (4) = Σ
k=1
+

1!

1
k
1
k=1

B2k 
1 3
2 3!
2k (2k +2)!
2k+2
 +Σ

k=1
 3! log -Σ  +Σ
 (3) -
2
2!

1
k
3
k =1
1!
2k+1
 (2)
3

B2k 
1 2
2 2!
2k (2k +1)!
k =1
B2k 
1 4
2 4!
2k (2k +3)!
2k+3

 (2)

j

n -2
e -k
j-1 
+ Σ(-1)
 (n - j)
j!
j=1
kn

 (n) = Σ
k=1
n-1
+ (-1)
 (n -1)! log -Σ  +Σ
 n-1
n -1
k=1
1
k

k=1
B2k 
1 n
2 n!
2k (2k + n -1)!

1  e -k 
Σ
21 k =1 k 2
(a)
2k+ n-1
Proof
From Fourier series (2.2') and Taylor series (1.2')
<1>
f

 2
=-
1
1
2
+

(
)
2!
21

 2
2
-
log 2
1!

 2
+
From this
 (2) = -2f
<1>

 2
2
+
2!

2
 2 -
 log 2
- 29 -
1!

+Σ
k=1
e -k
k2

  2  log 2 -Σ  +Σ

1
= -2
1!

1
1
k=1
=-
k=1
2
2!
+

 2
1
k =1
 log 2
2
-
B2k 

1
+Σ
k
k=1 2k (2k +1)!
 1! log -Σ 

22k B2k
2k (2k +1)!

1
k
2k+1
 2

+Σ
1!


k =1

2k+1
e -k
k2

1  2  e -k 
+
+
2
2 2! Σ
k=1 k
Next, substitute (a) for Taylor series (1.3') , then
0 x x
3
<2> 
   coth x dx

 
2
2 2
1
- 1
2
 2
= -f

 
1
 (2)+
2
2!
(1.3')

 2
3
log 2
1!

 2
2
1
+ 1
2

 2 Σ

k=1
e -k
k2

From this and Fourier series (2.3') , we obtain
<2>
f

 2
1
=- 1
2

 
1
 (2) +
2
3!

 2
3
log 2
2!

 2
2
1
1  e -k 
+ 2  (3)- 2 Σ 3
2
2 k=1 k
(b)
From this,
 (3) = 2 f
2 <2>
2
=2


22
+2
 (2) 2
3!
 2  
  2  log 2

1
2!

2
22
+  (2)3!
=
2
-Σ
k =1
3
 2 +
k=1
1
k

2!
 2log 2
3
 2 +

+Σ
k=1
 2log 2
2!
22k B2k
2k (2k +2)!

+Σ
k =1
e -k
k3

+Σ
k=1

e -k
k3
 2


2k+2

-k 

B2k

 e
1
1 3
+Σ
+
+  (2)
3
2 3! Σ
k
k=1 2k (2k +2)!
k=1 k
2k+2
log -Σ 
2
2!

2

Next, substitute (a), (b) for Taylor series (1.4') , then

   coth x dx = - f  2 
0
x
x



2
2
2
<3>
4
<3>
= -f

 2

 2
1
1
2!

2
 2
f<2>

 2
f<1>

 2
(1.4')'

1
e -k
 2  e -k 
3
2
2 Σ
22! 2 Σ
k=1 k
k=1 k
1
1 
 2

 (2) + 2
 (3) 2
2
23! 2
2
1
- 2
2
1
22!
1
+
1!

 
 
 

 
From this and Fourier series (2.4') , we obtain
- 30 -
 
4
<3>
f

 2
1
=22!

 2
2
1
22
 (2) +
1
1
- 3  (4) +
4!
2

 2

 2   (3)
4
log 2
3!

 2
3
1  e -k 
+ 3Σ 4
2 k=1 k
(c)
From this,
 (4) = -2
3
  2  log 2 -Σ  +Σ

1
3!
log 2
3!

3
3
k=1

 2
3
23
+
4!

 2
23  e -k 
23
+ 3Σ 4 22!
2 k=1 k
=-
 3! log -Σ
3
3
k =1

+Σ
k=1

1
k

 +Σ
k =1
k=1
2
 (2) +
23
22

2k+3
2
(n) for natural number n >2 .
- 31 -

 2
 2   (3)
B2k 
4
2k (2k +3)!
24!

e -k
 (2) +  (3)
4
2!
k
Hereafter, by induction we obtain
22k B2k
2k (2k +3)!
4
 2


1
k


2k+3
5.5 Termwise Higher Integral of csc x
log tan (x /2) of cscx is x =  /2 , we assume zeros of
the second or more order integral are also x =  /2 . Where, since the zero of cscx is i , these seem
Since the zero of the first order integral
to be a collateral higher integral.
5.5.0 Collateral Higher Integral of csc x

x
= log tan
csc x dx

2
x
  csc x dx
x


2
2
2
=

x
2
=
1
1- cosx
log
2
1+ cosx

non-elementary function
5.5.1 Termwise Higher Integral of Taylor Series of csc x
Formula 5.5.1
When B2k denote Bernoulli Numbers and functions
n 1
xn
=
f (x)
log x -Σ
n!
k=1 k
the following expressions hold for x >0 .

<n>

x

=f
are as follows
22k -2B2k 2k+n
+Σ
x
k=1 2k (2k +n )!

1

(x) x0!
2

<0>
csc x dx
2

(x) on 0< x < 
<n>
f

0
f<0>
n =1, 2, 3, 

 2
22k -2B2k 2k
= log x +Σ
x - log 2
2k (2k)!
k =1
x
1
1
 0 <1> 
xf
csc x dx 2 = f <1>(x) 

x


2
2
0!

x
x
x
  
3
csc x dx = f <2>(x) -

2
1

x2
0!


