Example 6.10.  (a)   Show that   [Graphics:Images/ContourIntegralMod_gr_325.gif]   but that   [Graphics:Images/ContourIntegralMod_gr_326.gif],  

where [Graphics:Images/ContourIntegralMod_gr_327.gif] is the semicircular path from [Graphics:Images/ContourIntegralMod_gr_328.gif], in the upper half plane, and [Graphics:Images/ContourIntegralMod_gr_329.gif] is the polygonal path from [Graphics:Images/ContourIntegralMod_gr_330.gif], respectively, shown in Figure 6.9.  

Figure 6.9  The two contours [Graphics:Images/ContourIntegralMod_gr_333.gif] and [Graphics:Images/ContourIntegralMod_gr_334.gif] joining  [Graphics:Images/ContourIntegralMod_gr_335.gif].  

Explore Solution 6.10 (a)

(a)  Use the semicircular path from -1 to 1 in the upper half plane,  set up the parameterization and compute the contour integral.  Enter the function  [Graphics:../Images/ContourIntegralMod_gr_363.gif].  

[Graphics:../Images/ContourIntegralMod_gr_364.gif]




[Graphics:../Images/ContourIntegralMod_gr_365.gif]

 

 

Method.  The integral can be computed using a complex integrand.

[Graphics:../Images/ContourIntegralMod_gr_366.gif]




[Graphics:../Images/ContourIntegralMod_gr_367.gif]

[Graphics:../Images/ContourIntegralMod_gr_368.gif]

[Graphics:../Images/ContourIntegralMod_gr_369.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) 2006 John H. Mathews, Russell W. Howell