14 Higher and Super Calculus of Logarithmic Integral etc.
14.1 Higher Integral of Exponential Integral
Exponential Integral is defined as follows.
x t
Ei(x) =

-
e
dt
t
(1.0)
Integrating both sides of (1.0) with respect to x repeatedly by ONLINE INTEGRATOR (Wolfram Mathematica)
and arranging the results, we obtain the following higher indefinite integrals.
Ei(x)dx = 1! xEi(x)- e 0!
1
Ei(x)dx = 2! x Ei(x)-e (0!x +1!)
1
Ei(x)dx = 3! x Ei(x)-e 0!x +1!x +2!
1
x
2
x
2
3
x
3
2


 Ei(x)dx n =
n -1
1
n-1- r
x nEi(x)-e xΣr! x
n!
r=0


Although these right sides are the lineal primitive functions of Ei(x), since both zeros of Ei(x) and
ex
are
- ,
zeros of the right sides are all - . Therfore, the lineal higher primitive function of Ei(x) can be expressed by
the higher integral with a fixed lower limit - .
Formula 14.1.1
When Exponential Integral is
 
x

-
x
Ei(x)dx n =
-

Ei(x)=
x
-
et
dt ,
t
the following expressions hold.
n -1
1
n-1- r
x nEi(x)-e xΣr! x
n!
r=0


Example : 3rd order integral of Ei (x)
-1-
(1.n)
-2-
14.2 Higher Integral of Cosine Integral
Cosine Integral is defined as follows.
x cos t

Ci(x)=

t
dt
Integrating both sides of this with respect to x repeatedly by ONLINE INTEGRATOR and arranging
the results, we obtain the following higher indefinite integrals.
Ci(x)dx
Ci(x)dx
Ci(x)dx
Ci(x)dx
1
1
x Ci(x) - 0!sin x
1!
1
2
2
=
x Ci(x) - 0!xsin x + 1!cos x
2!
1
3
3
2
=
x Ci(x) - 0!x -2!sin x + 1!x cos x
3!
1
4
4
3
2
=
x Ci(x) - 0!x -2!xsin x + 1!x -3!cos x 
4!
=


 Ci(x)dx n
=

(n -1)/2
1
Ci(x)x n - sin x Σ (-1)r(2r)! x n-1-2r
n!
r=0

(n -2)/2
+ cos x
Σ
r=0
(-1)r(2r +1)! x n-2-2r
Although these right sides are the lineal primitive functions of Ci(x), these zeros are all  . (See the above
figure.) Therfore, the lineal higher primitive function of Ci(x) can be expressed by the higher integral with a
fixed lower limit
.
Formula 14.2.1
When

 is floor function and Ci(x)=
 Ci(x)dx
 Ci(x)dx
x

x

x

=
n
=

x
cost
dt ,is Cosine Integral,
t
the following expressions hold.
1
1
x Ci(x) - 0!sin x
1!
(n-1)/2
1
n-1-2r
n
)
Ci(x x
sin x Σ (-1)r(2r)! x
!
n
r=0

-3-
(n-2)/2
+ cosx
Σ
r=0

(-1)r(2r +1)! x n-2-2r
n 2
Example : 4th order integral of Ci (x)
If both sides are illustrated, it is as follows. Since both sides overlap exactly, the left side (blue) is not visible.
-4-
14.3 Collateral Higher Integral of Sine Integral
Sine Integral is defined as follows.
x sin t

Si(x)=
t
0
dt
Integrating both sides of this with respect to x repeatedly by ONLINE INTEGRATOR and arranging
the results, we obtain the following higher indefinite integrals.
Si(x)dx
Si(x)dx
Si(x)dx
Si(x)dx
2
1
1
x Si(x) + 0!cos x
1!
1
2
=
x Si(x) + 0!x cos x + 1!sin x
2!
1
3
3
2
=
x Si(x) + 0!x -2!cos x +1! x sin x
3!
1
4
4
3
2
=
x Si(x) + 0!x -2!xcos x + 1!x -3!sin x
4!
=