 2

0
 f  2
<2>
1!

x-

2
1

x1!
2

2 2 2
-


 2
<0>
f

1
 f  2
1

x2!
2

1
<1>

2
f

 2
<0>

 
x

x


2
2
n
csc x dx = f
<n -1>
n -1
(x) -Σ
k=0
1

x2
k!


k
n -1-k >
f<

 2
Where
f<0>

 2  = log 2
Proof
 csc x dx = f
x

<0>
(x) - f<0>

 2

22k -2B2k 2k
= log x +Σ
x
2k (2k)!
k=1
2


x

2
- 32 -
2k
22k -2B2k 2k
 2 -2B2k

= log x +Σ
x - log 2 +Σ
2k (2k)!
2k (2k) !
k=1
k =1


22k -2B2k 2k


= log x +Σ
- log
x - log
2
4
2k (2k)!
k=1





Σ
k=1
2k
2 -2B2k
2k (2k) !

 2

 2
2k


2k
= -log

4
5.3.1 (2)

22k -2B2k 2k
= log x +Σ
x - log 2
2k (2k)!
k=1


x
x


2
2

x
2
csc x dx =
f<0>(x) - f<0>


 2  dx
2

1

(x) x1!
2

<1>
= f

1
<0>
f

 2 
x

2
= f<1>(x) - f<1>

 2
-
1

x1!
2


1
f<0>

 2
Hereafter, in a similar way we obtain the desired expressions..
Although Formula 5.5.1 is a general expression anyway, if it is expand actually, it is very complicated.
Then, this was devised and was made the same easy expression as sec x. It is as follows.
Formula 5.5.1'
When
hold for
E0=1, E2=-1, E4=5, E6=-61, E8=1385, 
0< x < .
  cscx dx
x
x


2
2

E2k
k =0
(2k + n)!
=Σ
n

x - 2 
are Euler Numbers,the following expressions
2k+ n
Proof
csc x =
1
=
sin x

secx =Σ
1

cos x -
E2k
k=0
(2k)!

E2k
k=0
(2k)!

2

x 2k

= sec x -

2
|x|<


2
From these
cscx =Σ

x - 2 
2k
0< x < 
Integrating both sides of this with respect to x from  /2 to x repeatedly, we obtain the desired expressions.
(1) Termwise Higher Definite Integral of Taylor Series of csc x
Substituting
x =0
for Formula 5.5.1 , we obtain the following expressions.
- 33 -
  csc x dx
x
0


2 2
0 x
 
x
1
csc x dx = 0!



2
2 2
3

0

0
<1>

1
+
1!
<2>

1
+
1!
1
2!
 2 f  2
1
=0!
2
 2
f
 2

1

 2
1

2

 2 f  2
<0>
 2
f<1>
(1.2')

 2
f<0>

 2
(1.3')

   csc x dx
0
x
x



2
2
2
n
n -1
= -Σ
k=0
(-1)k
k!

 2
k
n -1-k >
f<

 2
Where
(1.n')
f<0>

 2  = log 2
(2) Taylor Series of the 1st order Integral of csc x
The 1st order Integral of csc x is as follows.
2k
x
 2 -2B 2k


csc x dx = log x +Σ
2k (2k)!
k=1
x 2k - log 2 = log tan
2
From this, the following expression follows immediately.
2k
 2 -2B 2k
x
2k
Σ
k=1
= log tan
x
- log
x
2
0< x < 
0< x < 
2
2k (2k)!
x =1 for this , we obtain the following special value.
Substituting
x
2
(1.t)
22k -2B2k
1
= log tan + log 2
Σ
2
2k (2k)!
k=1

5.5.2 Termwise Higher Integral of Fourier Series of csc x
Formula 5.5.2
(-1)k
Let  (n ) =Σ
be Dirichlet Beta Function and  be floor function. Then the following
n
k=0 (2k +1)
expressions hold for 0< x <  .
x

1
1
= 2Σ
csc x dx
sin (2k +1)x 2
k =0 2k +1




2
x
  csc x dx
x


2
2
   csc x dx
x
x
x
  

= 2Σ
2
k =0
3

= 2Σ
k =0

1
2
2 +1)x 2 sin ( k
2
(2k +1)

3
1
2
+1
(
)
k
x
sin
2
(2k +1)3

2 2 2
- 34 -

2(2)

+
x0!
2

2(2)

+
x1!
2




0
1
  csc x dx
x

x


2
2
 
x

x


2
2
4

= 2Σ
k=0

csc x dx 5 = 2Σ
k=0
1
4
2 +1)x 4 sin ( k
2
(2k +1)


2(2)
 2 2(4)

+
xx2!
0!
2
2
1
5
2 +1)x 5 sin ( k
2
(2k +1)





1

2(2)

+
x3!
2


0

3
2(4)

x1!
2


  csc x dx
x
x


2
2
n

= 2Σ
k=0
1
n
2
+1
(
)
k
x
sin
2
(2k +1)n

n /2
k-1
+ Σ (-1)
k=1

2 (2k)

x2
(n -2k)!


