Solution 9.

See text and/or instructor's solution manual.

Answer.  [Graphics:../Images/IntegralRepresentationModHome_gr_348.gif].

                    [Graphics:../Images/IntegralRepresentationModHome_gr_349.gif]

                    The point  [Graphics:../Images/IntegralRepresentationModHome_gr_350.gif]  that lies inside the contour  [Graphics:../Images/IntegralRepresentationModHome_gr_351.gif].  

Solution.  The integrand  [Graphics:../Images/IntegralRepresentationModHome_gr_352.gif] is not defined at the point  [Graphics:../Images/IntegralRepresentationModHome_gr_353.gif]  which lies interior to the circle  [Graphics:../Images/IntegralRepresentationModHome_gr_354.gif],

and the integral  [Graphics:../Images/IntegralRepresentationModHome_gr_355.gif]  has the form  [Graphics:../Images/IntegralRepresentationModHome_gr_356.gif]  where   [Graphics:../Images/IntegralRepresentationModHome_gr_357.gif],  

so we can use Cauchy's Integral formula for derivatives (see Section 6.5).

Here we have   [Graphics:../Images/IntegralRepresentationModHome_gr_358.gif]   and   [Graphics:../Images/IntegralRepresentationModHome_gr_359.gif]   and calculation reveals that   [Graphics:../Images/IntegralRepresentationModHome_gr_360.gif].  

Applying Cauchy's Integral formula for derivatives   [Graphics:../Images/IntegralRepresentationModHome_gr_361.gif]   with   [Graphics:../Images/IntegralRepresentationModHome_gr_362.gif]   we write   

                              [Graphics:../Images/IntegralRepresentationModHome_gr_363.gif].  

Then multiplication by  [Graphics:../Images/IntegralRepresentationModHome_gr_364.gif]  establishes the desired result  

                              [Graphics:../Images/IntegralRepresentationModHome_gr_365.gif].

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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