1 THE COSINE INTEGRAL By Tom Irvine Email:

1 THE COSINE INTEGRAL By Tom Irvine Email:

THE COSINE INTEGRAL
By Tom Irvine
Email: [email protected]
June 18, 2005
________________________________________________________________________
f(x) = (1/x) cos x
10
8
f(x)
6
4
2
0
-2
0
1
2
3
4
5
6
7
8
9
10
x
Figure 1.
Integrate the following function.
b 1
∫a
x
(1)
cos x dx
1
The cosine function can be expanded in the following series.
cos x =
∞
 2k 
k x 
(
−
1
)
∑
 (2k )! 
k =0


(2)
By substitution,
b1
∫a x cos x dx =
 2k 
b1  ∞
k  x  dx

−
(
1
)
∑
∫a x 
 (2k )! 


 k =0
(3)
 x 2k −1  
b1
b  ∞
k

∫a x cos x dx = ∫a  ∑ (−1)  (2k )!   dx


 k =0
(4)
b
∞

 x 2k  
b 1
k

∫a x cos x dx = ln x + ∑ (−1)  (2k ) (2k ) !  
k =1

 a

b 1
cos x dx
a x
∫
(5)
∞

 b 2k  

= + ln b + ∑ (−1)k 
 (2k ) (2k ) !  

k =1


∞

 a 2k  

− ln a + ∑ (−1)k 


(
2
k
)
(
2
k
)
!

k =1


(6)
2
b 1
cos x dx
a x
∫
∞
 b 2k − a 2k 
b
k

= ln   + ∑ (−1) 
 (2k ) (2k ) ! 
 a  k =1


(7)
b 1
cos x dx
a x
∫
∞
2k
 2k
b 1
kb −a
= ln   +
−
(
1
)
∑
 (k ) (2k ) !
 a  2 k =1





(8)
Note that each integration limit must be greater than zero due to the natural log term.
For numerical computation, equation (8) is practical only if b < 10.
Furthermore, the cosine integral is defined in some references as
∞ 1
cos u du
x u
Ci( x ) = − ∫
The asymptotic expansion of this function is given in Appendix A.
3
(9)
APPENDIX A
An asymptotic solution is given in this section. The result is useful if x >> 1.
∞ 1
cos u du
x u
Ci( x ) = − ∫
(A-1)
Integrate by parts.
∞ 1
 ∞
1
Ci( x ) = − ∫ d sin u  + ∫ sin u d 
x u
 x
u
∞ 1
 ∞ 1
Ci( x ) = − ∫ d sin u  − ∫
sin u du
x u
 x u2
∞
1
Ci( x ) = − sin u
u
x
−
1
sin x −
x
Ci( x ) =
∞ 1
sin u du
2
∫x
u
(A-2)
(A-3)
(A-4)
∞ 1
sin u du
2
(A-5)
 1

 1 
∞
d cos u  − ∫ cos u d 
2
x
u

 u2 
(A-6)
∫x
u
Integrate by parts again.
∞
Ci( x ) =
1
sin x +
x
∫x
Ci( x ) =
1
sin x +
x
1
Ci( x ) =
u2
∞
cos u
x
1
1
sin x −
cos x
x
x2
4
∞ 1
cos u du
x u3
(A-7)
∞ 1
cos u du
x u3
(A-8)
+2 ∫
+2 ∫
Integrate by parts again.
Ci( x ) =
1
1
cos x
sin x −
x
x2
Ci( x ) =
1
1
cos x
sin x −
x
x2
Ci( x ) =

 1 
∞  1
∞
+ 2 ∫ d sin u du − 2 ∫ sin u d 
3
x u
x

 u3 
+
2
u3
∞
sin u
x
∞ 1
sin u du
x u4
+6 ∫
∞ 1
1
1
2
sin x −
cos x −
sin x + 6 ∫
sin u du
x u4
x
x2
x3
(A-9)
(A-10)
(A-11)
Integrate by parts again.
Ci( x ) =

 1 
∞  1
∞
2
1
1
sin x − 6 ∫ d cos u  + 6∫ cos u d 
cos x −
sin x −
2
3
4
x u
x
x
x
x

 u4 
(A-12)
∞
Ci( x ) =
 1 
∞
1
2
1
1
sin x − 6 cos u + 6∫ cos u d 
cos x −
sin x −
x
x
u4
x3
x2
 u4 
x
(A-13)
Ci( x ) =
∞ 1
1
1
2
6
sin x −
cos x −
sin x +
cos x − 24 ∫
cos u du
2
3
4
x
x
u5
x
x
x
(A-14)
5
Ci( x ) =
1
2
6
1
sin x −
cos x −
sin x +
cos x + R
x
x2
x3
x4
(A-15)
R is the remainder.
∞ 1
cos u du
x u5
R = −24 ∫
(A-16)
The solution may be restated as
Ci( x ) =
1
1!
2!
3!
sin x −
cos x −
sin x +
cos x + R
x
x2
x3
x4
(A-15)
R is the remainder.
∞ 1
cos u du
x u5
R = − 4! ∫
(A-16)
6