n-2k
Proof
csc x can be expanded to Fourier series in a broad meaning.as follows.
2i
2ie ix
= -2ie ix1 + e 2ix +
cscx = ix -ix = 2ix
e -e
1- e
e 4ix + e 6ix +
= -2 ie ix + e 3ix + e 5ix + e 7ix +
= 2(sin x + sin 3x + sin 5x ++) -2i(cosx + cos 3x + cos 5x + )
(2.0)
csc x is the locus of the median (central line) of this real number part. (See the figure)
However, calculus of the right side of (2.0) may be carried out, the real number does not turn into an imaginary number and the contrary does not exist, either. Therefore, limiting the range of csc x and the higher order
integral to the real number, we calculate only the real number part of the right side of (2.0). Although Dirichlet
Odd Beta is obtained from calculation of the imaginary number part, since it is long, it is omitted in this section
Now, integrate the real number part of both sides of (2.0) with respect to x from
- 35 -
 /2 to x .
Then

x

 2(sin x + sin 3x + sin 5x + sin 7x +)dx
x
csc x dx =
2

2


cos 1x cos 3x cos 5x cos 7x
+
+
+
+
= -2
1
3
5
7

x

2

cos 1x cos 3x cos 5x cos 8x
= -2
+
+
+
+
3
5
7
1
i.e.

x


cos(2k +1)x
2k +1
csc x dx = -2Σ
k=0
2



-cos x = sin x -
Here, let
x


. Then using this we obtain
2
1
1
sin (2k +1)x 2k +1
2


csc x dx = 2Σ
k =0

2
log tan(x /2) .
 /2 to x , then
This is consistent with the real number part of the Fourier series of
Next, integrate both sides of this with respect to x from

x
x


2
2
csc x dx 2 =
1
 2Σ 2k +1 sin (2k +1)x - 2  dx
x


1
k=0
2


= 2Σ
k=0
1
2
2
+1
(
)
k
x
sin
2
(2k +1)2


x

2

= 2Σ
k =0

 (n) =Σ
Here, let
k=0

x
x


2
2
1
2
2
+1
(
)
k
x
sin
2
(2k +1)2

(-1)k
(2k +1)n

k=0
Next, integrate both sides of this with respect to x from

x
x
x
  
csc x dx 3 =
k=0
(-1)k
(2k +1)2
, then we obtain
1
2
2
+1
(
)
k
x
sin
2
(2k +1)2

csc x dx 2 = 2Σ


+ 2Σ
 + 2(2)
 /2 to x , then
2
 2Σ (2k +1) sin (2k +1)x - 2  + 2(2) dx
x

2 2 2

1
2
k=0
2


= 2Σ
k=0

= 2Σ
k =0
1
3
2 +1)x 3 sin ( k
2
(2k +1)

3
1
2
+1
(
)
k
x
sin
2
(2k +1)3

Hereafter, in a similar way we obtain the desired expression..
- 36 -

 + 2(2)x - 2 

 + 2(2)x - 2 
x

2
(1) Termwise Higher Definite Integral of Fourier Series of csc x
Substituting
x =0
for Formula 5.5.2 , we obtain the following expressions.
2(2)
csc x dx =
0!


0

x
2 2
0 x
2
   csc x dx
x

 
2
2 2
3
   csc x dx
0

x
x
4


2 2
0 x
2
x

 2
(2.2')
2 (2)
1!

 2
1
=-

 2
2
=
2 (2)
2!

3

2 (2)
 csc x dx = 3!


2
2
 
0
5
+
 2
23-1
 (3)
22
(2.3')
2(4)
0!

 2
0
2(4)
+
1!

1
 2
(2.4')
25-1
 (5)
24
(2.5')
2

   csc x dx
0

2
x
x


2
2
n
n
= (-1)
n /2
k-1
(-1)
Σ
k =1
2(2k)
(n -2k)!

 2
n-2k
n 2n-1

- sin
 (n)
2 2n-1
(2.n')
Proof
Since
n
2
n
sin (2k +1)0 2


=0
 = 1
sin (2k +1)0 -
giving
x =0
x


2
2

csc x dx 2 = 2Σ
k=0
   csc x dx
0
for n =3, 5, 7, 
,
to Formula 5.5.2 and substituting these for them,

0
for n =2, 4, 6, 
x
x

 
2
2 2

= 2Σ
k=0
3
1
2
2 +1)02 sin ( k
2
(2k +1)


= 2Σ
k =0
 + 2(2) = 2(2)
1
3
2
+1
0(
)
k
sin
2
(2k +1)3

1
2(2)
3
1!
(2k +1)

 2
1

2 (2)

0+
1!
2
23-1
2 (2)
=
 (3) 2
1!
2
Hereafter, in a similar way we obtain the desired expressions..
(2) Fourier Series of the 1st order Integral of csc x
The 1st order Integral of csc x is as follows.
x
 cos(2k +1)x


csc x dx = -2Σ
k =0
2k +1
= log tan
2
From this, the following expression follows immediately.
- 37 -

x
2
0< x < 

 2

1
1

1
cos(2k +1)x
x
0< x < 
= - log tan
2k +1
2
2
Substituting x =1 for this, we obtain the following special values.
 cos(2k +1)
1
1
= - log tan
Σ
2
2k +1
2
k=0
Σ
k=0
(2.f)
5.5.3 Dirichlet Even Beta
Comparing Taylor series of csc x and Fourier series of csc x, we obtain Dirichlet Even Beta. Although the
general expression by the Euler Number is known about Dirichlet Odd Beta, the general expression of Dirichlet
Even Beta is not known. The following formula shows this by a little complicated Bernouilli series.
Formula 5.5.3
B2k , (x),  (x) denote Bernoulli Numbers, Dirichlet Beta Function, Riemann Zeta Function
<n>
(x) on 0< x <  are as follows
respectively, and functions f
2k
n 1
 2 -2B 2k
xn
<n>
+Σ
f (x)=
log x -Σ
x 2k+n
n =1, 2, 3,  ,
2
+
!
2
!
(
)
k k n
n
k=1 k
k=1
When


the following expressions hold.

1 1 (-1)k
 (2) = - Σ
2 k=0 k !
1 2 (-1)k
 (2) =
2Σ
k!
k=0
 2

 2
1 3 (-1)k
 (4) =
2Σ
k!
k=0
1 4 (-1)k
 (4) = - Σ
2 k=0 k !

