Module

for

Indented Contour Integrals

 

8.5  Indented Contour Integrals

    If f(x) is continuous on the interval  [Graphics:Images/IntegralsIndentedContourMod_gr_1.gif], but discontinuous at b, then the improper integral of f(x) over [Graphics:Images/IntegralsIndentedContourMod_gr_2.gif] is defined by  

            [Graphics:Images/IntegralsIndentedContourMod_gr_3.gif]

provided that the limit exists.  Similarly, if f(x) is continuous on the interval  [Graphics:Images/IntegralsIndentedContourMod_gr_4.gif], but discontinuous at b, then the improper integral of f(x) over [Graphics:Images/IntegralsIndentedContourMod_gr_5.gif] is defined by  

            [Graphics:Images/IntegralsIndentedContourMod_gr_6.gif]

provided that the limit exists.  For example,  

            [Graphics:Images/IntegralsIndentedContourMod_gr_7.gif].

    If we let f(x) be continuous for all values of x in the interval [Graphics:Images/IntegralsIndentedContourMod_gr_8.gif],  except at the value [Graphics:Images/IntegralsIndentedContourMod_gr_9.gif],  where [Graphics:Images/IntegralsIndentedContourMod_gr_10.gif].  The Cauchy principal value of f(x) over [Graphics:Images/IntegralsIndentedContourMod_gr_11.gif] is defined by
    
            [Graphics:Images/IntegralsIndentedContourMod_gr_12.gif],  
    
provided that the limit exists.

 

Example 8.19.  Evaluate  [Graphics:Images/IntegralsIndentedContourMod_gr_13.gif].  

[Graphics:Images/IntegralsIndentedContourMod_gr_14.gif]

Solution.  Evaluating the integrals and computing limits gives  

            [Graphics:Images/IntegralsIndentedContourMod_gr_15.gif]    

Explore Solution 8.19.

 

    In this section we show how to use residues to evaluate the Cauchy principal value of the integral of f(x) over [Graphics:Images/IntegralsIndentedContourMod_gr_42.gif] when the integrand f(x) has simple poles on the x axis.  We state our main results and then look at some examples before giving proofs.

 

Theorem 8.5.  Let [Graphics:Images/IntegralsIndentedContourMod_gr_43.gif] where P(x) and Q(x) are polynomials with real coefficients of degree m and n, respectively, and  [Graphics:Images/IntegralsIndentedContourMod_gr_44.gif].  If Q(x) has simple zeros at the points [Graphics:Images/IntegralsIndentedContourMod_gr_45.gif] on the x-axis, then  

(8-20)            [Graphics:Images/IntegralsIndentedContourMod_gr_46.gif],  

where [Graphics:Images/IntegralsIndentedContourMod_gr_47.gif] are the poles of f(z) that lie in the upper half-plane.  

Proof.

Proof of Theorem 8.5 is in the book.
Complex Analysis for Mathematics and Engineering

Figure 8.6  The poles [Graphics:Images/IntegralsIndentedContourMod_gr_48.gif] of [Graphics:Images/IntegralsIndentedContourMod_gr_49.gif] that lie on the x-axis and the poles [Graphics:Images/IntegralsIndentedContourMod_gr_50.gif] that lie above the semicircles [Graphics:Images/IntegralsIndentedContourMod_gr_51.gif].

 

Theorem 8.6.  Let P(x) and Q(x) be polynomials, of degree m and n, respectively, where [Graphics:Images/IntegralsIndentedContourMod_gr_52.gif] and let Q(x) have simple zeros at the points  [Graphics:Images/IntegralsIndentedContourMod_gr_53.gif]  on the x-axis.  If [Graphics:Images/IntegralsIndentedContourMod_gr_54.gif] is a positive real number and if  [Graphics:Images/IntegralsIndentedContourMod_gr_55.gif], then we can compute the Cauchy Principal Value (P.V.) of the following integrals

(8-21)            [Graphics:Images/IntegralsIndentedContourMod_gr_56.gif],  
    
        and  
    
(8-22)            [Graphics:Images/IntegralsIndentedContourMod_gr_57.gif],    

where [Graphics:Images/IntegralsIndentedContourMod_gr_58.gif] are the poles of f(z) that lie in the upper half plane.  

Proof.

