# General Cosine and Sine Integral of Powers by waabu

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General Cosine and Sine Integral of Powers
╬

Francis J. O’Brien, Jr., Ph.D.
Aquidneck Indian Council
Newport, RI

 Francis J. O’Brien, Jr.                                                                  April 22, 2013

Introduction
In this paper we show the derivation for two general algebraic-trigonometric integrals
using complex variables and the gamma function1. The new integral expressed in combined
form is:

   
         
m cos x 
n          cos 
    2       m 1
 x sin x n dx  n      ,   n ,  ,   0, n  0,
      
0                       sin   
  2 
        
where      is the gamma function. See L. Bers, Calculus (pp. 402-403) for elementary
properties.

Complex variables often simplify trigonometric integral evaluation2.

Proof Outline

Familiar elementary complex variable relations and operations:

   Euler’s Formula
eix  cos x  i sin x
e ix  cos x  i sin x

1
Submitted to 8th edition of Gradshyetn and Rhyzik (GR), Table of Integrals, Series and Products for Section
3.761, “Trigonometric Functions of More Complicated Arguments Combined with Powers”. NOTE: One
website, http://quickmath.com, was unable to do the definite form or the indefinite form (a
complicated indefinite form solution was given at the Mathematica website,
http://integrals.wolfram.com/index.jsp).
2
G. S. Carr, Formulas and Theorems in Pure Mathematics, solves similar integrals using complex
variables and gamma function; see Theorems 2577–2579. Theorem 2577 is derived below to
demonstrate complexification as a means to solve an integral computers cannot evaluate.

 Francis J. O’Brien, Jr., 2013 <> Aquidneck Indian Council<> All rights reserved.
Page 2 of 9

   By definition from Euler’s Formula,

eix  e ix
cos x                .
2

o If the power is 1 and one selects e ix or e  ix in Euler’s Formulas for analysis,
one may use real relations
                         
cos x  Re e  ix or Re eix
 Use      cos x  Re e ix  for convergence of integrals of negative
exponentials

   By definition from Euler’s Formula,
eix  e ix
sin x        .
2i
o If power is 1 and one selects e ix or e  ix in Euler’s Formulas for analysis, one
may use real relations

   sin x  Re
 i

 e ix
  Re  ie ix or Re  e

        
  ix
     i


 
  Re ie  ix
                                       
 e  ix 
 Use sin x  Re 

  Re ie  ix
i 
  for convergence of
        
integrals of negative exponentials

NOTE: the simplified real only formulas are valid for sine & cosine powers of 1. For integer
1  cos(2 x)
powers higher than 1 use identities such as sin 2 x               , then power of 1 real
2
relations. A future paper will demonstrate the technique.

   Euler’s identity, ei  1
i
o   e2      11 2  i
i
         
o   e 2  i  cos   i sin 
 2            2 
i
                     
o   e 2  i  cos   i sin 
 2             2 

 Francis J. O’Brien, Jr., 2013 <> Aquidneck Indian Council<> All rights reserved.
Page 3 of 9

   Complex conjugation

o    a  bi a  bi    a  bi  a  bi   a 2  b 2

o           1.
i  i   i 2

   Complexifying3 the integral by real parts only analysis

The following algebraic-exponential formula from Gradshteyn & Rhyzik is used to
simplify the complexified integrals:


             m 1
x
m  x n
e         dx              ,          ,  ,   0, n  0   Formula 3.326.2, GR (7th ed.)4

0                       n                n

Derivation

Inserting the above relations and operations, and simplifying, the complexified cosine
integral is:

  dx  Re  x meix dx  Re n i   ii 
                                                             
n
 x cos x
m                n

0                                0                         
 i           
    2                    
        Re e                   cos ,
n                    n       2 
         
m 1
           , Re ,   0, n  0
n

or, calculate sum of integrals by classical definition of cosine,

3
A good YouTube lecture from MIT is at “18.03 Differential Equations” (Lecture 6).
m  x n

4
Derived in “ x e       dx and related integrals, 2nd ed.”, Jan. 3, 2013.
http://www.docstoc.com/profile/waabu

 Francis J. O’Brien, Jr., 2013 <> Aquidneck Indian Council<> All rights reserved.
Page 4 of 9

 

x       cos x n dx
m

0
    i x n
m e
      n
 e  i x     
 Re  x                            dx
            2            
0                              
1          1 i             1  i              

2 n  


Re
 i i       
 Re
i  i            
        
       cos .
n        2 

____________

The complexified sine integral is similar. Not all the steps are specified:

 

x            sin x n dx
m

0

 Re i  x m e  ix dx
n

0
     
      sin  

 2 
n

or, calculate difference of integrals by classical definition of sine,

                
   i x n  e  i x n                     
Re  x    m e                                         dx    sin   
 
0


2i                                 
      n        2 

 cosx        dx

2
Example:
0

In the general solution,

 

      
x       cos x n dx 
m
cos ,
0                                  n      2 

 Francis J. O’Brien, Jr., 2013 <> Aquidneck Indian Council<> All rights reserved.
Page 5 of 9

    2   1
set m  0,   1, n  2,   1 / 2, cos          to give the solution by the “short cut” or
4   2     2
standard method:

 cosx        dx

2

0
 
    ix 2
e      e               ix 2 


dx  Re  e  ix dx
2
 Re                                
0
     2                        
         0

1
 
 
   cos 
2
2     4
2

4
1 
        0.62665706865,
2 2

1
where     .
2

Example: The cosine Fresnel integral is defined to be:

z
 2 
 cos 2 t

dt.

