Solution 12 (b).

See text and/or instructor's solution manual.

Answer.  [Graphics:../Images/IntegralRepresentationModHome_gr_557.gif][Graphics:../Images/IntegralRepresentationModHome_gr_558.gif].

                    [Graphics:../Images/IntegralRepresentationModHome_gr_559.gif]

                    The points  [Graphics:../Images/IntegralRepresentationModHome_gr_560.gif]  that lie inside the contour  [Graphics:../Images/IntegralRepresentationModHome_gr_561.gif].  

Solution.  For this part of the exercise there are two points [Graphics:../Images/IntegralRepresentationModHome_gr_562.gif] that lie inside the circle [Graphics:../Images/IntegralRepresentationModHome_gr_563.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_564.gif].

Therefore, we need to split the integrand into two parts  [Graphics:../Images/IntegralRepresentationModHome_gr_565.gif],  and the integral can be written as  

                    [Graphics:../Images/IntegralRepresentationModHome_gr_566.gif]

Then we have two integrals to compute:   [Graphics:../Images/IntegralRepresentationModHome_gr_567.gif]   and   [Graphics:../Images/IntegralRepresentationModHome_gr_568.gif]

Solution part (a).  Show that  [Graphics:../Images/IntegralRepresentationModHome_gr_569.gif].

                    [Graphics:../Images/IntegralRepresentationModHome_gr_570.gif]

                    The point  [Graphics:../Images/IntegralRepresentationModHome_gr_571.gif]  that lies inside the contour  [Graphics:../Images/IntegralRepresentationModHome_gr_572.gif].  

Here the integrand  [Graphics:../Images/IntegralRepresentationModHome_gr_573.gif] is not defined at the point  [Graphics:../Images/IntegralRepresentationModHome_gr_574.gif]  which lies interior to the circle  [Graphics:../Images/IntegralRepresentationModHome_gr_575.gif],

and the integral  [Graphics:../Images/IntegralRepresentationModHome_gr_576.gif]  has the form  [Graphics:../Images/IntegralRepresentationModHome_gr_577.gif]  so we can use the Cauchy Integral Formula (see Section 6.5).

Here we have  [Graphics:../Images/IntegralRepresentationModHome_gr_578.gif]  and calculation reveals that  [Graphics:../Images/IntegralRepresentationModHome_gr_579.gif].  

Applying the Cauchy's integral formula  [Graphics:../Images/IntegralRepresentationModHome_gr_580.gif]  we write   

                              [Graphics:../Images/IntegralRepresentationModHome_gr_581.gif].  

Then multiplication by  [Graphics:../Images/IntegralRepresentationModHome_gr_582.gif]  establishes the desired result  

                              [Graphics:../Images/IntegralRepresentationModHome_gr_583.gif].

Solution part (b).  Show that  [Graphics:../Images/IntegralRepresentationModHome_gr_584.gif].

                    [Graphics:../Images/IntegralRepresentationModHome_gr_585.gif]

                    The point  [Graphics:../Images/IntegralRepresentationModHome_gr_586.gif]  that lies inside the contour  [Graphics:../Images/IntegralRepresentationModHome_gr_587.gif].  

Here the integrand  [Graphics:../Images/IntegralRepresentationModHome_gr_588.gif] is not defined at the point  [Graphics:../Images/IntegralRepresentationModHome_gr_589.gif]  which lies interior to the circle  [Graphics:../Images/IntegralRepresentationModHome_gr_590.gif],

and the integral  [Graphics:../Images/IntegralRepresentationModHome_gr_591.gif]  has the form  [Graphics:../Images/IntegralRepresentationModHome_gr_592.gif]  so we can use the Cauchy Integral Formula (see Section 6.5).

Here we have  [Graphics:../Images/IntegralRepresentationModHome_gr_593.gif]  and calculation reveals that  [Graphics:../Images/IntegralRepresentationModHome_gr_594.gif].  

