Solution 15.

See text and/or instructor's solution manual.

Answer.   .

Solution.  The integrand is  ,  construct the complex function

using the substitutions    and  .

                       

The complex integrand is  ,  and we have  

                   .  

Factoring the denominator, we obtain

                    .   

Hence  ,  has simple zeros at     and     lies inside  .

Thus  ,  has simple poles at     and     lies inside  .

Using Theorem 8.1 (Cauchy's Residue Theorem), the value of the integral is computed   

                      

Here the denominator of  f(z) has a factor of the form  ,  and by Theorem 8.2  .  

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

This solution is complements of the authors.

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