Proof of Theorem 8.6 is in the book.
Complex Analysis for Mathematics and Engineering

 

Remark. The formulas in these theorems give the Cauchy principal value of the integral, which pays special attention to the manner in which any limits are taken.  They are similar to those in Section 8.3 and Section 8.4, except here we add one-half of the value of each residue at the points [Graphics:Images/IntegralsIndentedContourMod_gr_59.gif] on the x-axis.

 

Example 8.20.  Evaluate  [Graphics:Images/IntegralsIndentedContourMod_gr_60.gif]  by using complex analysis.  

[Graphics:Images/IntegralsIndentedContourMod_gr_61.gif]

Solution.  The integrand  

            [Graphics:Images/IntegralsIndentedContourMod_gr_62.gif]  

has simple poles at the points  [Graphics:Images/IntegralsIndentedContourMod_gr_63.gif]  on the x axis and  [Graphics:Images/IntegralsIndentedContourMod_gr_64.gif]  in the upper half-plane.  Computing the residues we get

            [Graphics:Images/IntegralsIndentedContourMod_gr_65.gif]  
            
            [Graphics:Images/IntegralsIndentedContourMod_gr_66.gif]

By Theorem 8.5,  

            [Graphics:Images/IntegralsIndentedContourMod_gr_67.gif]    

Explore Solution 8.20.

 

Example 8.21.  Evaluate  [Graphics:Images/IntegralsIndentedContourMod_gr_79.gif]  by using a computer algebra system.  

[Graphics:Images/IntegralsIndentedContourMod_gr_80.gif]

Solution.  Computer algebra systems such as Mathematica or MAPLE give the indefinite integral   

            [Graphics:Images/IntegralsIndentedContourMod_gr_81.gif].  

However, for real numbers, we should write the second term as  [Graphics:Images/IntegralsIndentedContourMod_gr_82.gif]  and use the equivalent formula:  

            [Graphics:Images/IntegralsIndentedContourMod_gr_83.gif]    

Figure 8.5  Graph of  [Graphics:Images/IntegralsIndentedContourMod_gr_84.gif].

    This antiderivative has the property that  [Graphics:Images/IntegralsIndentedContourMod_gr_85.gif],  as shown in Figure 8.5.  We also compute  

            [Graphics:Images/IntegralsIndentedContourMod_gr_86.gif]  

            [Graphics:Images/IntegralsIndentedContourMod_gr_87.gif]  

The Cauchy principal limit at  [Graphics:Images/IntegralsIndentedContourMod_gr_88.gif]  as  [Graphics:Images/IntegralsIndentedContourMod_gr_89.gif]  is  

[Graphics:Images/IntegralsIndentedContourMod_gr_90.gif]  

Therefore the Cauchy principal value of the improper integral is  

        [Graphics:Images/IntegralsIndentedContourMod_gr_91.gif]   

Explore Solution 8.21.

 

Example 8.22.  Show that  [Graphics:Images/IntegralsIndentedContourMod_gr_106.gif].    

[Graphics:Images/IntegralsIndentedContourMod_gr_107.gif]

Solution.  The integrand  [Graphics:Images/IntegralsIndentedContourMod_gr_108.gif]  has simple poles at the points  [Graphics:Images/IntegralsIndentedContourMod_gr_109.gif]  on the x axis and  [Graphics:Images/IntegralsIndentedContourMod_gr_110.gif]  in the upper half-plane.  Computing the residues we get  

            [Graphics:Images/IntegralsIndentedContourMod_gr_111.gif]  

            [Graphics:Images/IntegralsIndentedContourMod_gr_112.gif]

By Theorem 8.6,

            [Graphics:Images/IntegralsIndentedContourMod_gr_113.gif]   

Explore Solution 8.22.

 

    The proofs of Theorems 8.5 and 8.6 depend on the following result.

 

Lemma 8.2.  Let f(z) have a simple pole at the point  [Graphics:Images/IntegralsIndentedContourMod_gr_129.gif]  on the x-axis.  If the contour is  [Graphics:Images/IntegralsIndentedContourMod_gr_130.gif],  then

        [Graphics:Images/IntegralsIndentedContourMod_gr_131.gif].  

Proof.

Proof of Lemma 8.2 is in the book.
Complex Analysis for Mathematics and Engineering

 

Exercises for Section 8.5.  Indented Contour Integrals

 

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Residue Calculus

Contour Integrals

Cauchy Principal Value

 

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