Exercise 17.  Let  [Graphics:Images/FunTheoremCalculusModHome_gr_226.gif]  be the principal branch of the square root function.  

17 (a).  Evaluate  [Graphics:Images/FunTheoremCalculusModHome_gr_227.gif],  where C is the line segment joining  [Graphics:Images/FunTheoremCalculusModHome_gr_228.gif].  

Solution 17 (a).

See text and/or instructor's solution manual.

Answer.  [Graphics:../Images/FunTheoremCalculusModHome_gr_229.gif][Graphics:../Images/FunTheoremCalculusModHome_gr_230.gif].  

Solution.  The function  [Graphics:../Images/FunTheoremCalculusModHome_gr_231.gif]  is analytic everywhere except points   [Graphics:../Images/FunTheoremCalculusModHome_gr_232.gif],  

along the negative x-axis, where the principal branch of the square root function [Graphics:../Images/FunTheoremCalculusModHome_gr_233.gif] is discontinuous,

and  [Graphics:../Images/FunTheoremCalculusModHome_gr_234.gif]  has  [Graphics:../Images/FunTheoremCalculusModHome_gr_235.gif]  as an antiderivative.  

Letting D be the simply connected domain consisting of the entire complex plane except for the real numbers   [Graphics:../Images/FunTheoremCalculusModHome_gr_236.gif],  

along the negative x-axis, we see that  [Graphics:../Images/FunTheoremCalculusModHome_gr_237.gif]  and its listed antiderivative are analytic in D.  

Since the line segment from  [Graphics:../Images/FunTheoremCalculusModHome_gr_238.gif]  is contained in D, Theorem 6.9 gives

                    [Graphics:../Images/FunTheoremCalculusModHome_gr_239.gif]
                    

                    [Graphics:../Images/FunTheoremCalculusModHome_gr_240.gif]

                    The path of integration is the line segment from  [Graphics:../Images/FunTheoremCalculusModHome_gr_241.gif].  

                    Notice that  [Graphics:../Images/FunTheoremCalculusModHome_gr_242.gif]  is not analytic on the ray   [Graphics:../Images/FunTheoremCalculusModHome_gr_243.gif].

We are done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/FunTheoremCalculusModHome_gr_244.gif]

[Graphics:../Images/FunTheoremCalculusModHome_gr_245.gif]



[Graphics:../Images/FunTheoremCalculusModHome_gr_246.gif]

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This solution is complements of the authors.



































 

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