Module

for

Trigonometric Integrals via Contour Integrals

 

Evaluation of real integrals via complex analysis.

 

   Needless to say, when complex analysis methods are used to evaluate real integrals, something is going to get

messy, and Section 8.2 and Section 8.3 are no exceptions.  So prepare yourself for doing the computations and

learning some new theorems.

    In Section 8.2, the technique involves making special substitutions for the trigonometric functions and obtaining

a complex function that is the quotient of two polynomials.  The difficulties start with the algebraic skills necessary

to form the polynomials.  And then the difficulty is finding the roots of the polynomial in the denominator.  It is

a good idea to review the quadratic formula and factoring a polynomial before studying Section 8.2.  

Some textbooks study the material in Section 8.3 about rational integrals before studying the material in Section 8.2.  

The techniques will require careful attention to the details need to verify that the limit of a certain complex integral

taken around a semi-circle in the upper half plane will vanish as the radius increases.  The skills in Section 8.3

involve understanding proofs involving epsilon and limits tending to zero in the complex plane.

It is a toss-up to decide which section to study at this point in the book. Your background should help you decide.

 

8.2  Trigonometric Integrals

 

As indicated at the beginning of this chapter, our goal is to evaluate certain definite real integrals with the aid

of the Residue Theorem.  We can do this is by interpreting the definite integral as the parametric form of an

integral of an analytic function along a simple closed contour.

Suppose we wish to evaluate a real integral of the form  

(8-3)               [Graphics:Images/IntegralsTrigMod_gr_1.gif],    

where  [Graphics:Images/IntegralsTrigMod_gr_2.gif]  is a function of the two real variables  [Graphics:Images/IntegralsTrigMod_gr_3.gif].  We can transform the real integral into a

complex integral [Graphics:Images/IntegralsTrigMod_gr_4.gif],  where the contour of integration is the unit circle [Graphics:Images/IntegralsTrigMod_gr_5.gif],

and the parametrization of the circle  [Graphics:Images/IntegralsTrigMod_gr_41.gif]  is given by

(8-3)               [Graphics:Images/IntegralsTrigMod_gr_6.gif],      for     [Graphics:Images/IntegralsTrigMod_gr_7.gif].      

          The interval  [Graphics:Images/IntegralsTrigMod_gr_10.gif] for   [Graphics:Images/IntegralsTrigMod_gr_11.gif].                     The contour  [Graphics:Images/IntegralsTrigMod_gr_12.gif]  for   [Graphics:Images/IntegralsTrigMod_gr_13.gif].  

     Figure 8.2 a.  Transforming integration over a real interval [Graphics:Images/IntegralsTrigMod_gr_14.gif] to complex integration around the contour [Graphics:Images/IntegralsTrigMod_gr_15.gif].

 

Use the equation   [Graphics:Images/IntegralsTrigMod_gr_16.gif],   and obtain the following symbolic differentials.

(8-5)               [Graphics:Images/IntegralsTrigMod_gr_17.gif],     and   
(8-4)
(8-5)               [Graphics:Images/IntegralsTrigMod_gr_18.gif].  

Combining    [Graphics:Images/IntegralsTrigMod_gr_19.gif]    with    [Graphics:Images/IntegralsTrigMod_gr_20.gif],    we can obtain two useful substitutions   

(8-5)               [Graphics:Images/IntegralsTrigMod_gr_21.gif]     and     [Graphics:Images/IntegralsTrigMod_gr_22.gif].  

Using the substitutions for [Graphics:Images/IntegralsTrigMod_gr_23.gif][Graphics:Images/IntegralsTrigMod_gr_24.gif]   and  [Graphics:Images/IntegralsTrigMod_gr_25.gif]  in Expression (8-3) transforms the real integral into

a contour integral

                    [Graphics:Images/IntegralsTrigMod_gr_26.gif],  
or

                    [Graphics:Images/IntegralsTrigMod_gr_27.gif].    
                    
