Exercise 13.  .

Solution 13.

See text and/or instructor's solution manual.

Answer.   .

Solution.  The complex integrand is  .  

The denominator is   .

The zeros    lie on the real axis.

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

                    above the points    that lie on the real axis.

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( exp(I*z)/(z*(1-z^2)), z=-1 );
     > V1 := Re(V1)+I*Im(V1);

               

               

     > V2 := residue( exp(I*z)/(z*(1-z^2)), z=0 );

               V2 := 1

     > V3 := residue( exp(I*z)/(z*(1-z^2)), z=1 );
     > V3 := Re(V3)+I*Im(V3);

               

               

     > Pi*(Re(V1)+Re(V2)+Re(V3));

               

          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) sin(x)/(x*(1-x^2)) end proc;

                f := proc(x) sin(x)/(x*(1 - x^2)) 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 (which are left for the reader).

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

                      

      The indefinite integral will involve the special functions "Cosine Integral" and "Sine Integral."  

      We do not want to develop these special functions at this time and leave it for the reader to investigate their properties.  It is interesting to observe that the current software includes these functions.  For this type of problem the Residue Calculus can get the answer without resorting to special functions.

This solution is complements of the authors.

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