Solution 1 (c).

See text and/or instructor's solution manual.

Solution.  Here we have    and the denominator is zero when ,  where n is an integer.  

The integrand    is analytic everywhere except at the points  ,  all of which lie outside the contour  .

Hence, by the Cauchy-Goursat Theorem we have   .  

                    The points    and    which lie outside the contour  .
                    (Indeed, all of the points  ,  lie outside the contour  . )

We are done.   

Aside.  We can let Mathematica double check our work.

This solution is complements of the authors.

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