Here we have used the identity  

                    [Graphics:Images/IntegralsTrigMod_gr_28.gif],  

and the new complex integrand is    

                    [Graphics:Images/IntegralsTrigMod_gr_29.gif].  

 

Remark.  If   [Graphics:Images/IntegralsTrigMod_gr_30.gif]   involves  [Graphics:Images/IntegralsTrigMod_gr_31.gif]  then it is o. k. to use the substitutions  

                    [Graphics:Images/IntegralsTrigMod_gr_32.gif]     and     [Graphics:Images/IntegralsTrigMod_gr_33.gif].  

 

Two graphical visualizations for the "area under the curve."

 

                  The real integral   [Graphics:Images/IntegralsTrigMod_gr_34.gif]   can be visualized in  [Graphics:Images/IntegralsTrigMod_gr_35.gif] space

          as the area under the curve   [Graphics:Images/IntegralsTrigMod_gr_36.gif]   over the interval   [Graphics:Images/IntegralsTrigMod_gr_37.gif].  

          The complex integral   [Graphics:Images/IntegralsTrigMod_gr_38.gif]   can be visualized in  [Graphics:Images/IntegralsTrigMod_gr_39.gif]  space as  

          the area under the parametric curve  [Graphics:Images/IntegralsTrigMod_gr_40.gif]  over the contour  [Graphics:Images/IntegralsTrigMod_gr_41.gif]

Area under the curve  [Graphics:Images/IntegralsTrigMod_gr_44.gif]  over  [Graphics:Images/IntegralsTrigMod_gr_45.gif],     Area under the curve  [Graphics:Images/IntegralsTrigMod_gr_46.gif],  

                    [Graphics:Images/IntegralsTrigMod_gr_47.gif].                                                                     [Graphics:Images/IntegralsTrigMod_gr_48.gif].  

               Figure 8.2 b.  Comparison of the  [Graphics:Images/IntegralsTrigMod_gr_49.gif]  and  [Graphics:Images/IntegralsTrigMod_gr_50.gif]  visualizations for the "area under the curve."  

 

      In the optional explorations for Examples 8.11 - 8.12 we show the surfaces over the unit disk  [Graphics:Images/IntegralsTrigMod_gr_51.gif],

in the complex plane, and we will be able to visualize the poles of the complex function  [Graphics:Images/IntegralsTrigMod_gr_52.gif]  that lie

inside the unit circle  [Graphics:Images/IntegralsTrigMod_gr_53.gif].  

 

Theorem (Contour Integration for Trigonometric Integrals). Use the above substitutions to construct [Graphics:Images/IntegralsTrigMod_gr_54.gif].

Then [Graphics:Images/IntegralsTrigMod_gr_54.gif] is analytic inside and on the unit circle [Graphics:Images/IntegralsTrigMod_gr_55.gif], except at the poles [Graphics:Images/IntegralsTrigMod_gr_56.gif] that lie inside

the unit circle  [Graphics:Images/IntegralsTrigMod_gr_57.gif].  We can compute the integral by using the Cauchy Residue Theorem,  

(8-6)               [Graphics:Images/IntegralsTrigMod_gr_58.gif],     
(8-6)               and  

(8-6)               [Graphics:Images/IntegralsTrigMod_gr_59.gif],  

where   [Graphics:Images/IntegralsTrigMod_gr_60.gif]   are the poles of   [Graphics:Images/IntegralsTrigMod_gr_61.gif]  that lie inside  [Graphics:Images/IntegralsTrigMod_gr_62.gif].  

                              [Graphics:Images/IntegralsTrigMod_gr_63.gif]  

                    Figure 8.2 c.  The poles inside  [Graphics:Images/IntegralsTrigMod_gr_64.gif] that are used to compute  [Graphics:Images/IntegralsTrigMod_gr_65.gif].  

 

Example 8.10.  Evaluate  [Graphics:Images/IntegralsTrigMod_gr_66.gif]  by using complex analysis and the Cauchy Residue Theorem.  

