Exercise 3.  Consider the integral  [Graphics:Images/ContourIntegralModHome_gr_126.gif],  where C is the positively oriented upper semicircle of radius 1, centered at 0.  

3 (b).  Compute the integral exactly by selecting a parametrization for C and applying Theorem 6.1.

Solution 3 (b).

See text and/or instructor's solution manual.

Answer.  [Graphics:../Images/ContourIntegralModHome_gr_137.gif].  

Solution Method I.  A parametrization for C is   [Graphics:../Images/ContourIntegralModHome_gr_138.gif]   for   [Graphics:../Images/ContourIntegralModHome_gr_139.gif].

                    [Graphics:../Images/ContourIntegralModHome_gr_140.gif]

                    The contour   [Graphics:../Images/ContourIntegralModHome_gr_141.gif]   for   [Graphics:../Images/ContourIntegralModHome_gr_142.gif].

The function is  [Graphics:../Images/ContourIntegralModHome_gr_143.gif]  and the curve is  [Graphics:../Images/ContourIntegralModHome_gr_144.gif]  and we obtain

                               [Graphics:../Images/ContourIntegralModHome_gr_145.gif]  and  [Graphics:../Images/ContourIntegralModHome_gr_146.gif],  

Use these substitutions and get  


                               [Graphics:../Images/ContourIntegralModHome_gr_147.gif]  

The last two real integrals are computed using

                               [Graphics:../Images/ContourIntegralModHome_gr_148.gif]  
                    
                               and

                               [Graphics:../Images/ContourIntegralModHome_gr_149.gif]  

We are done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/ContourIntegralModHome_gr_150.gif]



[Graphics:../Images/ContourIntegralModHome_gr_151.gif]

[Graphics:../Images/ContourIntegralModHome_gr_152.gif]



[Graphics:../Images/ContourIntegralModHome_gr_153.gif]

[Graphics:../Images/ContourIntegralModHome_gr_154.gif]



[Graphics:../Images/ContourIntegralModHome_gr_155.gif]

[Graphics:../Images/ContourIntegralModHome_gr_156.gif]



[Graphics:../Images/ContourIntegralModHome_gr_157.gif]

[Graphics:../Images/ContourIntegralModHome_gr_158.gif]



[Graphics:../Images/ContourIntegralModHome_gr_159.gif]

[Graphics:../Images/ContourIntegralModHome_gr_160.gif]

We are really done.   

Solution Method II.  We could use a method that uses the following complex computations.

Solution.  A parametrization for C is   [Graphics:../Images/ContourIntegralModHome_gr_161.gif]   for   [Graphics:../Images/ContourIntegralModHome_gr_162.gif].

                    [Graphics:../Images/ContourIntegralModHome_gr_163.gif]

          The contour   [Graphics:../Images/ContourIntegralModHome_gr_164.gif]   for   [Graphics:../Images/ContourIntegralModHome_gr_165.gif].

The function is  [Graphics:../Images/ContourIntegralModHome_gr_166.gif]  and the curve is  [Graphics:../Images/ContourIntegralModHome_gr_167.gif]  for  [Graphics:../Images/ContourIntegralModHome_gr_168.gif].   Then we obtain

                    [Graphics:../Images/ContourIntegralModHome_gr_169.gif]   and   [Graphics:../Images/ContourIntegralModHome_gr_170.gif],  
                    
                    then

                     [Graphics:../Images/ContourIntegralModHome_gr_171.gif]  

Here we have used the calculations indicated by equation (6-8) in Section 6.1:

                    [Graphics:../Images/ContourIntegralModHome_gr_172.gif]    

We are really really done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/ContourIntegralModHome_gr_173.gif]


[Graphics:../Images/ContourIntegralModHome_gr_174.gif]

[Graphics:../Images/ContourIntegralModHome_gr_175.gif]



[Graphics:../Images/ContourIntegralModHome_gr_176.gif]

[Graphics:../Images/ContourIntegralModHome_gr_177.gif]



[Graphics:../Images/ContourIntegralModHome_gr_178.gif]

[Graphics:../Images/ContourIntegralModHome_gr_179.gif]



[Graphics:../Images/ContourIntegralModHome_gr_180.gif]

[Graphics:../Images/ContourIntegralModHome_gr_181.gif]



[Graphics:../Images/ContourIntegralModHome_gr_182.gif]

[Graphics:../Images/ContourIntegralModHome_gr_183.gif]

We are really really really done.   

Remark concerning parts (a) and (b).  The approximation in part (a) was   [Graphics:../Images/ContourIntegralModHome_gr_184.gif]

and the exact value in part (b) is   [Graphics:../Images/ContourIntegralModHome_gr_185.gif]

We are really really really  really done.   

Aside.  After we have developed the Cauchy-Goursat Theorem in Section 6.3 and the Fundamental Theorem of Calculus in Section 6.4

we will be able to compute the integral in an easy fashion:

                    [Graphics:../Images/ContourIntegralModHome_gr_186.gif].  

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/ContourIntegralModHome_gr_187.gif]

[Graphics:../Images/ContourIntegralModHome_gr_188.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 



This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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