Module

for

Improper Integrals of Rational Functions

 

 

8.3  Improper Integrals of Rational Functions

 

The Cauchy principal value of an improper integral.

 

    In calculus we studied improper integrals, and now we will extend this concept.  First, we shall assume that 

[Graphics:Images/IntegralsRationalMod_gr_1.gif] is a continuous function, and is defined on the semi-infinite interval  [Graphics:Images/IntegralsRationalMod_gr_2.gif].  Also, we assume that

for any real number [Graphics:Images/IntegralsRationalMod_gr_3.gif],  satisfying  [Graphics:Images/IntegralsRationalMod_gr_4.gif],  the integral  [Graphics:Images/IntegralsRationalMod_gr_5.gif]  is properly defined.  

          [Graphics:Images/IntegralsRationalMod_gr_6.gif]  

                                        [Graphics:Images/IntegralsRationalMod_gr_7.gif]   is the area over the interval  [Graphics:Images/IntegralsRationalMod_gr_8.gif].  

Next, we can extend the integral to the entire interval  [Graphics:Images/IntegralsRationalMod_gr_9.gif],  and write   [Graphics:Images/IntegralsRationalMod_gr_10.gif],  

and this can be accomplished if we can take the limit when [Graphics:Images/IntegralsRationalMod_gr_3.gif] goes to infinity.  

        Recall from calculus that the improper integral of  [Graphics:Images/IntegralsRationalMod_gr_11.gif]  over  [[Graphics:Images/IntegralsRationalMod_gr_12.gif])  is defined by

(8-7)               [Graphics:Images/IntegralsRationalMod_gr_13.gif],  
    
provided that the limit exists.  

          [Graphics:Images/IntegralsRationalMod_gr_14.gif]  

                              [Graphics:Images/IntegralsRationalMod_gr_15.gif]   is the area over the interval   [[Graphics:Images/IntegralsRationalMod_gr_16.gif]).  

        If  [Graphics:Images/IntegralsRationalMod_gr_17.gif]  is defined for all real [Graphics:Images/IntegralsRationalMod_gr_18.gif], then the integral of  [Graphics:Images/IntegralsRationalMod_gr_19.gif]  over  [Graphics:Images/IntegralsRationalMod_gr_20.gif]  is defined by

(8-7)               [Graphics:Images/IntegralsRationalMod_gr_21.gif],  

provided both limits exist.  

          [Graphics:Images/IntegralsRationalMod_gr_22.gif]  

                    [Graphics:Images/IntegralsRationalMod_gr_23.gif]   is the area over the interval  [Graphics:Images/IntegralsRationalMod_gr_24.gif].

     If the two limits in Equation (8-7) exists, then we can obtain the value of the integral by taking a single limit,

(8-8)               [Graphics:Images/IntegralsRationalMod_gr_25.gif].  

          [Graphics:Images/IntegralsRationalMod_gr_26.gif]  

                         [Graphics:Images/IntegralsRationalMod_gr_27.gif]   is the area over the interval [Graphics:Images/IntegralsRationalMod_gr_28.gif].

 

        For a wide class of functions the limit on the right side of Equation (8-8) exists, but the right side of

Equation (8-7) doesn't exist. It is our goal in this section to explain how the definition in Equation  (8-8)

leads to the construction of a certain a contour and then we will show how to use the power of complex

numbers and the residue calculus to evaluate the improper integral   [Graphics:Images/IntegralsRationalMod_gr_29.gif].  

 

Definition 8.2  (Cauchy Principal Value - P. V.). Let [Graphics:Images/IntegralsRationalMod_gr_30.gif] be a continuous real valued function for all [Graphics:Images/IntegralsRationalMod_gr_31.gif].

The Cauchy principal value ([Graphics:Images/IntegralsRationalMod_gr_32.gif]) of the integral   [Graphics:Images/IntegralsRationalMod_gr_33.gif]   is defined by

                    [Graphics:Images/IntegralsRationalMod_gr_34.gif],  

provided the limit exists.

 

Example 8.13.   [Graphics:Images/IntegralsRationalMod_gr_35.gif]   does not exist,     and     [Graphics:Images/IntegralsRationalMod_gr_36.gif].   

                    [Graphics:Images/IntegralsRationalMod_gr_37.gif]  

Solution.  

        If we attempt to use Equation (8-7) then we obtain

                    [Graphics:Images/IntegralsRationalMod_gr_38.gif]  

and the last computation   "[Graphics:Images/IntegralsRationalMod_gr_39.gif]"   is undefined .  

Thus, the improper integral   [Graphics:Images/IntegralsRationalMod_gr_40.gif]   does not exist.  

        If we use Equation (8-8) then we obtain

                    [Graphics:Images/IntegralsRationalMod_gr_41.gif]   

This computation is well defined and is known as the Cauchy principal value ([Graphics:Images/IntegralsRationalMod_gr_42.gif]) of   [Graphics:Images/IntegralsRationalMod_gr_43.gif].   

