Solution 17.

See text and/or instructor's solution manual.

Answer.

                      [Graphics:../Images/IntegralRepresentationModHome_gr_867.gif]  

 

                    [Graphics:../Images/IntegralRepresentationModHome_gr_868.gif]

                    The points  [Graphics:../Images/IntegralRepresentationModHome_gr_869.gif]  that lie inside the contour  [Graphics:../Images/IntegralRepresentationModHome_gr_870.gif].  

Solution.  For this part of the exercise there are two points [Graphics:../Images/IntegralRepresentationModHome_gr_871.gif] that lie inside the circle [Graphics:../Images/IntegralRepresentationModHome_gr_872.gif].

Both Cauchy's Integral formula and Cauchy's Integral formula for derivatives apply when only one point lies inside the circle [Graphics:../Images/IntegralRepresentationModHome_gr_873.gif].

Therefore, we need to split the integrand into two parts  [Graphics:../Images/IntegralRepresentationModHome_gr_874.gif],  and the integral can be written as  

                    [Graphics:../Images/IntegralRepresentationModHome_gr_875.gif]

Then we have two integrals to compute:   [Graphics:../Images/IntegralRepresentationModHome_gr_876.gif]   and   [Graphics:../Images/IntegralRepresentationModHome_gr_877.gif].  

Solution part (a).  Show that  [Graphics:../Images/IntegralRepresentationModHome_gr_878.gif].

                    [Graphics:../Images/IntegralRepresentationModHome_gr_879.gif]

                    The point  [Graphics:../Images/IntegralRepresentationModHome_gr_880.gif]  that lies inside the contour  [Graphics:../Images/IntegralRepresentationModHome_gr_881.gif].  

Here the integrand  [Graphics:../Images/IntegralRepresentationModHome_gr_882.gif]  is not defined at the point  [Graphics:../Images/IntegralRepresentationModHome_gr_883.gif]  which lies interior to the circle  [Graphics:../Images/IntegralRepresentationModHome_gr_884.gif],

and the integral  [Graphics:../Images/IntegralRepresentationModHome_gr_885.gif]  has the form  [Graphics:../Images/IntegralRepresentationModHome_gr_886.gif]  where  [Graphics:../Images/IntegralRepresentationModHome_gr_887.gif]  so we can use the Cauchy Integral Formula (see Section 6.5).

Here we have  [Graphics:../Images/IntegralRepresentationModHome_gr_888.gif]  and calculation reveals that  [Graphics:../Images/IntegralRepresentationModHome_gr_889.gif].  

Applying the Cauchy's integral formula  [Graphics:../Images/IntegralRepresentationModHome_gr_890.gif].  

Then multiplication by  [Graphics:../Images/IntegralRepresentationModHome_gr_891.gif]  establishes the desired result  

                              [Graphics:../Images/IntegralRepresentationModHome_gr_892.gif].

Solution part (b).  Show that  [Graphics:../Images/IntegralRepresentationModHome_gr_893.gif].

                    [Graphics:../Images/IntegralRepresentationModHome_gr_894.gif]

                    The point  [Graphics:../Images/IntegralRepresentationModHome_gr_895.gif]  that lies inside the contour  [Graphics:../Images/IntegralRepresentationModHome_gr_896.gif].  

Here the integrand  [Graphics:../Images/IntegralRepresentationModHome_gr_897.gif] is not defined at the point  [Graphics:../Images/IntegralRepresentationModHome_gr_898.gif]  which lies interior to the circle  [Graphics:../Images/IntegralRepresentationModHome_gr_899.gif],

and the integral  [Graphics:../Images/IntegralRepresentationModHome_gr_900.gif]  has the form  [Graphics:../Images/IntegralRepresentationModHome_gr_901.gif]  where  [Graphics:../Images/IntegralRepresentationModHome_gr_902.gif]  so we can use the Cauchy Integral Formula (see Section 6.5).

Here we have  [Graphics:../Images/IntegralRepresentationModHome_gr_903.gif]  and calculation reveals that  [Graphics:../Images/IntegralRepresentationModHome_gr_904.gif].  

Applying the Cauchy's integral formula  [Graphics:../Images/IntegralRepresentationModHome_gr_905.gif].  

Then multiplication by  [Graphics:../Images/IntegralRepresentationModHome_gr_906.gif]  establishes the desired result  

                              [Graphics:../Images/IntegralRepresentationModHome_gr_907.gif].

Now we combine the two results from Solution part (a) and solution part (b).  

Therefore

                     [Graphics:../Images/IntegralRepresentationModHome_gr_908.gif]   

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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