Exercise 3.   .

Solution 3.

See text and/or instructor's solution manual.

Answer.   .  

Solution.  The complex integrand is  .  

The denominator is   .

The complex zeros are  ,  and the real zero is  .

                    The indented contour C that consists of the semi-circle , and segments along the real axis and a small semi-circle

                    above the point    that lies on the real axis.  The point    lies inside  C.  

Using Theorem 8.1 (Cauchy's Residue Theorem), and Theorem 8.5 (Contour Integration for Indented Contour Integrals), the value of the integral is computed   

                      

We are done.   

Aside.  We can let Mathematica double check our work.

Maple can check our work too!

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

               

               

     > V2 := residue( z/(z^3+1), z=-1 );

               

     > simplify( 2*Pi*I*V1 + Pi*I*V2 );

               

          A portion of the "area under the curve"  .  

We are really done.   

Aside.  Both and are capable of finding the definite integral, but the "Principal Value" option must be used.

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

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

     > simplify(convert(int(f(x), x=-infinity..infinity, 'CauchyPrincipalValue'),exp));

                  

We are really really done.

The above computation illustrates how the Residue Calculus can be used to find Indented Contour Integrals.

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

                    The indefinite integral  .  From the graph we see that  

                    .

Caveat.  If you do not trust the graphical explanation then you must trust the Residue Calculus solution.  

Or you could compute the Cauchy Principal value, but this will require some difficult limits.

                    The Cauchy Principal Value (P.V.) of the integral is    and it can be computed as follows:

                      

The details are tedious and we let Mathematica do the work.

Alternately, we have the computation:

This solution is complements of the authors.

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