 Si(x)dx n
=

(n -1)/2
1
Si(x)x n + cos x Σ (-1)r(2r)! x n-1-2r
n!
r=0
(n -2)/2
+ sin x
Σ
r=0

n-2-2r
(-1)r(2r +1)! x
Although these right sides are the lineal primitive functions of Si(x), those zeros are all 0 at the time of even
order and are not 0 at the time of odd order. That is, the lineal higher primitive function of Si(x) can not be
expressed by the higher integral with a fixed lower limit.(See the above figure.) Therfore, the higher integra
of Si(x) with a fixed lower limit 0 is not lineal but collateral. However, the idea which makes 0 a common lower
limit is natural. It is because the Si(x) itself is defined by the integral with a lower limit 0.
Collateral Higher Integral of Sine Integral
Collateral Higher Integrals of Si(x) are obtained by compensating the above lineal higher primitive functions
with Constant-of-integration Polynomials .
Formula 14.3.1
When

 is floor function and Si(x)=
0
x
sin t
dt ,is Sine Integral,
t
-5-
the following expressions hold.

1
x
Si(x)dx = 2! x Si(x) + 0!x cos x + 1!sin x - 11!
1
x
x
+
 Si(x)dx = 3! x Si(x) + 0!x -2!cos x +1! x sin x - 12!
30!
1
)

Si(
x
dx
=
x Si(x) + 0!x -2!xcos x + 1!x -3!sin x

4!
0
x
x
1
1
=
x Si(x) + 0!cos x 1!
10!
Si(x)dx
0
1
x x
2
2
0 0
2
x x x
3
3
0
2
0 0 0
x
0
x
4
4
3
2
0
3
1
x
x
+
13! 31!


x

0
0
x
Si(x)dx n
=
(n-1)/2
1
n-1-2r
Si(x)x n + cosx Σ (-1)r(2r)! x
!
n
r=0


(n-2)/2
+ sin x
(-1)r(2r +1)! x n-2-2r
Σ
r=0
-
Σ
r=0
n-1-2r
x
(-1)
(2r +1)(n -1-2r)!
(n-1)/2
r
Example : Collateral the 4th order integral of Si (x)
If both sides are illustrated, it is as follows. Since both sides overlap exactly, the left side (blue) is not visible.
-6-
14.4 Higher Integral of Logarithmic Integral
Logarithmic Integral is defined as follows.
x 1

li(x)=
0
log t
dt
(1.0)
6
4
2
2
4
6
8
10
-2
-4
-6
First, we prepare two Lemmas.
Lemma 14.4.1
When Exponential Integral is

x
Ei(x)=
-
et
dt ,
t
the following expressions hold.
Ei(2log x)dx = xEi(2log x) - Ei(3log x)
Ei(3log x)dx = xEi(3log x) - Ei(4log x)

Ei(n log x)dx = xEi(n log x) - Ei (n +1)log x


(1.n)
Proof
Let
t
2
t
1
x
dx = dt = e 2 dt .
2
2
2log x = t . Then x = e ,

1
Ei(2log x)dx =
2
Ei(t)e
t
2
Hence
dt
Calculating the integral of the right side by ONLINE INTEGRATOR, we obtain
t
t
3
2
2
Ei(t)e
dt = 2Ei(t)e - 2Ei
 2 t
Using this,

t
2
Ei(2log x)dx = Ei(t)e - Ei
Next, let
  = xEi(2log x) - Ei(3log x)
3
t
2
3log x = t . Then we obtain the following expresssion by the same calculation.
Ei(3log x)dx = 
t
1
1
Ei(t) e 3 dt =
3
3
-7-
Ei(t)e
t
3
t
3
dt = Ei(t)e - Ei
 
4
t
3
= xEi(3log x) - Ei(4log x)
Hereafter, by induction, we obtain the desired expresson.
Note
log x -
Since
at the time
x +0 , x =0
is clearly a zero of these functions. Then, (1.n) can be
written as follows.
x
 Ei(n log x)dx = xEi(n log x) - Ei (n +1)log x


(1.n')
0
Lemma 14.4.2

et
When Exponential Integral is Ei(x)=
dt , the following expressions hold.
- t
1
1 n+1
x nEi(log x)dx =
x Ei(log x)Ei(n +2)log x
n +1
n +1
x

