Exercise 3.  Evaluate  

3 (g).  .

Solution 3 (g).

See text and/or instructor's solution manual.

Answer.   .  

Solution.   Consider the denominator    at the point   ,  here we have

              

          and

              

Applying Definition 7.6 we can conclude that    has a zero of order  3  at the point  .  

        Use the techniques for determining the order of poles: apply  Corollary 7.5  in Section 7.4 to conclude that

the integrand   has a pole of order  3  at the point  .  

                         The point    that lies inside the contour  .  

        Now apply Theorem 8.1 the Cauchy's Residue Theorem.

The point    is the singularity that lie inside the contour  .  

Since   is a pole of order  3, by Theorem 8.2 the residue is calculated as follows:

                       
                    
                    

The "Residue Calculus" gives the value of the integral:

                                

We are done.   

Aside.  We can let Mathematica double check our work.

Maple can check our work too!

     > (1/2!)*limit( diff((z-0)^3*1/(z*sin(z)^2),z$2), z=0 );

               

We are really done.   

Aside.  Both and are capable of finding residues.

     > residue( 1/(z*sin(z)^2), z=0 );

               

     > 2*Pi*I*residue(1/(z*sin(z)^2),z=0);

               

This solution is complements of the authors.

(c) 2008 John H. Mathews, Russell W. Howell