Exercise 1. Use Equations (6-1) and (6-2) to find
1 (e).
.
Solution 1 (e).
See text and/or instructor's solution manual.
Answer
.
Solution. Given
the real functions u(t) and v(t)
with
and
, we
use the formula
![]()
.
First, expand the integrand into it's real and
imaginary parts
![]()
.
Here we have
, and
.
Now integrate u(t) and v(t) and
obtain
, and
.
Compute values for U(t) and V(t)
Compute the real definite integrals
![[Graphics:../Images/ComplexIntegralModHome_gr_180.gif]](../Images/ComplexIntegralModHome_gr_180.gif)
![[Graphics:../Images/ComplexIntegralModHome_gr_181.gif]](../Images/ComplexIntegralModHome_gr_181.gif)
Evaluate the complex definite integral
Some points
in
the interval of integration and their images
, for
.
The right endpoint is
and the last image point is
.
We are done.
Aside. We can let Mathematica double check our work.
The details for this computation are:
![[Graphics:../Images/ComplexIntegralModHome_gr_199.gif]](../Images/ComplexIntegralModHome_gr_199.gif)
Some points
in
the interval of integration and their images
, for
.
The right endpoint is
and the last image point is
.
We are done.
Aside. After we have developed
the topics of contour integrals (in Section
6.2), the independence of path for integration of an
analytic function (in Section
6.3),
and established the Fundamental Theorem of Calculus (in Section
6.4) then we will be able to revisit this integral and use
the more straightforward computation:
Remark. It is
important to understand whether you are permitted to use only real
variables in computing an integral,
or whether you are permitted to use a complex variable (
)
and complex functions in computing the integral.
The three popular software packages Maple, Matlab and Mathematica use
complex variable based computations.
This solution is complements of the authors.
(c) 2008 John H. Mathews, Russell W. Howell