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

[Graphics:Images/IntegralsIndentedContourMod_gr_14.gif]

Explore Solution 8.19.

We must guide Mathematica and write an appropriate form of the anti-derivative.

[Graphics:../Images/IntegralsIndentedContourMod_gr_16.gif]


[Graphics:../Images/IntegralsIndentedContourMod_gr_17.gif]

 

 

The integrand is discontinuous at x = 0 where it has a singularity.  Indeed, Mathematica has a branch cut for  [Graphics:../Images/IntegralsIndentedContourMod_gr_18.gif]  along the negative x-axis and maps these points onto the ray [Graphics:../Images/IntegralsIndentedContourMod_gr_19.gif].  Hence the function  [Graphics:../Images/IntegralsIndentedContourMod_gr_20.gif]  does not have the graph as taught in elementary calculus.  We must use the following function  h[x].

[Graphics:../Images/IntegralsIndentedContourMod_gr_21.gif]



[Graphics:../Images/IntegralsIndentedContourMod_gr_22.gif]

[Graphics:../Images/IntegralsIndentedContourMod_gr_23.gif]

 

 

Now we pursue the Cauchy principal value of the integral.

[Graphics:../Images/IntegralsIndentedContourMod_gr_24.gif]




[Graphics:../Images/IntegralsIndentedContourMod_gr_25.gif]

 

 

 

 

Warning. The following computations will not produce the principal value of the real integral.  Indeed, Mathematica has a branch cut for  [Graphics:../Images/IntegralsIndentedContourMod_gr_26.gif]  along the negative x-axis and maps these points onto the ray [Graphics:../Images/IntegralsIndentedContourMod_gr_27.gif],  in particular [Graphics:../Images/IntegralsIndentedContourMod_gr_28.gif].  

[Graphics:../Images/IntegralsIndentedContourMod_gr_29.gif]




[Graphics:../Images/IntegralsIndentedContourMod_gr_30.gif]

 

 

 

 

Correction. The following choice for the form for the indefinite integral will produce the desired computation.

[Graphics:../Images/IntegralsIndentedContourMod_gr_31.gif]




[Graphics:../Images/IntegralsIndentedContourMod_gr_32.gif]

[Graphics:../Images/IntegralsIndentedContourMod_gr_33.gif]

[Graphics:../Images/IntegralsIndentedContourMod_gr_34.gif]

 

 

 

Which is also the correct answer.

Remark. The PrincipalValue option does not give and answer to this problem.

[Graphics:../Images/IntegralsIndentedContourMod_gr_35.gif]


[Graphics:../Images/IntegralsIndentedContourMod_gr_36.gif]

[Graphics:../Images/IntegralsIndentedContourMod_gr_37.gif]

 

 

We are done.

Aside.  Observe that  [Graphics:../Images/IntegralsIndentedContourMod_gr_38.gif]

[Graphics:../Images/IntegralsIndentedContourMod_gr_39.gif]





[Graphics:../Images/IntegralsIndentedContourMod_gr_40.gif]

[Graphics:../Images/IntegralsIndentedContourMod_gr_41.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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