Exercise 1.   [Graphics:Images/IntegralsTrigImproperModHome_gr_1.gif].  

Solution 1 (a).

See text and/or instructor's solution manual.

Answer.   [Graphics:../Images/IntegralsTrigImproperModHome_gr_3.gif].  

Solution.  The complex integrand is  [Graphics:../Images/IntegralsTrigImproperModHome_gr_4.gif].  

Factor the denominator and get

                    [Graphics:../Images/IntegralsTrigImproperModHome_gr_5.gif].  

The zeros are  [Graphics:../Images/IntegralsTrigImproperModHome_gr_6.gif]  and  [Graphics:../Images/IntegralsTrigImproperModHome_gr_7.gif]  lies in the upper half-plane.

It follows that  [Graphics:../Images/IntegralsTrigImproperModHome_gr_8.gif]  has simple poles at  [Graphics:../Images/IntegralsTrigImproperModHome_gr_9.gif],  

and the pole  [Graphics:../Images/IntegralsTrigImproperModHome_gr_10.gif]  lies in the upper half plane.

                              [Graphics:../Images/IntegralsTrigImproperModHome_gr_11.gif]

                    The contour  [Graphics:../Images/IntegralsTrigImproperModHome_gr_12.gif]  consisting of the semi-circle  [Graphics:../Images/IntegralsTrigImproperModHome_gr_13.gif]  and the interval [Graphics:../Images/IntegralsTrigImproperModHome_gr_14.gif].  

                    The point  [Graphics:../Images/IntegralsTrigImproperModHome_gr_15.gif]  lies in the upper half plane.

Using Theorem 8.1 (Cauchy's Residue Theorem), and Theorem 8.4 (Contour Integration for Improper Trig. Integrals), the value of the integral is computed   

                    [Graphics:../Images/IntegralsTrigImproperModHome_gr_16.gif]  

Here the denominator of  f(z) has a factor of the form  [Graphics:../Images/IntegralsTrigImproperModHome_gr_17.gif],  and  [Graphics:../Images/IntegralsTrigImproperModHome_gr_18.gif].    

In this exercise, the limit can be calculated as follows:

                    [Graphics:../Images/IntegralsTrigImproperModHome_gr_19.gif]   

We are done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/IntegralsTrigImproperModHome_gr_20.gif]

[Graphics:../Images/IntegralsTrigImproperModHome_gr_21.gif]


[Graphics:../Images/IntegralsTrigImproperModHome_gr_22.gif]

[Graphics:../Images/IntegralsTrigImproperModHome_gr_23.gif]

Maple can check our work too!

     > residue( exp(I*z)/(z^2+9), z=3*I );

                            
[Graphics:../Images/IntegralsTrigImproperModHome_gr_24.gif]

     > -2*Pi*Im( residue(exp(I*z)/(z^2+9),z=3*I) );

                            
[Graphics:../Images/IntegralsTrigImproperModHome_gr_25.gif]

                    [Graphics:../Images/IntegralsTrigImproperModHome_gr_26.gif]

                    A portion of the "area under the curve"  [Graphics:../Images/IntegralsTrigImproperModHome_gr_27.gif].  

We are really done.

Aside.  Both [Graphics:../Images/IntegralsTrigImproperModHome_gr_28.gif] and [Graphics:../Images/IntegralsTrigImproperModHome_gr_29.gif] are capable of finding the definite integral.

[Graphics:../Images/IntegralsTrigImproperModHome_gr_30.gif]

[Graphics:../Images/IntegralsTrigImproperModHome_gr_31.gif]


     > f := proc(x) cos(x)/(x^2+9) end proc;

                f := proc(x) cos(x)/(x^2 + 9) end proc

     > simplify(convert(int(f(x), x=-infinity..infinity),exp));

                            [Graphics:../Images/IntegralsTrigImproperModHome_gr_32.gif]

We are really really done.

Aside.  Both [Graphics:../Images/IntegralsTrigImproperModHome_gr_33.gif] and [Graphics:../Images/IntegralsTrigImproperModHome_gr_34.gif] are capable of finding the indefinite integral.  

                    [Graphics:../Images/IntegralsTrigImproperModHome_gr_35.gif]

                    The indefinite integral  [Graphics:../Images/IntegralsTrigImproperModHome_gr_36.gif].  From the graph we see that  

                    [Graphics:../Images/IntegralsTrigImproperModHome_gr_37.gif].  

      The indefinite integral will involve the special functions "Cosine Integral" and "Sine Integral."   We do not want to develop these special functions at this time and leave it for the reader to investigate their properties.  It is interesting to observe that the current software includes these functions.   

[Graphics:../Images/IntegralsTrigImproperModHome_gr_38.gif]

[Graphics:../Images/IntegralsTrigImproperModHome_gr_39.gif]

 

 

     > int( cos(x)/(x^2+9) , x );

               [Graphics:../Images/IntegralsTrigImproperModHome_gr_40.gif]  

Remark.  We encourage the reader to explore the following limit values using software such as [Graphics:../Images/IntegralsTrigImproperModHome_gr_41.gif] and/or [Graphics:../Images/IntegralsTrigImproperModHome_gr_42.gif].

[Graphics:../Images/IntegralsTrigImproperModHome_gr_43.gif]

[Graphics:../Images/IntegralsTrigImproperModHome_gr_44.gif]


[Graphics:../Images/IntegralsTrigImproperModHome_gr_45.gif]

[Graphics:../Images/IntegralsTrigImproperModHome_gr_46.gif]


[Graphics:../Images/IntegralsTrigImproperModHome_gr_47.gif]

[Graphics:../Images/IntegralsTrigImproperModHome_gr_48.gif]



















This solution is complements of the authors.





































 

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