Exercise 3.  Evaluate  

3 (c).  .

Solution 3 (c).

See text and/or instructor's solution manual.

Answer.   .  

Solution.  Factor the denominator and get

                    ,  

and conclude that    has simple zeros at  ,    and  .   

Use the techniques for determining the order of poles: apply  Corollaries 7.5 and 7.8  in Section 7.4 to conclude that

the integrand    has simple poles at  ,    and  .   

                The points  ,    and    that lie inside the contour  .  

        Now apply Theorem 8.1 the Cauchy's Residue Theorem.

The points  ,    and    are the singularities that lie inside the contour  .  

Since   is a simple pole, by Theorem 8.2 the residue is calculated as follows:  

                       

Similarly   is a simple pole, and by Theorem 8.2 the residue is calculated as follows:  

                       

Similarly   is a simple pole, and 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!

     > limit( (z-0)*exp(z)/(z^3+z), z=0 );

               1

     > limit( (z-I)*exp(z)/(z^3+z), z=I );

               

     > limit( (z-(-I))*exp(z)/(z^3+z), z=-I );

               

We are really done.   

Aside.  Both and are capable of finding residues.

     > V1 := residue( exp(z)/(z^3+z), z=0 );

               V1 := 1

     > V2 := residue( exp(z)/(z^3+z), z=I );

               

     > V3 := residue( exp(z)/(z^3+z), z=-I );

               

     > val:=2*Pi*I*(V1+V2+V3);

               

     > simplify(val);

               

     > val:=2*Pi*I*(residue(exp(z)/(z^3+z),z=0)
                  + residue(exp(z)/(z^3+z),z=I)
                  + residue(exp(z)/(z^3+z),z=-I));

               

This solution is complements of the authors.

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