1 5 (-1)k
 (6) = - Σ
2 k =0 k !
1 6 (-1)k
 (6) =
2Σ
k!
k=0
k
f <1-k >
k-1
1
2
2
k
<3-k >
 (2)


2
 2  + 2!  2 
(2) 

+
 2  3!  2  +
25-1  (5)

24


2
f
<4-k >
k
 (4)

2
 2  f  2  + 2!  2   (4) 


f
+
 2
 2  3!  2 
<5-k >
k-1
 (2)
 2
 (2) 
5!  2 
-
+
k
 2 f
(-1)n 2n (-1)k
 (2n) = 2 Σ
k!
k=0

 2

 2  -Σ(-1)
<2n-1-k >
n -1
k=1
k-1
f

 2  -Σ(-1)
<2n-k >
n -1
k
k =1
+
Proof

0
x

1
csc x dx = 0!

2
2
2

 2
0
<1>
f
k

 2
- 38 -
1
+
1!

 2
4
27-1 (7)
26


4
4!
2
<6-k >
(-1)n 2n -1 (-1)k
 (2n) =
2 Σ
k!
k=0

- f<1>
<2-k >
k-1

 log 2
 2  2
 2 
2 -1  (3)

+
 2 2 
=
3
f
 2 f

 2

 (2n -2k)


(2k) !
 2
2k
 (2n -2k)

2k
 
2
(2k +1) !
22n+1 -1 (2n +1)
1
f<0>
22n

 2

(1.2')

(-1)k
= -Σ
k!
k=0
x
2(2)
csc x dx 2 =
0!

2
2

k
 2 f
1
<1-k >

 2

0
1 1 (-1)k
  (2) = - Σ
2 k=0 k !
   csc x dx
0
x
x

 
2
2 2
x
 2
3-1
= -Σ
k =0
k
f <1-k >
(-1)k
k!

 2

 2 f
x


2
2 2
<3-1-k >

k-1
 2
f
<3-1-k >
 2

 2

 2
3
1 3-1 (-1)k
  (2) = Σ
2 k=0 k !
 log 2
1
2
=
k
2 (2)
23-1
 (3) csc x dx =
2
1!
2

 
0
3

(2.2')

 2
- f<1>

 2 
(1.3')
1
(2.3')
23-1  (3)
+

22
Hereafter, in a similar way we obtain the desired expressions.
5.5.4 Riemann Odd Zeta
Riemann Odd Zeta can be conversely obtained from the half of Formula 5.5.3 . The following formula shows
the relation between Riemann Odd Zeta and Dirichlet Even Beta by a little complicated expression.
Formula 5.5.4
B2k , (x),  (x) denote Bernoulli Numbers, Dirichlet Beta Function, Riemann Zeta Function
<n>
(x) on 0< x <  are as follows
respectively, and functions f
2k
n 1
 2 -2B 2k
xn
<n>
+Σ
f (x)=
log x -Σ
x 2k+n
n =1, 2, 3,  ,
n!
k=1 k
k=1 2k (2k +n )!
When


the following expressions hold.
3
2
 (2)
 (3) = 3
2 -1 1!
25
 (5) = 5
2 -1


 2
 (4)
1!
1
22 2 (-1)k
- 3 Σ
2 -1 k=0 k !

 2
 (2)
1
-
3!
24 4 (-1)k
+ 5 Σ
2 -1 k =0 k !
27
 (7) = 7
2 -1

(6)
1!

 2
1
-
26 6 (-1)k
- 7 Σ
2 -1 k=0 k !

 2
3!
 2

- 39 -
 2 f
<2-k >

 2
3
k
f <4-k >

 2

k
 2 

 (4)

 2
3
+
k
f <6-k >

 (2)
5!

 2

 2 
5
 (2n +1) =
22n+1
n -1
22n+1 -1
n
+ (-1)
(-1)
Σ
k=0
22n
22n+1 -1
k
2n
(2n -2k)
Σ
k=0
(2k +1)!
(-1)k
k!
- 40 -


 2
 2
2k+1
k
f <2n -k >

 2
5.6 Termwise Higher Integral of csch x
csch x
log tanhx /2
x = , the latter seems to be
a lineal primitive function with a fixed lower limit. On the other hand, the Fourier series of csch x is known for
x >0 . Then, we assume zeros of the second or more order primitive function are also x = .
Since both zeros of
and the primitive function
are
5.6.0 Higher Integral of csch x

 csch x dx
x
= log tanh
csch x dx
x
x
2
=
x
2
non-elementary function
5.6.1 Termwise Higher Integral of Taylor Series of csch x
The Taylor series of
csch x
is known as follows.
2k
 2 -2 B 2k
1
-Σ
csch x =
x 2k-1
x k =1 2k (2k -1)!
However, this can not be integrated with the lower limit
0< x < 
x = .
5.6.2 Termwise Higher Integral of Fourier Series of csch x
Formula 5.6.2
The following expressions hold for
x >0 .
 csch x dx = -2Σ
  csch x dx = 2Σ (e2k +1)
   csch x dx = -2Σ (e2k +1)

x
x
2
 
x
k =0

-(2k +1)x
2
k =0
x
x
3
  
x
e -(2k +1)
2k +1

x

-(2k +1)x
k =0
3

  csch x dx
x

x

n

= (-1) 2Σ
n
k=0
x
e -(2k +1)
(2k +1)n
Proof
csch x can be expanded to a Fourier series as follows. (See the figure of the following page.)
2
2e -x
=
=
= 2e -x1 + e -2x + e -4x + e -6x +
csch x
-x
-2x
x
e -e
1- e
= 2e -1x + e -3x + e -5x + e -7x +
(2.0)
= 2(cosh 1x +cosh 3x +cosh 5x +) - 2(sinh 1x +sinh 3x +sinh 5x +)
= 2(cosh ix +cosh 3ix +cosh 5ix +)- 2i(sinh ix +sinh 3ix +sinh 5ix +)
sinh x = -isin ix 
  cosh x = cosix ,
- 41 -
Integrate the both sides of (2.0) with respect to x from

x

 2e
x
csch x dx =
-1x

= -2


to x , then
+ e -3x + e -5x + e -7x +dx

e -x e -3x e -5x e -7x
+
+
+
+
1
3
5
7
x

i.e.

