Exercise 1.  .  

Solution 1.

See text and/or instructor's solution manual.

Answer.   .  

Solution.  The complex integrand is  .  

Factor the denominator and get

                    .  

Hence  ,  has zeros of order  2  at     and     lies in the upper half-plane.

It follows that    has poles of order  2  at  ,  

and the pole at    lies in the upper half-plane.

                    The contour    consisting of the semi-circle    and the interval .  

                    The point    lies in the upper half-plane.

Using Theorem 8.1 (Cauchy's Residue Theorem), and Theorem 8.3 (Contour Integration for Rational Functions), the value of the integral is computed   

                      

Here the denominator of  f(z) has a factor of the form  ,  and  .  

In this exercise, the limit can be calculated as follows:

                      

We are done.   

Aside.  We can let Mathematica double check our work.

Maple can check our work too!

     > residue( z^2/(z^2+16)^2, z=4*I );

               

     > 2*Pi*I*residue(z^2/(z^2+16)^2,z=4*I);

               

                    A portion of the area under the curve  .

We are really done.   

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

     > int( x^2/(x^2+16)^2, x=-infinity..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 Cauchy Principal Value (P.V.) 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