Therefore,

                    [Graphics:Images/IntegralsRationalMod_gr_44.gif].    

Explore Solution 8.13.

 

Extra Example 1.   [Graphics:Images/IntegralsRationalMod_gr_78.gif]    and also    [Graphics:Images/IntegralsRationalMod_gr_79.gif].  

                    [Graphics:Images/IntegralsRationalMod_gr_80.gif]  

Solution.  

        If we attempt to use Equation (8-7) then we obtain
        
                    [Graphics:Images/IntegralsRationalMod_gr_81.gif]   

and we say that   [Graphics:Images/IntegralsRationalMod_gr_82.gif]   diverges to  "[Graphics:Images/IntegralsRationalMod_gr_83.gif]."  

If computations with the extended real numbers are permitted, then

                    [Graphics:Images/IntegralsRationalMod_gr_84.gif].  

        If we attempt to use Equation (8-8) then we obtain

                    [Graphics:Images/IntegralsRationalMod_gr_85.gif]  

and we still say that   [Graphics:Images/IntegralsRationalMod_gr_86.gif]   diverges to  "[Graphics:Images/IntegralsRationalMod_gr_87.gif]."  

If computations with the extended real numbers are permitted, then

                    [Graphics:Images/IntegralsRationalMod_gr_88.gif].  

Explore Extra Solution 1.

 

Example 8.14.  Find the Cauchy principal value of   [Graphics:Images/IntegralsRationalMod_gr_128.gif].

[Graphics:Images/IntegralsRationalMod_gr_129.gif][Graphics:Images/IntegralsRationalMod_gr_130.gif]

                    A portion of the area under the curve   [Graphics:Images/IntegralsRationalMod_gr_131.gif],   over the interval  [Graphics:Images/IntegralsRationalMod_gr_132.gif].

Solution.  

                    [Graphics:Images/IntegralsRationalMod_gr_133.gif]   

Explore Solution 8.14.

 


     If [Graphics:Images/IntegralsRationalMod_gr_259.gif], where [Graphics:Images/IntegralsRationalMod_gr_260.gif] and [Graphics:Images/IntegralsRationalMod_gr_261.gif] are polynomials, then [Graphics:Images/IntegralsRationalMod_gr_262.gif] is called a rational function. Techniques

in calculus were developed to integrate rational functions.  We now show how the Cauchy's Residue Theorem

can be used to obtain the Cauchy Principal Value, ([Graphics:Images/IntegralsRationalMod_gr_263.gif]) of the integral of  [Graphics:Images/IntegralsRationalMod_gr_264.gif]  over  [Graphics:Images/IntegralsRationalMod_gr_265.gif].  

 

How to evaluate an improper integral with a contour integral.

 

     The improper integral [Graphics:Images/IntegralsRationalMod_gr_266.gif] involves a real function and the contour integral [Graphics:Images/IntegralsRationalMod_gr_267.gif] involves

a complex function. How can they be related? It's really simple (no pun intended), we just used our standard

"trick" and replaced the variable [Graphics:Images/IntegralsRationalMod_gr_268.gif] with the variable [Graphics:Images/IntegralsRationalMod_gr_269.gif].  But something strange is happening here, integration

over [Graphics:Images/IntegralsRationalMod_gr_270.gif] has been replaced with integration along the contour [Graphics:Images/IntegralsRationalMod_gr_271.gif]. What is the connection between  

                     [Graphics:Images/IntegralsRationalMod_gr_272.gif]    and    [Graphics:Images/IntegralsRationalMod_gr_273.gif]  ?

     We must learn how to place the real integral in its proper complex variable context, and this is accomplished

by forming a very special contour in the complex plane.  The secret is the construction of the contour  

                    [Graphics:Images/IntegralsRationalMod_gr_274.gif].  

Here we have


                    [Graphics:Images/IntegralsRationalMod_gr_275.gif] which is a line segment in the real [Graphics:Images/IntegralsRationalMod_gr_276.gif]-axis,
                    
and

                    [Graphics:Images/IntegralsRationalMod_gr_277.gif] is a semi-circle of radius [Graphics:Images/IntegralsRationalMod_gr_278.gif] in the upper half-plane.

 

Theorem 8.3 (Contour Integration for Rational Functions). Let [Graphics:Images/IntegralsRationalMod_gr_279.gif] where [Graphics:Images/IntegralsRationalMod_gr_280.gif] and [Graphics:Images/IntegralsRationalMod_gr_281.gif]

are polynomials, of degree  [Graphics:Images/IntegralsRationalMod_gr_282.gif],  respectively.   If   [Graphics:Images/IntegralsRationalMod_gr_283.gif]   for all real [Graphics:Images/IntegralsRationalMod_gr_284.gif] and   [Graphics:Images/IntegralsRationalMod_gr_285.gif],   then

the Cauchy Principal Value (or [Graphics:Images/IntegralsRationalMod_gr_286.gif]) of the integral is

                    [Graphics:Images/IntegralsRationalMod_gr_287.gif].  

where  [Graphics:Images/IntegralsRationalMod_gr_288.gif]  are the poles of  [Graphics:Images/IntegralsRationalMod_gr_289.gif]  that lie in the upper half-plane, as shown in Figure 8.4.  