x

x
e -(2k +1)
2k +1

csch x dx = -2Σ
k=1
This is consistent with the real number part of the Fourier series of
Next, integrate both sides of this with respect to x from
  csch x dx =  -2Σ
x
x
x
2
 


k=0
x
e -(2k +1)
2k +1

log tanh
x
2
.
to x , then
 dx = 2Σ

k =0
x
e -(2k +1)
(2k +1)2

i.e.
  csch x dx
x
x
2
 

= 2Σ
k=0
x
e -(2k +1)
(2k +1)2
Hereafter, in a similar way we obtein the desired expression.
5.6.3 Dirichlet Lambda Function
Substituting
x =0
for Formula 5.6.2 , we obtain Dirichlet Lambda Function immediately.
Formula 5.6.3

When
(n) =Σ
k=0
1
(2k +1)n
, the following expressions hold.
  csch x dx = 2(2)
   csch x dx = -2(3)
0
x
 
0 x
2
x
3
  
- 42 -
x


   csch x dx
0
x
 

x

n
= (-1)n2 (n)
5.6.4 Riemann Zeta Function
Applying
(n) =
2n
2n - 1
(n) for Formula 5.6.3 , we obtain Riemann Zeta Function.
Formula 5.6.4
2n-1
 (n) = (-1) n
2 -1
n
   csch x dx
0
x
 
x

- 43 -
n
5.7 Termwise Higher Integral of sec x
1
1+ sin x
of secx is x = 0 , we assume the zeros of
log
2
1- sin x
the second or more order integral are also x = 0 . Where, since the zero of secx is i , these seem
Since the zero of the first order integral
to be a collateral higher integral.
5.7.0 Collateral Higher Integral of sec x

 secx dx
x
1
1+ sin x
log
2
1- sin x
sec x dx =
0
x x
2
=
non-elementary function
0 0
5.7.1 Termwise Higher Integral of Taylor Series of sec x
Formula 5.7.1
E0=1, E2=-1, E4=5, E6=-61, E8=1385, 
When
hold for
are Euler Numbers, the following expressions
|x|<  /2 .

 sec x dx
 sec x dx

x
=Σ
sec x dx
(2k +1)!

E2k
=Σ
x 2k+2
k =0 (2k +2)!

E2k
=Σ
x 2k+3
k =0 (2k +3)!
2
0 0
x x x
x 2k+1
k =0
0
x x
E2k
3
0 0 0

 
x

x
0
0

E2k
k =0
(2k + n)!
secx dx n =Σ
x 2k+
n
Proof
secx
can be expanded to Taylor series as follows.

secx =Σ
k=0
E2k
(2k)!
x 2k
|x| <

2
Integrating both sides of this with respect to x from 0 to x repeatedly, we obtain the desired expression.
(1) Termwise Higher Definite Integral of Taylor Series of sec x
Substituting

x =  /2
  sec x dx
2 x
0
for Formula 5.7.1 , we obtain the following expressions.
0
E2k
k=0
(2k +2)!

  sec x dx
2 x x
0
0 0
3


=Σ
2

E2k
k =0
(2k +3)!
=Σ
 2

 2
2k+2
(1.2')
2k+3

- 44 -
(1.3')

  secx dx
2
0
x
0
x
n
0


E2k
k =0
(2k + n)!
=Σ
 2
2k+ n
(1.n')
(2) Taylor Series of the 1st order Integral of sec x
The 1st order Integral of sec x is as follows.
x

E2k
2k+1
 sec x dx =Σ (2k +1)! x
=
k=0
0
1
1+ sin x
log
2
1- sin x
|x|<

2
From this, the following expression follows immediately.
E2k

x
Σ
k=0 (2k +1)!
2k+1
=
1
1+ sin x
log
2
1- sin x
|x|<

2
(1,t)
x =1 for this , we obtain the following special value.
1
1+ sin 1
E2k
= log
2
1- sin 1
(2k +1)!
Substituting

Σ
k=0
Reference : Euler Number , Tangent number, Bernoulli Number
Euler Number is defined as follows.
Ek k

|x| <
x
2
k=0 k !
E2n-1 = 0 n =1, 2, 3,  .

sech x =Σ
As a result,
When Euler numbers , Tangent numbers and Bernoulli Numbers (all non-zero) are listed, it is as follows.
E6=-61,
E8=1385, E10=-50521, 
T1=1, T3=2, T5=16,
T7=272,
1
1
1
, B 6=
,
B0=1, B2= , B4=6
30
42
T9=7936, T11= 353792, 
1
5
B8=, B10=
,
30
66
E0=1, E2=-1, E4=5,
According to Mr. Sugimoto ( http://homepage3.nift/y_sugi/gf/gf17.htm ) , the following equations holds
between Euler Numbers and Tangent Numbers.
n
E2n =Σ(-1)k
k =1


