Module

for

Integrands with Branch Points

 

8.6  Integrands with Branch Points

    We now show how to evaluate certain improper real integrals involving the integrand  [Graphics:Images/IntegralsBranchPointsMod_gr_1.gif].  Since the complex function  [Graphics:Images/IntegralsBranchPointsMod_gr_2.gif]  is multivalued, so we must first specify the branch to be used.

    Let [Graphics:Images/IntegralsBranchPointsMod_gr_3.gif] be a real number with [Graphics:Images/IntegralsBranchPointsMod_gr_4.gif].  In this section we use the branch of [Graphics:Images/IntegralsBranchPointsMod_gr_5.gif] corresponding to the branch of the logarithm [Graphics:Images/IntegralsBranchPointsMod_gr_6.gif] (see Equation (5-20)) as follows:

            [Graphics:Images/IntegralsBranchPointsMod_gr_7.gif],  

where  [Graphics:Images/IntegralsBranchPointsMod_gr_8.gif]  and  [Graphics:Images/IntegralsBranchPointsMod_gr_9.gif].  Note that this is not the traditional principal branch of [Graphics:Images/IntegralsBranchPointsMod_gr_10.gif] and that, as defined, the function [Graphics:Images/IntegralsBranchPointsMod_gr_11.gif] is analytic in the domain [Graphics:Images/IntegralsBranchPointsMod_gr_12.gif].

 

Theorem 8.7.  Let P(x) and Q(x) be polynomials of degree m and n, respectively, where  [Graphics:Images/IntegralsBranchPointsMod_gr_13.gif].  If  [Graphics:Images/IntegralsBranchPointsMod_gr_14.gif],  and Q(x) has a zero of order at most 1 at the origin, and [Graphics:Images/IntegralsBranchPointsMod_gr_15.gif], then  

            [Graphics:Images/IntegralsBranchPointsMod_gr_16.gif],  

where  [Graphics:Images/IntegralsBranchPointsMod_gr_17.gif]  are the non-zero poles of  [Graphics:Images/IntegralsBranchPointsMod_gr_18.gif].  

Figure 8.7  The contour C that encloses the nonzero poles  [Graphics:Images/IntegralsBranchPointsMod_gr_19.gif]  of  [Graphics:Images/IntegralsBranchPointsMod_gr_20.gif].

Proof.

Proof of Theorem 8.7 is in the book.
Complex Analysis for Mathematics and Engineering

 

Example 8.23.  Evaluate  [Graphics:Images/IntegralsBranchPointsMod_gr_21.gif],  where  [Graphics:Images/IntegralsBranchPointsMod_gr_22.gif].  

[Graphics:Images/IntegralsBranchPointsMod_gr_23.gif]

Solution.  The function  [Graphics:Images/IntegralsBranchPointsMod_gr_24.gif]  has a nonzero pole at the point [Graphics:Images/IntegralsBranchPointsMod_gr_25.gif], and the denominator has a zero of order at most 1 (in fact, exactly 1) at the origin.  Calculating the residue we get  

            [Graphics:Images/IntegralsBranchPointsMod_gr_26.gif]

Using Theorem 8.7, we have  

            [Graphics:Images/IntegralsBranchPointsMod_gr_27.gif]   

Explore Solution 8.23.

 

    We can apply the preceding ideas to other multivalued functions.  

 

Example 8.24.  Evaluate  [Graphics:Images/IntegralsBranchPointsMod_gr_43.gif],  where  [Graphics:Images/IntegralsBranchPointsMod_gr_44.gif].  

[Graphics:Images/IntegralsBranchPointsMod_gr_45.gif]

Solution.  We use the function  [Graphics:Images/IntegralsBranchPointsMod_gr_46.gif].  Recall that  

            [Graphics:Images/IntegralsBranchPointsMod_gr_47.gif],  


where  [Graphics:Images/IntegralsBranchPointsMod_gr_48.gif]  and  [Graphics:Images/IntegralsBranchPointsMod_gr_49.gif].  The path C of integration will consist of the segments [Graphics:Images/IntegralsBranchPointsMod_gr_50.gif] of the x axis together with the upper semicircles  [Graphics:Images/IntegralsBranchPointsMod_gr_51.gif]  and  [Graphics:Images/IntegralsBranchPointsMod_gr_52.gif],  for  [Graphics:Images/IntegralsBranchPointsMod_gr_53.gif],  as shown in Figure 8.8.

Figure 8.8  The contour C for the integrand  [Graphics:Images/IntegralsBranchPointsMod_gr_54.gif].

    We chose the branch  [Graphics:Images/IntegralsBranchPointsMod_gr_55.gif]  because it is analytic on C and its interior, hence so is the function f(z) which has a simple pole in the upper half-plane at the point [Graphics:Images/IntegralsBranchPointsMod_gr_56.gif].

            [Graphics:Images/IntegralsBranchPointsMod_gr_57.gif]

This choice enables us to apply the residue theorem properly (see the hypotheses of Theorem 8.1), and we get  

            [Graphics:Images/IntegralsBranchPointsMod_gr_58.gif]  

Keeping in mind the branch of logarithm that we're using, we then have  

(8-33)            [Graphics:Images/IntegralsBranchPointsMod_gr_59.gif]   

If  [Graphics:Images/IntegralsBranchPointsMod_gr_60.gif],  then by the ML inequality (Theorem 6.3)

            [Graphics:Images/IntegralsBranchPointsMod_gr_61.gif]

and L'Hôpital's rule yields  [Graphics:Images/IntegralsBranchPointsMod_gr_62.gif].  A similar computation shows that  [Graphics:Images/IntegralsBranchPointsMod_gr_63.gif].  
We use these results when we take limits Equations (8-33) to get  

            [Graphics:Images/IntegralsBranchPointsMod_gr_64.gif]   

Equating the real parts in this equation gives  

            [Graphics:Images/IntegralsBranchPointsMod_gr_65.gif]

Explore Solution 8.24.

 

Exercises for Section 8.6.  Integrands with Branch Points

 

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