Exercise 5.   .   

Hint:  Use the complex integrand  .  

Solution 5.

See text and/or instructor's solution manual.

Answer.   .  

Solution.  We use the complex function  .   The path C of integration will consist of

the segments of the x-axis together with the upper semicircle  ,  for  ,  as shown below.  

                    The contour    consisting of the interval ,     and the semi-circle  .  

                    The point    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  .  

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:

                         (see the complete details below).

Taking the limit as    we obtain  

                    .  

On the line segment    use the change of variable    to write

                     ,  

Then use this substitute it in left side of the previous equation to obtain  

                    

Now we can write  

                    

Taking the real part results, we obtain the desired result

                    .

We are done.   

      For curiosity sake, the imaginary part is

                    ,

this also follows immediately from Example 8.14 in Section 8.3.  

Aside.  We can let Mathematica double check our work.

Maple can check our work too!

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

               

     > V := evalc( 2*Pi*I*V1 );

               

     > Re(V);

               

     > Im(V);

               

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  .  

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  .  

We are done with the details.

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

     > int( log(x^2+1)/(x^2+1), 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 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 function "PolyLog."  

      We do not want to develop this special 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