0

It cannot be solved in closed form unless z   in which case the general integral

 

    
x       cos x n dx 
m
cos  holds (for                              .
   2 
0                        n
                            2   1
Set m  0,       , n  2,   1 / 2, cos          :
2                        4   2     2


 2      1
 cos 2 t

 dt  .
     2
0

The sine Fresnel integral

 Francis J. O’Brien, Jr., 2013 <> Aquidneck Indian Council<> All rights reserved.
Page 6 of 9

z
 2 
lim
z 
 sin 2 t

dt

0

gives the same solution.

                                  
NOTE: Theorem 2577 (p. 384) in G.S Carr,               x
n 1  ax
e        cosbx dx and    x
n 1  ax
e       sin bx dx .
0                                  0
Online integrators cannot solve these definite integrals; e.g., http://www.quickmath.com/

These integrals are useful because they use clever manipulations of complex variables
and serve as a general form from which specific cases can be derived.

Figure 1 provides the essential information needed for complex variables to evaluate
the integrals.

 i
a  bi  re

 in
r                                                a  bi n
r e
n

b = r sin                           2
a b  r
2       2

                                                                 b
tan  
a
a = r cos 
 1 b 
  tan         a
Figure 1. Polar form of complex number a  bi                                            
on a right triangle. The form a  bi would
lie in Quadrant IV.

NOTE: it is more traditional to use x & y vs. a & b for polar coordinates.


Re-expressing in complex variable form and applying Form. 3.326.2,  x m e  x dx, the cosine
n

0
integral is:

 Francis J. O’Brien, Jr., 2013 <> Aquidneck Indian Council<> All rights reserved.
Page 7 of 9

                                                                         
n 1  a  bi x              n 
x
n 1  ax
e       cosbx dx         x
n 1  ax  ibx
e      e           dx    x       e               dx 
0                                    0                                     0                               a  bi n

NOTE: With complex variables we have reduced a complicated algebraic-exponential-
trigonometric function to an algebraic-exponential function which is more manageable.

The solution must be given in terms of the real number parameters a, b & n. Thus,
conjugating, inserting the auxiliary polar equations, and simplifying provides a string of
reductions:

                                             
n 1  a  bi x
 x e cosbxdx  Re  x e                dx  n  Re
n 1  ax                                                                                      1
0                                              0                                            a  bi n
 n  Re
1     a  bi n               n  Re
a  bi n
a  bi n a  bi n                            a  bi a  bi n
 n  Re
1
rei n
a 2  b2 n
 n  Re
rn
2n
cos n  i sin n   n cosnn
r                                                    r
tan   b a
n                       b 
                              cos n tan 1 .
                               a 
n
2   2 2                 
a b


n                           b 
Thus,   x
n 1  ax
e       cosbx dx                             cos n  tan 1 .
a 2  b2                         a 
n
0                                                     2


          

NOTE: Solution for            x
n 1  ax
e          sin bx dx is similar, defining sin(bx )  Re ie  ibx .
0
NOTE: If b  0, Theor. 2577 gives


n                   n 
x
n 1  ax
e      dx                cos n0  
n
0                              a                        an
since arctan(0)  0, cos(0)  1.

 Francis J. O’Brien, Jr., 2013 <> Aquidneck Indian Council<> All rights reserved.
Page 8 of 9

NOTE: If a  0, Theor. 2577 gives


x
n 1
cosbx dx
0

 Re  x n 1e  bi x dx
0
n                    b 
       cos n  lim tan 1  
bn         a  0         a 


n 
b n

cos n tan 1   ,       

and recalling the limit of arctangent at infinity, lim tan 1 x                     1.57 , then
x             2


n   
x
n 1
cosbx dx          cos n 
0                             bn     2

Figure 2. Plot of y  tan 1 x .

NOTE: If n  1, the first step in the derivation of Theor. 2577 gives an immediate answer:


1             ( a  bi )
cosbx dx  Re
 ax                                                            a
e                              a  bi 
 Re                   
a  bi a  bi  a  b 2
2
0
 Francis J. O’Brien, Jr., 2013 <> Aquidneck Indian Council<> All rights reserved.
Page 9 of 9

where  1 =1.

Clearly, Theorem 2577 is a useful integral. Why can’t computers solve it?

A future paper will demonstrate the solution to integrals of the form

             cos n  x 
             
 x m 1e x              dx
0             sin n  x  
             

The approach is modeled on the solution of Theorem 2577.

 Francis J. O’Brien, Jr., 2013 <> Aquidneck Indian Council<> All rights reserved.

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