2n
+1
T
2k -1 2k-1
n =1, 2, 3, 
On the other hand, between Tangent numbers and Bernoulli Numbers, the following equations hold.
2n
2n
n-1 2 2 -1
T2n-1 = (-1)
B2n
n =1, 2, 3, 
2n
(1)
(2)
Therefore, the following equations have to hold between Euler Numbers and Bernoulli Numbers.
2k
2k
2n
n
E2n = 1 -Σ
k =1
2 2 -1
B2k
2k
2k -1


n =1, 2, 3, 
(3)
In fact, for example,

 
 
 
43 1 6
1615 1 6
6463 1 6
+
=1 2 6  1 4 30  3 6 42  5  = -61
E6 = 1 -
23
23
23
2424-1
2626-1
2222-1
B2
B4
B6
+
+
4
6
2
1
3
5
On the contrary, the following equations between Bernoulli Numbers and Euler Numbers are shown by
Mr. Nishimura. ( http://www1.ocn.ne.jp/~oboetene/zeta.pdf ).
- 45 -
B2n =


n -1 2n -1
2n
E2k
Σ
2k
22n22n-1 k =0
n =1, 2, 3, 
(4)
The following example shows the formula is right.
   2  
5
1
5 =
      4  42
23-1
23
B6 =
23-1
E0 +
0
223223-1
5
5
23
11+
=
0
2
6463
E 
23-1
E2 +
4
4
In addition, Mr. Nishimura expressed Even Zeta with the Euler numbers using this equation.
5.7.2 Termwise Higher Integral of Fourier Series of sec x
Formula 5.7.2
When
E2k ,  (n) , 
denote Euler Numbers, Dirichlet Beta Function, floor function respectively,
the following expressions hold for |x |<  /2 .
1
(-1)k
cos (2k +1)x 2
k=0 2k +1
0
x x

2
(-1)k
2
= 2Σ
2
+1
(
)
secx dx
cos
k
x
2
2
k =0 (2k +1)
0 0
k
x x x

3
(-1)
2 +1)x sec x dx 3 = 2Σ
3 cos ( k
2
k=0 (2k +1)
0 0 0




x
= 2Σ
sec x dx
 
x

0
x

sec x dx 4 = 2Σ
k =0
  sec x dx
0
x
5
0

= 2Σ
k=0

  sec x dx
x
0
x
0



0
x

n

= 2Σ
k =0


2 (2) 0
x
0!
2 (2) 1
+
x
1!
+
4
(-1)k
2 +1)x 4 cos ( k
2
(2k +1)
2 (2) 2 2 (4) 0
+
x x
2!
0!


5
(-1)k
2 +1)x 5 cos ( k
2
(2k +1)
2 (2) 3 2 (4) 1
+
x x
3!
1!


(-1)k
n
2
+1
(
)
cos
k
x
2
(2k +1)n

n /2

+ Σ (-1)k-1
k=1
2 (2k) n-2k
x
(n -2k)!
Proof
sec x can be expanded to Fourier series in a broad meaning.as follows.
2
2e ix
= 2e ix - e 3ix + e 5ix secx = ix -ix =
2ix
1+ e
e +e
= 2(cosx - cos 3x + cos 5x - cos 7x +-)
+ 2i(sin x - sin 3x + sin 5x - sin 7x +-)
- 46 -
e 7ix +-
(2.0)
sec x is the locus of the median (central line) of this real number part. (See the figure)
y
40
30
20
10
-1.4 -1.2 -1.0 -0.8 -0.6 -0.4 -0.2 0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
x
1/cos(x)
2*cos(x) - 2*cos(3*x) + 2*cos(5*x) - 2*cos(7*x) +
However calculus of the right side of (2.0) may be carried out, the real number does not turn into an imaginary
number and the contrary does not exist, either. Therefore, limiting the range of sec x and the higher order
integral to the real number, we calculate only the real number part of the right side of (2.0) .
Although the Dirichlet Odd Beta is obtained from calculation of the imaginary number part, since it is long, it is
omitted in this section.
Now, integrate the real number part of both sides of (2.0) with respect to x from 0 to x, then
x
x
 sec x dx =  2(cosx - cos 3x + cos 5x - cos 7x +-)dx
0
0




sin 1x sin 3x sin 5x sin 8x
+
+-
=2
1
2
3
4
sin 1x sin 3x sin 5x sin 8x
=2
+
+-
1
2
3
4
x
0
i.e.
 sec x dx = 2Σ

x
k =0
0
Here, since
(-1)k
sin (2k +1)x
2k +1 

sin x = cos x -
 sec x dx = 2Σ

x
k =0
0

2

, we obtain
1
(-1)k
cos (2k +1)x 2k +1
2


This is consistent with the real number part of the Fourier series of
1
1+ sin x
log
2
1- sin x
Next, integrate both sides of this with respect to x from 0 to x , then

x x
0 0
1
(-1)k
2Σ
secx dx =
cos (2k +1)x 2
k =0 2k +1
0

2
(-1)k
= 2Σ
2
+1
(
)
cos
k
x
2
2
k =0 (2k +1)

x
2



 dx


- 47 -
x
0
.