(2.n)
Calculation
Calculating by ONLINE INTEGRATOR , we obtain (2.n) immediatly.
Note
log x -
Since
at the time
written as follows.
x
n

x +0 , x =0
is crealy a zero of these functions. Then, (2.n) can be
1
1 n+1
x Ei(log x)Ei(n +2)log x
n +1
n +1
x Ei(log x)dx =
0
(2.n')
Formula 14.4.3
When

li(x)=
0
x
1
dt
log t
,

x
0
x
li(x)dx n =
0
Ei(x)=
-
the following expression holds for


x
et
dt
t
x 0 .
1 n
(-1)r nＣ r x n- r Ei(r +1)log x
Σ
n ! r=0
Proof
Let
t = log x . Then [0 , x][- , t] , dx = e tdt . Hence
x 1
t et
dx =
dt = [Ei (t )]t- = Ei (log x ) = li(x)
log
x
t
0
-


Next, let

x x
0 0
1
dx 2 =
log x
 li(x)dx
x
0
Calculating the integral of the right side by ONLINE INTEGRATOR, we obtain
li(x)dx = xli (x )- Ei (2log x )
-8-
(3.n)
Since the zero of this right side is x=0 obviously,
x
 li(x)dx = xEi(log x) - Ei(2log x)
0
Next, integrating both sides of this with respect to x and applying Lemma 14.4.1 , 14.4.2
to the result, we obtain the following
x x
x
x
2
 li(x)dx =  xEi(log x)dx -  Ei(2log x)dx
0 0
0
0
1
1 2
x Ei(log x)- Ei(3log x)- xEi(2log x)- Ei(3log x)
2
2
1
= x 2Ei(log x)-2xEi(2log x)+Ei(3log x)
2
=
Next, integrating both sides of this with respect to x and applying Lemma 14.4.1 , 14.4.2 to the result,
we obtain the following

x x x
3
li(x)dx =
0 0 0
1
2

x
2

x Ei(log x) dx -
0
0
x
x Ei(2log x) dx +
1
2
 Ei(3log x)dx
x
0
1
1 3
1
1
x Ei(log x)Ei(4log x)- x 2Ei(2log x)+ Ei(4log x)
3!
3!
2
2
1
1
+ xEi(3log x) - Ei(4log x)
2
2
1
3
2
=
x Ei(log x)-3x Ei(2log x)+3xEi(3log x)-Ei(4log x)
3!
=
Hereafter, by induction, we obtain the desired expresson .
Example : 2nd order integral of li(x)
-9-
14.5 Higher Integral of Double Logarithmic Function
Double Logarithmic Function is defined as follows.
f(x) = log|log x|
(1.0)
Integrating both sides of this with respect to x repeatedly and arranging the results, we obtain the following
higher indefinite integrals. Where,
li(x)=Ei(log x)
is Logarithmic Integral mentioned in the previous.
log|log x|dx = 1! xlog|logx| - li(x)
1
log|log x|dx = 2! x log|logx| -2xli(x)+ Ei(2log x)
1
=
|
|
log
log
x
dx
x log|logx| -3x li(x)+3xEi(2log x)- Ei(3log x)

3!
1

2

2
3
3
2


 log|log x|dx n =
n
1
n-r
x nlog|logx| + Σ(-1)rnＣ r x Ei(rlog x)
n!
r=1


Although these right sides are the lineal primitive functions of
log|logx | , since both zeros of x nlog|logx |
Ei(nlog x ) are 0 , zeros of the right sides are all 0 . Therfore, the lineal higher primitive function of
log|logx | can be expressed by the higher integral with a fixed lower limit 0 .
and
Formula 14.5.1
When

x
Ei(x)=
-

x

0
et
dt ,
t
x
the following expressions hold for
log|log x|dx n =
0
x 0 .
n
1
n-r
x n log|logx| + Σ(-1)rnＣ r x Ei(rlog x)
n!
r=1


