Solution 13 (b).
See text and/or instructor's solution manual.
Answer
.
Solution. Factor
the denominator of the integrand
to
see that
.
The integrand
is
analytic everywhere except at the points
which
lie inside the contour
.
Now use partial fractions and express the integrand
as
, and
split up the integral into two parts
![[Graphics:../Images/CauchyGoursatModHome_gr_530.gif]](../Images/CauchyGoursatModHome_gr_530.gif)
Notice that the contour
is
negatively oriented and the contour
is
positively oriented.
Use the Cauchy-Goursat
Theorem to determine that
and
.
Now use Corollary
6.1 to evaluate the last two
integrals:
and ![]()
Therefore, we have
![[Graphics:../Images/CauchyGoursatModHome_gr_538.gif]](../Images/CauchyGoursatModHome_gr_538.gif)
Figure
6.28 The points
and
which
lie inside the contour
.
We are done.
Aside. In
Section
6.5 we will learn to evaluate contour integrals with the
Cauchy Integral Formula.
The more advanced
methods are easier to use but require special attention to
details.
Techniques for evaluating integrals using contours is an important
area of complex analysis,
and we will explore this in detail starting with the Residue Calculus
in Section
8.1.
Once the Residue
Calculus has been introduced in Section
8.1, computing this integral will be a trivial:
.
We are really 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