= 2Σ
k =0
(n) =Σ
k =0
 secx dx
x x

(-1)k
(2k +1)n

Here, let
2
(-1)k
2
+1
(
)
cos
k
x
2
(2k +1)2
= 2Σ
k =0
0 0
k =0
(-1)k
cos(-)
(2k +1)2
, then we obtain

2
 - 2Σ

2
(-1)k
2 +1)x 2 cos ( k
2
(2k +1)


+
2 (2) 0
x
0!
Next, integrate both sides of this with respect to x from 0 to x , then

x x x
3
2 (2) 0
2
(-1)k
+
2
+1
(
)
cos
x dx
k
x
2
0!
2
k=0 (2k +1)
0
x

2 (2) 1
3
(-1)k
+
2 +1)x = 2Σ
x
3 cos ( k
1!
2
k =0 (2k +1)
0
k

2 (2) 1
3
(-1)
+
2 +1)x = 2Σ
x
3 cos ( k
1!
2
k =0 (2k +1)
0 0 0

x
sec x dx =


2Σ








Hereafter, in a similar way , we obtain the desired expressions.
(1) Termwise Higher Definite Integral of Fourier Series of sec x
Substituting
x = /2
for Formula 5.7.2 , we obtain the following expressions.

  secx dx
2 x
0
2
0

 
2 x x
0
0 0

2 x
x
0


2 x
0 0
x
0

 2
2 (2)
sec x dx =
1!
3

0 0
2 (2)
=
0!
 2
2 (2)
sec x dx =
2!
4

(2.2')
1

 2
2 (2)
sec x dx =
3!
5
0

23-1
 (3)
22
2
 2
(2.3')
2 (4)
0!

 2
0
2 (4)
1!

1
3
 2
(2.4')
25-1
+
 (5)
24
(2.5')


 
2 x

0 0
0
x
n /2
sec x dx = Σ (-1)
n
k =1
k-1
2 (2k)
(n -2k)!

 2
n-2k
+ sin
n 2n-1

 (n)
2 2n-1
(2.n')
Proof
Since
n
2 2
 n
cos (2k +1) 2 2


cos (2k +1)

-
=0
 = (-1)
for n =2, 4, 6, 
k+1
- 48 -
for
n =1, 3, 5, 
k =0, 1, 2, 
,
giving
x = /2 to
Formula 5.7.2 and substituting these for them,

  secx dx
2 x
0


= 2Σ
2
k=0
0
  sec x dx
2 x x
0
3
0 0
(-1)k
 2

2
+1
(
)
cos
k
2 2
(2k +1)2

= 2Σ
k=0

 + 2(2) = 2(2)
(-1)k
 3
2 +1) 3 cos ( k
2 2
(2k +1)

2 (2) 
(-1)k (-1)k+1
+
2
1!
k=0
(2k +1)3
23-1
2 (2)  1
= - 2  (3) +
2
1!
2

 

= 2Σ
2 (2)
+
1!

 2
1
1
 
Hereafter, in a similar way we obtein the desired expressions.
(2) Fourier Series of the 1st order Integral of sec x
The 1st order Integral of sec x is as follows.
x

k sin(2k +1)x
 sec x dx = 2Σ(-1)
2k +1
k =0
0
=
1
1+ sin x
log
2
1- sin x

|x|<
2
From this, the following expression follows immediately.

(-1)k
Σ
k=0
Substituting
x =1

(-1)k
Σ
k=0
1
1+ sin x
sin(2k +1)x
= log
2k +1
4
1- sin x
|x|<

(2.f)
2
for this, we obtain the following special value.
1
1+ sin 1
sin(2k +1)
= log
2k +1
4
1- sin 1
5.7.3 Dirichlet Even Beta
Comparing Taylor series of sec x and Fourier series of sec x, we obtain Dirichlet Even Beta. Although the
general expression by the Euler Number is known about Dirichlet Odd Beta, the general expression of
Dirichlet Even Beta is not known. The following formula shows this by the easy Euler series.
Formula 5.7.3
When
E2k ,  (x),  (x) denote Euler Numbers, Dirichlet Beta Function,
Riemann Zeta Function
respectively, the following expressions hold.
(2) =
(2) =
1  E2k
2Σ
k =0 (2k +2)!
1  E2k
2Σ
k =0 (2k +3)!
1  E2k
(4) = - Σ
2 k=0 (2k +4)!
1  E2k
(4) = - Σ
2 k=0 (2k +5)!
(6) =
1  E2k
2Σ
k =0 (2k +6)!

 2
2k+2

 2
2k+2

 2
2k+4

2k+4
 2

 2
23-1  (3)
+

22
+
+
2k+6
+
- 49 -
(2)

2!
 2
2
(2)

3!
 2
2
(4)

2
2!
 2
25-1  (5)
+

24
-
(2)
4!

 2
4

1  E2k
2Σ
k =0 (2k +7)!
(6) =
 2
(4)
2k+6
+
3!

 2
-

(-1)n-1 
E2k
Σ
2
k=0 (2k +2n ) !
 2
(-1)n-1 
E2k
 (2n) =
Σ
2
k =0 (2 k +2 n +1) !
 (2n -2k)
k=1
(2k) !
+Σ(-1)k-1
 2
2k+2n+1
n
+Σ(-1)k-1

  sec x dx
2 x
0

0
  secx dx
2 x
0
2
E2k
k=0
(2k +2)!
=Σ
2


 2
1  E2k
  (2) = Σ
2 k =0 (2k +2)!
  sec x dx
0
0 0
 2
2k

 2
(2k +1)!

22n+1-1 (2n +1)

22n
2k+2
(1.2')
 2
2k+2


E2k
k =0
(2k +3)!
=Σ
2k
(2.2')

3

2 (2)
0!
=
0
2 x x
4
 (2n -2k )
k =1
+
Proof
 
n
2k+2n


2
5!
7
2 -1  (7)
+

26

 (2n) =
(2)
2
 2
2k+3
(1.3')

23-1
2 (2)  1
sec x dx = - 2  (3) +
2
1!
0 0 0
2
1  E2k
 2k+2 23-1  (3)
  (2) = Σ
+
2 k=0 (2k +3)! 2

22
 
2 x x
 
3
(2.3')
 
Hereafter, in a similar way we obtein the desired expressions.
c.f.
Mr.Sugioka expressed Dirichlet Even Beta by definite Reamann Odd Zeta and infinite Reiamnn Even Zeta
using his " Taylor System ".
22-1
 (4) =
-
22
3
22
+
 2 3!
1
24
5
4
2
(3)
log 2
24-1
 (6) =
 (4),  (6) which he calculated are as follows.
(5)
 2 5!
log 2
1
1
1!
-
 2
22-1
22
1
1!
-


2
2
1!
(2)
 2
22-1
1
22 5!
1
-
(2)
-
22-1
22
1!
2
24-1
24
(3)
-
1
3!
24-1
2 7!
- 50 -
4
2
(4)
3!
24 7!