Applying the Cauchy's integral formula  [Graphics:../Images/IntegralRepresentationModHome_gr_595.gif]  we write   

                              [Graphics:../Images/IntegralRepresentationModHome_gr_596.gif].  

Then multiplication by  [Graphics:../Images/IntegralRepresentationModHome_gr_597.gif]  establishes the desired result  

                              [Graphics:../Images/IntegralRepresentationModHome_gr_598.gif].

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

Therefore  [Graphics:../Images/IntegralRepresentationModHome_gr_599.gif][Graphics:../Images/IntegralRepresentationModHome_gr_600.gif].

We are done.   

Aside.  Be patient, it is our goal to develop an elegant and efficient method to compute this type of contour integral.  

The method is known as the "Residue Calculus," and is introduced in Section 8.1.  

Looking Forward.  Cauchy's Residue Theorem (see Theorem 8.1) will use the notion of  [Graphics:../Images/IntegralRepresentationModHome_gr_601.gif] given in Definition 8.1.

For this problems we have chosen the details so that the integrand  [Graphics:../Images/IntegralRepresentationModHome_gr_602.gif]  is analytic inside C and on C,  

except at the point  [Graphics:../Images/IntegralRepresentationModHome_gr_603.gif]  that lie inside C.  Then Cauchy's Residue Theorem states

                              [Graphics:../Images/IntegralRepresentationModHome_gr_604.gif].  

Using the Residue Calculus.  Let us announce that in Section 8.1, we can compute  [Graphics:../Images/IntegralRepresentationModHome_gr_605.gif]  with  [Graphics:../Images/IntegralRepresentationModHome_gr_606.gif],  and  [Graphics:../Images/IntegralRepresentationModHome_gr_607.gif],  and

                    [Graphics:../Images/IntegralRepresentationModHome_gr_608.gif]   and   [Graphics:../Images/IntegralRepresentationModHome_gr_609.gif],   and   [Graphics:../Images/IntegralRepresentationModHome_gr_610.gif].
                              
Then the "Residue Calculus" gives a quick way to find the answer:

                    [Graphics:../Images/IntegralRepresentationModHome_gr_611.gif]   
            
After the concept of  [Graphics:../Images/IntegralRepresentationModHome_gr_612.gif] is developed in Section 8.1, you will enjoy calculating integrals the "easy way."

We are really done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/IntegralRepresentationModHome_gr_613.gif]

[Graphics:../Images/IntegralRepresentationModHome_gr_614.gif]



[Graphics:../Images/IntegralRepresentationModHome_gr_615.gif]

[Graphics:../Images/IntegralRepresentationModHome_gr_616.gif]



[Graphics:../Images/IntegralRepresentationModHome_gr_617.gif]

[Graphics:../Images/IntegralRepresentationModHome_gr_618.gif]



[Graphics:../Images/IntegralRepresentationModHome_gr_619.gif]

[Graphics:../Images/IntegralRepresentationModHome_gr_620.gif]

Preview of calculating a residue.

In Section 8.1 we will show that if the denominator of F(z) has a factor of the form  [Graphics:../Images/IntegralRepresentationModHome_gr_621.gif],  then  [Graphics:../Images/IntegralRepresentationModHome_gr_622.gif].  

In this exercise, the the limit is calculated as follows:

                    [Graphics:../Images/IntegralRepresentationModHome_gr_623.gif]  

          and

                    [Graphics:../Images/IntegralRepresentationModHome_gr_624.gif]  

We are really really done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/IntegralRepresentationModHome_gr_625.gif]

[Graphics:../Images/IntegralRepresentationModHome_gr_626.gif]



[Graphics:../Images/IntegralRepresentationModHome_gr_627.gif]

[Graphics:../Images/IntegralRepresentationModHome_gr_628.gif]

Remark.  Since   [Graphics:../Images/IntegralRepresentationModHome_gr_629.gif]   and   [Graphics:../Images/IntegralRepresentationModHome_gr_630.gif]  

the calculation with Cauchy's integral formula is the same as the residue calculation using a limit.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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