Exercise 7.  [Graphics:Images/FunTheoremCalculusModHome_gr_79.gif],  where C is the line segment from  [Graphics:Images/FunTheoremCalculusModHome_gr_80.gif].  

Solution 7.

See text and/or instructor's solution manual.

Answer.  [Graphics:../Images/FunTheoremCalculusModHome_gr_81.gif].  

Solution.  The function  [Graphics:../Images/FunTheoremCalculusModHome_gr_82.gif]  is analytic on the entire complex plane, except at the origin,

and  [Graphics:../Images/FunTheoremCalculusModHome_gr_83.gif]  is analytic in the half upper half-plane [Graphics:../Images/FunTheoremCalculusModHome_gr_84.gif] that contains the  line segment joining  [Graphics:../Images/FunTheoremCalculusModHome_gr_85.gif],  

and  [Graphics:../Images/FunTheoremCalculusModHome_gr_86.gif]  has  [Graphics:../Images/FunTheoremCalculusModHome_gr_87.gif]  is an antiderivative.

Thus,  we can use Theorem 6.9 to obtain  

                              [Graphics:../Images/FunTheoremCalculusModHome_gr_88.gif]
                              

                    [Graphics:../Images/FunTheoremCalculusModHome_gr_89.gif]

                    The path of integration is the line segment from  [Graphics:../Images/FunTheoremCalculusModHome_gr_90.gif].  

We are done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/FunTheoremCalculusModHome_gr_91.gif]

[Graphics:../Images/FunTheoremCalculusModHome_gr_92.gif]




















This solution is complements of the authors.



































 

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