Exercise 16.   .

Hint.  Use the contour    shown in the figure below,

and establish that  .

Solution 16.

See text and/or instructor's solution manual.

Answer.   .  

Solution.  The complex integrand is  .  

The denominator is   .

The zeros of    are  .

The poles of    are  .

      We will use the contour    consisting of the semi-circle    and the interval  ,   

and the segment  .   The pole    lies in inside  .

                    The point    lies in inside  .

      Using Theorem 8.1 (Cauchy's Residue Theorem), we obtain

                      

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

                    

      On the segment    we use the change of variable    and     for   ,  

and we can calculate the contour integral over    in the following manner:   

                    

      A details in the proof of Theorem 8.3 in Section 8.3 can be used to conclude that

                    .  
                    
Now use the value contour integral that we obtained by the residue calculus

                    

An easy computation will now finish our work

                     

We are done.   

Aside.  We can let Mathematica double check our work.

Maple can check our work too!

     > V1 := residue( 1/(z^3+1), z=(1+I*sqrt(3))/2 );
     > V1 := Re(V1)+I*Im(V1);

               

               

     > V := (2*Pi*I/(1-exp(2*Pi*I/3)))*V1;
     > V := Re(V)+I*Im(V);

               

               

          A portion of the area under the curve  .

We are really done.   

Aside.  Both and are capable of finding the definite integral.

     > f := proc(x) 1/(x^3+1) end proc;

               f := proc(x) 1/(x^3 + 1) end proc

     > int(f(x), x=0..infinity);

               

We are really really done.   

The intent of this exercise is to illustrate how the Residue Calculus is used to find improper integrals.

Aside.  For comparison, we show how to find the value of the integral using the indefinite integral and limits.

                    The indefinite integral  .

                    The value of the integral is    and it can be computed as follows:

                      

This solution is complements of the authors.

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