8.2  Trigonometric Integrals

    As indicated at the beginning of this chapter, we can evaluate certain definite real integrals with the aid of the residue theorem.  One way to 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 an integral of the form  

(8-3)            ,    

where F(u,v) is a function of the two real variables u and v.  Consider the unit circle with parametrization  

            ,    for    ;   

which gives us the following symbolic differentials.

            ,    and   
(8-4)
            .  

Combining      with   ,   we can obtain  

(8-5)                and    .  

Using the substitutions for  ,   , and    in Expression (8-3) transforms the definite integral into a contour integral

            ,  

where the new integrand is  .  

    Suppose that f(z) is analytic inside and on the unit circle , except at the points    that lie interior to  .  Then the residue theorem gives

(8-6)            .  

The situation is illustrated in Figure 8.2.

Figure 8.2  The change of variables from a definite integral on to a contour integral around C.

Some of these problems can be solved using Mathematica's table of integrals.

We need to use the following substitution procedure.

Example 8.10.    Evaluate    by using complex analysis.  

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

            

where  .  The singularities of f(z) are poles located at the points where    or equivalently, where  .  Using the quadratic formula, we see that the singular points satisfy the relation    which implies that    or  .  Hence the only singularities that lie inside the unit circle are simple poles corresponding to the solutions of  ,  which are the two points    and  .  We use Theorem 8.2 and L'Hôpital's rule to get the residues at and :  

               

As    and  ,  the residues are given by  .  We now use Equation (8-6) to compute the value of the integral:

               

Explore Solution 8.10.

Example 8.11.  Evaluate    by using a computer algebra system.  

Solution.  We can obtain the antiderivative of by using software such as Mathematica or MAPLE.  It is   .  
Since    and    are not defined, the computations for both and are indeterminate.  The graph shown in Figure 8.3 reveals another problem:  
The integrand is a continuous function for all t, but the function has a discontinuity at .  This condition 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.  The problem is that is not an antiderivative of for all t in the interval .  Oddly, it is the antiderivative at all points except , , and , which you can verify by computing and showing that it equals whenever is defined.

Figure 8.3  Graph of  .

    The integration algorithm used by computer algebra systems here (the Risch-Norman algorithm) gives the antiderivative  ,  and we must take great care in using this information.  We get the proper value of the integral by using on the open subintervals and where it is continuous, and taking appropriate limits:   

              

Explore Solution 8.11.

Example 8.12.    Evaluate  .  

Solution.  For values of z that lie on the unit circle , we have

               and   

We solve for    to obtain the substitutions  

                and    .  

Using the identity for along with Substitutions (8-4) and (8-5), we rewrite the integral as  


            

where  .  The singularities of lying inside are poles located at the points and .  We use Theorem 8.2 to get the residues:

                
        
               

Therefore we conclude that  

                

Explore Solution 8.12.

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(c) 2006 John H. Mathews, Russell W. Howell