Exercise 11.  [Graphics:Images/FunTheoremCalculusModHome_gr_132.gif],  where C is the line segment from  [Graphics:Images/FunTheoremCalculusModHome_gr_133.gif].  

Solution 11.

See text and/or instructor's solution manual.

Answer.  [Graphics:../Images/FunTheoremCalculusModHome_gr_134.gif][Graphics:../Images/FunTheoremCalculusModHome_gr_135.gif].  

Solution.  The function  [Graphics:../Images/FunTheoremCalculusModHome_gr_136.gif]  is analytic on the entire complex plane, except points on the negative x-axis and at the origin,

and  [Graphics:../Images/FunTheoremCalculusModHome_gr_137.gif]  is analytic in the right half-plane [Graphics:../Images/FunTheoremCalculusModHome_gr_138.gif] that contains the  line segment joining  [Graphics:../Images/FunTheoremCalculusModHome_gr_139.gif].  

and  [Graphics:../Images/FunTheoremCalculusModHome_gr_140.gif]  has  [Graphics:../Images/FunTheoremCalculusModHome_gr_141.gif]  is an antiderivative.

Thus,  we can use Theorem 6.9 to obtain  

                              [Graphics:../Images/FunTheoremCalculusModHome_gr_142.gif]
                              

                    [Graphics:../Images/FunTheoremCalculusModHome_gr_143.gif]

                    The path of integration is the line segment from  [Graphics:../Images/FunTheoremCalculusModHome_gr_144.gif].  

                    Notice that  [Graphics:../Images/FunTheoremCalculusModHome_gr_145.gif]  is not analytic on it's branch cut   [Graphics:../Images/FunTheoremCalculusModHome_gr_146.gif].

We are done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/FunTheoremCalculusModHome_gr_147.gif]

[Graphics:../Images/FunTheoremCalculusModHome_gr_148.gif]



[Graphics:../Images/FunTheoremCalculusModHome_gr_149.gif]

[Graphics:../Images/FunTheoremCalculusModHome_gr_150.gif]




















This solution is complements of the authors.



































 

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