                    [Graphics:Images/IntegralsRationalMod_gr_290.gif]  

                    Figure 8.4  The poles  [Graphics:Images/IntegralsRationalMod_gr_291.gif]  of  [Graphics:Images/IntegralsRationalMod_gr_292.gif]  that lie in the upper half-plane.

                    Remark.  The residues at the poles in the lower-half plane are not used in the computation. 

Proof.

 

Example 8.15.  Use the residue calculus to compute   [Graphics:Images/IntegralsRationalMod_gr_293.gif].  

          [Graphics:Images/IntegralsRationalMod_gr_294.gif][Graphics:Images/IntegralsRationalMod_gr_295.gif]

               A portion of the area under the curve [Graphics:Images/IntegralsRationalMod_gr_296.gif], over the interval [Graphics:Images/IntegralsRationalMod_gr_297.gif].

Solution.  

        We write the integrand as  [Graphics:Images/IntegralsRationalMod_gr_298.gif].  

We see that  [Graphics:Images/IntegralsRationalMod_gr_299.gif]  has simple poles at the points  [Graphics:Images/IntegralsRationalMod_gr_300.gif]  and  [Graphics:Images/IntegralsRationalMod_gr_301.gif],  and that the

poles at  [Graphics:Images/IntegralsRationalMod_gr_302.gif]  and  [Graphics:Images/IntegralsRationalMod_gr_303.gif],  are the only singularities of  [Graphics:Images/IntegralsRationalMod_gr_304.gif]  in the upper half-plane.  

                              [Graphics:Images/IntegralsRationalMod_gr_305.gif]  

                    The contour [Graphics:Images/IntegralsRationalMod_gr_306.gif] consisting of the semi-circle [Graphics:Images/IntegralsRationalMod_gr_307.gif] and the interval [Graphics:Images/IntegralsRationalMod_gr_308.gif].

                    The points  [Graphics:Images/IntegralsRationalMod_gr_309.gif]  lie in the upper half-plane.  

Computing the residues, we obtain  

                    [Graphics:Images/IntegralsRationalMod_gr_310.gif]  

Similarly,

                    [Graphics:Images/IntegralsRationalMod_gr_311.gif]   

Using Theorem 8.3, we conclude that  

                    [Graphics:Images/IntegralsRationalMod_gr_312.gif]  

Explore Solution 8.15.

 

Example 8.16.  Use the residue calculus to compute   [Graphics:Images/IntegralsRationalMod_gr_448.gif].  

          [Graphics:Images/IntegralsRationalMod_gr_449.gif][Graphics:Images/IntegralsRationalMod_gr_450.gif]

                    A portion of the area under the curve   [Graphics:Images/IntegralsRationalMod_gr_451.gif],   over the interval  [Graphics:Images/IntegralsRationalMod_gr_452.gif].  

Solution.  

        The integrand [Graphics:Images/IntegralsRationalMod_gr_453.gif] has a poles of order [Graphics:Images/IntegralsRationalMod_gr_454.gif] at the points [Graphics:Images/IntegralsRationalMod_gr_455.gif],

and the poles a  [Graphics:Images/IntegralsRationalMod_gr_456.gif]  is the only singularity of  [Graphics:Images/IntegralsRationalMod_gr_457.gif]  in the upper half-plane.  

 

                              [Graphics:Images/IntegralsRationalMod_gr_458.gif]  

                    The contour [Graphics:Images/IntegralsRationalMod_gr_459.gif] consisting of the semi-circle [Graphics:Images/IntegralsRationalMod_gr_460.gif] and the interval [Graphics:Images/IntegralsRationalMod_gr_461.gif].

                    The point  [Graphics:Images/IntegralsRationalMod_gr_462.gif]  lies in the upper half-plane.

 

Computing the residue at   [Graphics:Images/IntegralsRationalMod_gr_463.gif], we get  

                    [Graphics:Images/IntegralsRationalMod_gr_464.gif]   

Using Theorem 8.3, we conclude that  

                    [Graphics:Images/IntegralsRationalMod_gr_465.gif]  

Explore Solution 8.16.

 

 

Tutorial Exercises for Section 8.3.  Improper Integrals of Rational Functions

 

 

Library Research Experience for Undergraduates

Residue Calculus  

Contour Integrals  

Cauchy Principal Value  

Hilbert Transformation  

The z-Transform  

 

 

 

 

The Next Module is

Improper Integrals Involving Trigonometric Functions

 

 

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This material is coordinated with our book Complex Analysis for Mathematics and Engineering.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) 2012 John H. Mathews, Russell W. Howell