 2
(4)
26-1
26
(6)
5!
26 9!

-
3
3!
4
-
2 9!
-
26-1
6
2
(6)
5!
6
2 11!

-
5.7.4 Riemann Odd Zeta
Riemann Odd Zeta can be conversely obtained from the half of Formula 5.7.3 . The following formula shows
the relation between Riemann Odd Zeta and Dirichlet Even Beta by the easy expression.
Formula 5.7.4
When
E2k , (x),  (x) denote Euler Numbers, Dirichlet Beta Function, Riemann Zeta Function
respectively, the following expressions hold.
23  (2)
 (3) = 3
2 -1 1!
25
 (5) = 5
2 -1


 2
(4)
1!
1
22  E2k
- 3 Σ
2 -1 k=0 (2k +3)!

 2
(2)
1
-
3!


 2
2k+3
 2 
3
24  E2k
+ 5 Σ
2 -1 k =0 (2k +5)!
27
 (7) = 7
2 -1

(6)
1!

 2
(4)
1
-
3!

 2
3
+
(2)
5!


 2
2k+5

2k+7
 2 
5
26  E2k
- 7 Σ
2 -1 k =0 (2k +7)!
 2

 (2n +1) =
22n+1
22n+1 -1
n -1
(-1)
Σ
k=0
k
(2n -2k)
(2k +1)!
n
+ (-1)
22n
22n+1 -1
- 51 -

 2
2k+1
E2k

Σ
k=0 (2k +2n +1)!  2 

2k+2n+1
5.8 Termwise Higher Integral of sech x
Since both zeros of
sech x
and the primitive function
2tan e x-
-1
are
x = ,
the latter seems to be
a lineal primitive function with a fixed lower limit. Then, we assume zeros of the second or more order primitive
x = .
function are also
5.8.0 Higher Integral of sech x
 sech x dx = 2tan e  - 
 sech x dx = non-elementary function
x
-1

x x
x
2
 
5.8.1 Termwise Higher Integral of Taylor Series of sech x
When
E2k denote Euler Numbers, the Taylor series of sech x is known as follows.

sech x =Σ
k=0
E2k 2k
x
(2k)!
|x|<

2
However, this can not be integrated with the lower limit
x = .
5.8.2 Termwise Higher Integral of Fourier Series of sech x
Formula 5.8.2
The following expressions hold for
x >0 .
 sech x dx
  sech x dx = 2Σ(-1) (e2k +1)
e
=
-1
(
)
-2
sech
x
dx
Σ

(2k +1)

x

x
= -2Σ(-1)k
k =0
x
 
x
-(2k +1)x

2
k
2
k=0
x
x
3
  
x
e -(2k +1)
2k +1
-(2k +1)x

k
k=0
3

 
x

x

x
e -(2k +1)
sech x dx = (-1) 2Σ(-1)
k=0
(2k +1)n
n
n

k
Proof
sech x can be expanded to a Fourier series as follows. (See the figure of the following page.)
2
2e -x
=
=
= 2e -x 1 - e -2x + e -4x - e -6x +-
sech x
-x
-2x
x
e +e
1 +e
= 2e -1x - e -3x + e -5x - e -7x +-
= 2(cos 1ix - cos 3ix + cos 5ix - cos 7ix +- )
+ 2i(sin 1ix - sin 3ix + sin 5ix - sin 7ix +- )
- 52 -
(2.0)
y
1.0
0.8
0.6
0.4
0.2
0.0
0
1
2
3
4
5
x
1/cosh(x)
2*(exp(-x) - exp(-201*x))/(exp(-2*x) + 1)
Integrate the both sides of (2.0) with respect to x from

x

 2e
x
sech x dx =
-x

= -2


to x , then
- e -3x + e -5x - e -7x +-dx

e -x e -3x e -5x e -7x
+
+-
1
3
5
7
x
0
i.e.

x

x
e -(2k +1)
sech x dx = -2Σ(-1)
2k +1
k =0

k
This is consistent with the real number part of the Fourier series of
Next, integrate both sides of this with respect to x from

x x
 


x
-(2k +1)
2
k e
sech x dx = 2Σ(-1)
k=0
(2k +1)2

2tan e x-
-1
.
to x , then

x

x
e -(2k +1)
= 2Σ(-1)
k=0
(2k +1)2

k
Hereafter, in a similar way we obtein the desired expressions.
5.8.3 Dirichlet Beta Function
Substituting
x =0
for Formula 5.8.2 , we obtain Dirichlet Beta Function immediately.
Formula 5.8.3

When
(n) =Σ
k=0
(-1)k
(2k +1)n
is the Dirichlet Beta Function, the following expressions hold.
 sech x dx = -2(1)
  sech x dx = 2(2)
   sech x dx = -2(3)
0

0
x
 
0 x
2
x
3
  
- 53 -

   sech x dx
0
x
 

x

n
= (-1)n2 (n)
Note
1
1
1
1

- 1 + 1 - 1 +- =
1
4
1
3
5
7
1
1
1
1
(2) = 2 - 2 + 2 - 2 +- = 0.9159655941
3
5
7
1
(1) =
: Madhava series
: Catalan's constant
2006.10.05
2012.05.29 Updated
K. Kono
Alien's Mathematics
- 54 -