          [Graphics:Images/IntegralsTrigMod_gr_67.gif][Graphics:Images/IntegralsTrigMod_gr_68.gif]

Solution.  Using Substitutions (8-4) and (8-5), we transform the integral to  

                    [Graphics:Images/IntegralsTrigMod_gr_69.gif]

where   [Graphics:Images/IntegralsTrigMod_gr_70.gif].  

The singularities of  [Graphics:Images/IntegralsTrigMod_gr_71.gif]  are poles located at the points where  [Graphics:Images/IntegralsTrigMod_gr_72.gif]  or equivalently,

where  [Graphics:Images/IntegralsTrigMod_gr_73.gif].  Using the quadratic formula, we find that the singular points satisfy

[Graphics:Images/IntegralsTrigMod_gr_74.gif]  which implies that [Graphics:Images/IntegralsTrigMod_gr_75.gif] or [Graphics:Images/IntegralsTrigMod_gr_76.gif].  Hence the only singularities

that lie inside the unit circle are simple poles corresponding to the solutions of  [Graphics:Images/IntegralsTrigMod_gr_77.gif], which are the two

points  [Graphics:Images/IntegralsTrigMod_gr_78.gif]  and  [Graphics:Images/IntegralsTrigMod_gr_79.gif].   Now we use Theorem 8.2 and L'Hôpital's rule to calculate the

residues at the points [Graphics:Images/IntegralsTrigMod_gr_80.gif] and [Graphics:Images/IntegralsTrigMod_gr_81.gif].  

                    [Graphics:Images/IntegralsTrigMod_gr_82.gif]   


Since  [Graphics:Images/IntegralsTrigMod_gr_83.gif]  and  [Graphics:Images/IntegralsTrigMod_gr_84.gif],  the residues are given by   

                    [Graphics:Images/IntegralsTrigMod_gr_85.gif].  

We now use Equation (8-6) to compute the value of the integral:

                    [Graphics:Images/IntegralsTrigMod_gr_86.gif]   

Explore Solutions 8.10 and 8.11.

 

Example 8.11.  Evaluate   [Graphics:Images/IntegralsTrigMod_gr_290.gif]   by using a computer algebra system.   

[Graphics:Images/IntegralsTrigMod_gr_291.gif][Graphics:Images/IntegralsTrigMod_gr_292.gif]

Solution.  We can obtain antiderivative(s) of  [Graphics:Images/IntegralsTrigMod_gr_293.gif]  by using software such as [Graphics:Images/IntegralsTrigMod_gr_294.gif] or [Graphics:Images/IntegralsTrigMod_gr_295.gif].  

One of the possibilities that have been obtained is

                    [Graphics:Images/IntegralsTrigMod_gr_296.gif].  

Since [Graphics:Images/IntegralsTrigMod_gr_297.gif] and [Graphics:Images/IntegralsTrigMod_gr_298.gif] are not defined, the computations for both [Graphics:Images/IntegralsTrigMod_gr_299.gif] and [Graphics:Images/IntegralsTrigMod_gr_300.gif] are indeterminate.

The graph  [Graphics:Images/IntegralsTrigMod_gr_301.gif]  shown below in Figure 8.3 reveals another problem.  The integrand  [Graphics:Images/IntegralsTrigMod_gr_302.gif]

is a continuous function for all  [Graphics:Images/IntegralsTrigMod_gr_303.gif],  but the function [Graphics:Images/IntegralsTrigMod_gr_304.gif] has a discontinuity at [Graphics:Images/IntegralsTrigMod_gr_305.gif].  On first glance this

appears to be a violation of the fundamental theorem of calculus, which asserts that the integral of a continuous

function must be differentiable and hence continuous.  But the problem is that  [Graphics:Images/IntegralsTrigMod_gr_306.gif]  is not an antiderivative

of  [Graphics:Images/IntegralsTrigMod_gr_307.gif]  for all values of  [Graphics:Images/IntegralsTrigMod_gr_308.gif]  in the interval  [Graphics:Images/IntegralsTrigMod_gr_309.gif].   Oddly, it is the antiderivative at all points

except  [Graphics:Images/IntegralsTrigMod_gr_310.gif],  [Graphics:Images/IntegralsTrigMod_gr_311.gif],  and  [Graphics:Images/IntegralsTrigMod_gr_312.gif],  and you can easily verify that it's derivative  [Graphics:Images/IntegralsTrigMod_gr_313.gif] equals  [Graphics:Images/IntegralsTrigMod_gr_314.gif]

wherever  [Graphics:Images/IntegralsTrigMod_gr_315.gif]  is defined.

