Exercise 13.  ,  where C is the line segment from  .  

Solution 13.

See text and/or instructor's solution manual.

Answer.  

                      

Solution.  The function    is analytic everywhere except at  ,  

and    has    as an antiderivative.  

Letting D be the simply connected domain consisting of the entire complex plane except for the real numbers  ,  

we see that    and its listed antiderivative are analytic in D.

Since the line segment from    is contained in D, Theorem 6.9 gives

                    
                    

                    The path of integration is the line segment from  .  

                    Notice that    is not analytic on the ray   .

We are done.   

Aside.  We can let Mathematica double check our work.

This solution is complements of the authors.

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