Example 8.12.    Evaluate  [Graphics:Images/IntegralsTrigMod_gr_105.gif].  

[Graphics:Images/IntegralsTrigMod_gr_106.gif]

Explore Solution 8.12.

Enter the function  [Graphics:../Images/IntegralsTrigMod_gr_123.gif] and use substitution to obtain the complex function f[z].  

[Graphics:../Images/IntegralsTrigMod_gr_124.gif]




[Graphics:../Images/IntegralsTrigMod_gr_125.gif]

 

 

Locate the singularities of  [Graphics:../Images/IntegralsTrigMod_gr_126.gif].  

[Graphics:../Images/IntegralsTrigMod_gr_127.gif]



[Graphics:../Images/IntegralsTrigMod_gr_128.gif]

 

 

 

Find out which singularities lie inside the unit circle  [Graphics:../Images/IntegralsTrigMod_gr_129.gif].  

[Graphics:../Images/IntegralsTrigMod_gr_130.gif]



[Graphics:../Images/IntegralsTrigMod_gr_131.gif]

[Graphics:../Images/IntegralsTrigMod_gr_132.gif]

 

 

Compute the residues at  [Graphics:../Images/IntegralsTrigMod_gr_133.gif], and use the residue calculus to compute the value of the integral.

[Graphics:../Images/IntegralsTrigMod_gr_134.gif]





[Graphics:../Images/IntegralsTrigMod_gr_135.gif]

[Graphics:../Images/IntegralsTrigMod_gr_136.gif]

[Graphics:../Images/IntegralsTrigMod_gr_137.gif]

 

 

 

 

Mathematica  is capable of finding some of these difficult integrals.  The function  [Graphics:../Images/IntegralsTrigMod_gr_138.gif]  can be integrated.

[Graphics:../Images/IntegralsTrigMod_gr_139.gif]


[Graphics:../Images/IntegralsTrigMod_gr_140.gif]

 

 

However, direct substitution using [Graphics:../Images/IntegralsTrigMod_gr_141.gif] and the endpoints of the interval seems to be erroneous.  This is due to the fact that in the interior of the interval, [Graphics:../Images/IntegralsTrigMod_gr_142.gif] is undefined. Hence [Graphics:../Images/IntegralsTrigMod_gr_143.gif]is discontinuous.

[Graphics:../Images/IntegralsTrigMod_gr_144.gif]


[Graphics:../Images/IntegralsTrigMod_gr_145.gif]

 

 

This is due to the fact that  [Graphics:../Images/IntegralsTrigMod_gr_146.gif] is undefined.  A graph of  [Graphics:../Images/IntegralsTrigMod_gr_147.gif]  reveals that it has a discontinuity at  [Graphics:../Images/IntegralsTrigMod_gr_148.gif]  in the middle of the interval  [Graphics:../Images/IntegralsTrigMod_gr_149.gif].  This is a violation of the Fundamental Theorem of Calculus, which asserts that the integral of a continuous function must be continuous.

[Graphics:../Images/IntegralsTrigMod_gr_150.gif]



[Graphics:../Images/IntegralsTrigMod_gr_151.gif]

[Graphics:../Images/IntegralsTrigMod_gr_152.gif]

 

 

 

The indefinite integrand  [Graphics:../Images/IntegralsTrigMod_gr_153.gif]  can be used piecewise over  [Graphics:../Images/IntegralsTrigMod_gr_154.gif]  where it is continuous, by using the left and right hand limits at the endpoints.

[Graphics:../Images/IntegralsTrigMod_gr_155.gif]



[Graphics:../Images/IntegralsTrigMod_gr_156.gif]

 

 

The above answer is correct too.  Limits were required because  [Graphics:../Images/IntegralsTrigMod_gr_157.gif]  is not continuous at  [Graphics:../Images/IntegralsTrigMod_gr_158.gif].  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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