Example : 2nd order integral of log|logx|
When the one arbitrary point
x =1.6
is given) , the values of the both sides are as follows.
- 10 -
(1.n)
14.6 Super Calculus of Logarithmic Integral
Among the higher integrals mentioned in previous sections, Higher Integral of Logarithmic Integral is
extensible even to Super Calculus. It is because this higher integral is expressed with binomial coefficients.
14.6.1 Super Integral of Logarithmic Integral
Formula 14.6.1
When

li(x)=
0
x
1
dt ,
log t

x
Ei(x)=
-
et
dt
t
p 0 and x 0 .
x
x
p

1
p- r
 li(x)dx p =
(-1)r
x Ei(r +1)log x
Σ
(1+ p) r=0
r
0
0
the following expression holds for


Proof
First, replace
n!, nＣ r
with
(1+n) ,
r 
n
respectively in Formula 14.4.3 . Next, analytically
continuing the index of the integration operator to [0 , p ] from[1 , n ], we obtain the desired formula.
Example : 3/2th order integral of li(x)
We calculated the function values on arbitrary one point
x =4
according to the formula and Riemann-
Liouville integral. As the result, two values were almost corresponding.
Compared with the figure of the 2nd order integral in 14.4 , we can find that this curvature is loose.
14.6.2 Super Derivative of Logarithmic Integral
Formula 14.6.2
When
- 11 -

x
li(x)=
0
1
dt ,
log t

x
Ei(x)=
-
et
dt
t
p >0 , p  1, 2, 3,  and x 0 .
-p

1
r
(p)
r
=
-1
(
)
x -p- Ei(r +1)log x
li(x)
Σ
(1- p) r=0
r
the following expression holds for
 
Proof
p  -1, -2, -3,  , x 0 . Therefore, in Formula 14.6.1 , replacing the
<p > with the differentiation operator (p ) = <-p > , we obtain the desired expression.
In fact, Formula 14.6.1 holds for
integration operator
Example : 1/2th order derivative of li(x)
We calculated the the super differential coefficients on arbitrary one point
x =2
according to the formula
and Riemann-Liouville differintegral. As the result, two values were almost corresponding.
- 12 -
14.7 Super Calculus of Double Logarithmic Function
Among the higher integrals mentioned in previous sections, Higher Integral of Double Logarithmic Function is
extensible even to Super Calculus. It is because this higher integral is expressed with binomial coefficients.
14.7.1 Super Integral of Double Logarithmic Function
Formula 14.7.1
When

x
Ei(x)=
-

x

0
et
dt ,
t
x
the following expressions hold for
log|log x|dx p =
0
n!, nＣ r
with


(1+n) ,
r 
n
respectively in Formula 14.5.1 . Next, analytically
continuing the index of the integration operator to [0 , p ] from[1 , n ], we obtain the desired formula.
Example : 5/3th order integral of log|logx|
We calculated the function values on arbitrary one point
x =1.5
according to the formula and Riemann-
Liouville integral. As the result, two values were almost corresponding.
14.7.2 Super Derivative of Double Logarithmic Function
Formula 14.7.2
When

x
Ei(x)=
-

p

1
p-r
x Ei(rlog x)
x plog|logx| + Σ(-1)r
(1+ p)
r=1
r
Proof
First, replace
p >0 , x 0.
et
dt , the following expression holds for p >0 , p  1, 2, 3,  , x 0 ,
t
- 13 -
the following expression holds.
(p)
log|log x|
=

 
-p

1
r
-1
+
(
)
|
log|
logx
Σ
r=1
r
x p(1-p)
Ei(rlog x)
xr

Proof
p  -1, -2, -3,  , x 0 . Therefore, in Formula 14.7.1 , replacing the
<p> with the differentiation operator (p) = <-p > , we obtain the desired expressions.
In fact, Formula 14.7.1 holds for
integration operator
Example : 0.3th order derivative of log|logx|
We calculated the the super differential coefficients on arbitrary one point
x =0.5
according to the formula
and Riemann-Liouville differintegral. As the result, two values were almost corresponding.
Since the number of the order of the differentiation is small, it resembles the figure of
log|logx | well.
2007.10.05
K. Kono
Alien's Mathematics
- 14 -