Exercise 13.   .     

Solution 13.

See text and/or instructor's solution manual.

Answer.   .  

Solution.  The complex integrand can be written in the form  

                    ,    

and    where    and  .  

The cube root function    has a logarithmic singularity at the origin  ,  because it's complex expression involves

the branch of the logarithm (see Equation (5-20) in Section 5.2) as follows:

                    ,  

where    and    (notice that the branch cut is the positive x-axis).  

Therefore,    has a logarithmic singularity at the origin  ,  and a pole of order  2  at  .

     It will suffice to consider the pole    and the contour  ,  

which consists of the semi-circles   and the segments   on the x-axis.

                    The contour    consisting of the intervals and the semi-circles  .  

                    The point    lies inside the contour  .

Method I.  The integrand ,  with  ,  satisfies the hypothesis of Theorem 8.7 ,

and the value of the integral is computed using the residue at the pole .

                      

The residue is calculated by using Theorem 8.1 (Cauchy's Residue Theorem).  

Here the denominator of  f(z) has a factor of the form  ,  and  .

                      

We are done.   

Aside.  We can let Mathematica double check our work.

Maple can check our work too!

     > V1 := limit( diff((z+1)^2*z^(1/3)/(z+1)^2,z), z=-1);

               

     > evalc( 2*Pi*I/(1-exp(I*2*Pi/3))*V1 );

               

We are really done.   

Method II.  If more details are desired, then consider evaluating the contour integral of    over  .  

Applying the Cauchy Residue Theorem (see Theorem 8.1 in Section 8.1), we obtain:

                    ,

                    and

                    .

The parameterizations for   are   ,  

and   .     

Along    we have  ,   and

                    .  

Now substitute this in the previous equation and write

                    .

      The ML-Inequality and a proof similar to the one used to establish equation (8-11) can be used to prove that:

                         and          (see the complete details below).

Taking the limits as    we obtain

                    ,

                    then

                    ,  

                    then

                     ,  

                    then

                      .

Equating the real parts we get:

                      ,  
                      
which is easy to solve to obtain the desired conclusion

                      ,  

Or, we can equate the imaginary parts and get:

                    
                      
which is also easy to solve to obtain the desired conclusion

                      .  

                    A portion of the "area under the curve"  .  

We are really really done.   

The complete details.

     For completeness, we now give the details for establishing    and  .  

Use the ML-Inequality (Theorem 6.3 in Section 6.2) and a proof similar to the one used to establish equation (8-11) of Theorem 8.3 to prove that:

                    .  

For this example, for  ,  we have the relations  

                      

Now taking limits, we get

                    .

Therefore, it follows that  .  

A similar proof shows that:

                    .  

For this example, for  ,  we have the relations  

                      

Now taking limits, we get

                    .

Therefore, it follows that  .  

We are done with the details.

Aside.  Both and are capable of finding the definite integral.

     > int( x^(1/3)/(x+1)^2, x=0..infinity );

               

We are really really really done.

The above computation illustrates how the Residue Calculus can be used to find Contour Integrals with Integrands with Branch Points.  

Aside.  For comparison, we show how to find the value of the integral using the indefinite integral.

                    The indefinite integral  .   From the graph we see that  

                    .

Caveat.  If you do not trust the graphical explanation then you must trust the Residue Calculus solution.  

Or you could compute the Cauchy Principal value, but this will require some limits.

                    The Cauchy Principal Value (P.V.) of the integral is    and it can be computed as follows:

                      

This solution is complements of the authors.

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