                    [Graphics:Images/IntegralsTrigMod_gr_316.gif]  

                    Figure 8.3   Graph of   [Graphics:Images/IntegralsTrigMod_gr_317.gif].

 

      The integration algorithm used by computer algebra systems (the Risch-Norman algorithm) gives us

the antiderivative  

                    [Graphics:Images/IntegralsTrigMod_gr_318.gif],  

and we must take great care in using this formula. We get the proper value of the integral by using [Graphics:Images/IntegralsTrigMod_gr_319.gif] 

on the open subintervals  [Graphics:Images/IntegralsTrigMod_gr_320.gif]  and  [Graphics:Images/IntegralsTrigMod_gr_321.gif]  where it is continuous, and by taking the appropriate

limits at  [Graphics:Images/IntegralsTrigMod_gr_322.gif].  

                    [Graphics:Images/IntegralsTrigMod_gr_323.gif]  

Explore Solutions 8.10 and 8.11.

 

Example 8.12.  Evaluate  [Graphics:Images/IntegralsTrigMod_gr_324.gif]  by using complex analysis and the Cauchy Residue Theorem.  

          [Graphics:Images/IntegralsTrigMod_gr_325.gif][Graphics:Images/IntegralsTrigMod_gr_326.gif]

Solution.  For values of [Graphics:Images/IntegralsTrigMod_gr_327.gif] that lie on the unit circle  [Graphics:Images/IntegralsTrigMod_gr_328.gif],  we have

                     [Graphics:Images/IntegralsTrigMod_gr_329.gif]     and     [Graphics:Images/IntegralsTrigMod_gr_330.gif]

We can solve for  [Graphics:Images/IntegralsTrigMod_gr_331.gif]  to obtain the substitutions  

                     [Graphics:Images/IntegralsTrigMod_gr_332.gif]     and     [Graphics:Images/IntegralsTrigMod_gr_333.gif].  

Using the identity for  [Graphics:Images/IntegralsTrigMod_gr_334.gif]  along with Substitutions (8-4) and (8-5), we rewrite the integral as  


                    [Graphics:Images/IntegralsTrigMod_gr_335.gif]  

where  [Graphics:Images/IntegralsTrigMod_gr_336.gif].  

The singularities of  [Graphics:Images/IntegralsTrigMod_gr_337.gif]  lying inside  [Graphics:Images/IntegralsTrigMod_gr_338.gif]  are poles located at the points   [Graphics:Images/IntegralsTrigMod_gr_339.gif]   and   [Graphics:Images/IntegralsTrigMod_gr_340.gif].  

We use Theorem 8.2 to calculate the residues.

                    [Graphics:Images/IntegralsTrigMod_gr_341.gif]    

and

                    [Graphics:Images/IntegralsTrigMod_gr_342.gif]   

Therefore, we conclude that  


                    [Graphics:Images/IntegralsTrigMod_gr_343.gif]    

Explore Solution 8.12.

 

 

Tutorial Exercises for Section 8.2.  Trigonometric Integrals

 

 

Library Research Experience for Undergraduates

Residue Calculus  

Contour Integrals  

Cauchy Principal Value  

 

 

 

 

The Next Module is

Improper Integrals of Rational Functions

 

 

 Return to the Complex Analysis Modules

 

 

Return to the Complex Analysis Project

 

 

 

 

 

 

 

 

 

 

 

 

 

  

 

This material is coordinated with our book Complex Analysis for Mathematics and Engineering.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) 2012 John H. Mathews, Russell W. Howell