Exercise 15.      and    .

Remark.  We have already calculated the value for    in Exercise 14, by using a different method.  

Solution 15.

See text and/or instructor's solution manual.

Answer.      and   .  

Solution.  We use the function   ,   where the branch of the logarithm is    

                    ,   and the branch of the square root is  

                    ,  

where    and  ,   (see Equation (5-20) in Section 5.2),  (notice that the branch cut is the negative y-axis).  

      The path  C  of integration will consist of the segments    and   of the x-axis

together with the upper semicircles    and  ,  for  ,  as shown in the figure below.

                    The indented contour    for the integrand  

                    consists of the large semi-circle , and a small semi-circle , and segments    along the real axis.  

                    The point    that lies inside the contour  .  

      Factor the denominator and get

                    .  

The zeros of   are  .

It follows that    has simple poles at  .

      Now evaluate the contour integral of    over  .  

We chose the branch    because it is analytic on C and its interior.  

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

              

Here the denominator of    has a factor of the form  ,  and  .  

              

      We can write the contour integral in the following form:

                    ,  
                    
                    or
                    
                    .  

      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

                    ,
                    
                    then

                    ,
                    
                    then

                    .   

It is easier to see what is happening if we make the substitutions     and   ,  then the previous equation is

                    
                    
which we can write as a system of equations

                    

This can be solved using Cramer's rule  

                       
                    
                       

Therefore,  we have shown that    

                    ,    and   

                    .  

Aside.  We can let Mathematica double check our work.

Maple can check our work too!

     > V1 := evalc( residue( (z^(1/3)*log(z))/(z^2+1), z=I ));

               

     > evalc( 2*Pi*I*V1 );

               

    > restart;
     > solve( { 3/2*V1 - sqrt(3)/2*Pi*V2 = -1/4*Pi^2,
     >          sqrt(3)/2*V1 + 1/2*Pi*V2 = sqrt(3)/4*Pi^2 } );

               

We are really done.   

                    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.

Remark.  Mathematica 4.0 can't get the correct value for the definite integral.

Remark.  Mathematica 6.0 will get the correct value for the definite integral.

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

               

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 very difficult limits.

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

                      

      The indefinite integral will involve the special functions "HypergeometricPFQ"  and "Hypergeometric2F1".

      We do not want to develop these specials functions at this time and leave it for the reader to investigate.  It is interesting to observe that the current software includes this function.  For this type of problem the Residue Calculus can get the answer without resorting to special functions.

This solution is complements of the authors.

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