Exercise 23.  Consider the following computation for evaluating ,  
where C is the line segment from  .  

.  

Is this a valid computation ?  Why ?  Justify your answer.   

Solution 23.

See text and/or instructor's solution manual.

Answer.  The computation is not valid because    is not defined at the point .  

Solution.  The function    is analytic on the entire complex plane, except at the origin, and

the function    is analytic on the entire complex plane, except points on the negative x-axis and at the origin,

and    has    is an antiderivative.

Hence, the anti-derivative is not defined at the point on the line segment from .  

Therefore the computation is not valid.  

                    The path of integration is the line segment C from  .  

                    Notice that    is not analytic on it's branch cut   .

We are Done.

Caveat.  A older version of  Mathematica might give the wrong answer  .

The current version Mathematica 6.0 will give the correct response:

                    .
          
Maple gives the correct answer:

                    

                     

For this example the reader is responsible to determine if the answer is justified.  

Use A Different Path of Integration.  

Aside.  Suppose that the contour is changed and C is the portion of the circle    from  .  

                    The path of integration C is the portion of the circle    from  .  

                    Notice that    is not analytic on it's branch cut   .

                    Observe that    is analytic on a domain containing the contour C.

        The stated computation is valid if the path of integration C is the portion of the circle    from  .  

Notice that this path does not cross the branch cut    !

Use A Different Branch of the Logarithm.  

Aside.  If a different branch of the logarithm is used,  ,  then the computation would be valid.  

According to equation (5-20),   ,    where   ,  and   .  

Then we have      where     and  .  

Notice that   is not analytic on it's branch cut   .

Using this branch,  ,  of the logarithm, we have  

                    ,    and  

                    ,   

and a valid computation would be:

                      

                    The path of integration is the line segment C from  .  

                    Notice that    is not analytic on it's branch cut   .

Remark.  The new versions of Mathematica and Matlab both calculate the value    for   .

How do you think they are programmed to do it?

Remark.  It is interesting to follow the trends in computer algebra software and notice that

they keep improving their computations for definite integrals.  

However, we must be vigilant and be aware that sometimes computers might make a mistake.

This solution is complements of the authors.

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