Example 3
.
Evaluate
Z
2
π
0
1
sin
θ
+ 2
dθ
.
Answer
.
We put
z
=
e
jθ
and get
Z
2
π
0
1
sin
θ
+ 2
dθ
=
Z
C
1
(
z

1
z
)
1
2
j
+ 2
dz
jz
where
C
is the unit circle. Now
Z
C
dz
2
jz
+ (
z
2

1)
1
2
=
Z
C
2
dz
z
2

1 + 4
jz
.
The roots of the bottom are
z
=

4
j
±
√

16 + 4
2
= (

2
±
√
3)
j,
so
z
2
+ 4
jz

1 = (
z
+ (2 +
√
3)
j
)(
z
+ (2

√
3)
j
)
.
C
153
Observe that (

2 +
√
3)
j
is inside
C
, but (

2

√
3)
j
is not, since
(

(

2 +
√
3)
j

= 2

√
3
<
1
,

(

2

√
3)
j

= 2 +
√
3
>
1
.
We also note that the pole at (

2 +
√
3)
j
is
simple
. Then
Z
C
2
z
2
+ 4
jz

1
dz
= 2
πj
2
2
√
3
j
=
2
π
√
3
.
As a final comment, we recall that
f
(
x
) is even iff
f
(
x
) =
f
(

x
). Suppose
f
(
x
) is even, then
Z
∞
0
f
(
x
)
dx
=
1
2
Z
∞
∞
f
(
x
)
dx.
We can thus evaluate
R
∞
0
f
(
x
)
dx
by evaluating
R
∞
∞
f
(
x
)
dx
and dividing by 2. As
an example consider
R
∞
0
1
1+
x
2
dx
.
Here
f
(
x
) =
1
1+
x
2
for
f
(
x
) =
f
(

x
) and
f
is
even
. We thus have
R
∞
0
1
1+
x
2
dx
=
1
2
·
π
from Example 1. Note also that
f
(
x
) is
odd
if
f
(
x
) =

f
(

x
), so
Z
∞
∞
f
(
x
)
dx
=
lim
R
→∞
Z
R

R
f
(
x
)
dx
but
Z
R

R
f
(
x
)
dx
=
Z
0

R
f
(
x
)
dx
+
Z
R
0
f
(
x
)
dx
=

Z
R
0
f
(
x
)
dx
+
Z
R
0
f
(
x
)
dx
= 0
,
and
R
∞
∞
f
(
x
)
dx
= 0! Unfortunately, we can’t do
R
∞
0
f
(
x
)
dx
if
f
(
x
) is
odd
.
154
III.h
Inverting the Laplace Transform and Applications
We now come one of the main reasons for all the work we have done: inverting
the Laplace Transform. The basic idea—constructing suitable paths in the complex
plane—is similar to the previous section.
You may recall that given
y
=
f
(
t
)
(usually
t
is actually time),
L
(
f
)(
s
)—the Laplace Transform of
f
—is given by
L
(
f
)(
s
) =
Z
∞
0
e

st
f
(
t
)
dt.
In elementary courses,
s
is thought of as being a real number, but in practical
problems this is
not
the case and you should think of
s
as being complex. Indeed,
s
=
jw
with
w
= frequency is of particular significance.
So we wish to think as
L
(
f
)(
s
) being given for any complex
s
.
The formulas you hopefully recall hold
unchanged for
s
complex. In particular,
L
(
e
at
)(
s
) = 1
/
(
s

a
) if Re (
s
)
>
Re (
a
).
Now let
f
(
s
) be the transform of some function
h
(
t
). We know
f
(
s
) and wish
to find
h
(
t
), i.e., we wish to invert the transform. In practice,
f
(
s
) will have some
poles at some points
s
1
, . . ., s
m
of the complex plane. As is often the case in practice
we start by assuming
s
1
, . . ., s
m
all lie in the left half plane. (We shall deal with
the other case later.) We construct the path
C
=
C
1
+
C
2
with
C
2
a semicircle of
radius
R
as shown.
C
2
C
R
1
s
1
s
2
s
m
s
3
s
j R
 j R
Think of
R
as being large. If
s
is any point inside the semicircle (as shown), then
155
since
f
is analytic inside and on
C
we have
f
(
s
) =
1
2
πj
Z
C
f
(
z
)
z

s
dz.
Think of
s
as being temporarily fixed, and assume

f
(
z
)
 →
0 as

z
 → ∞
. This
is the case for most “practical” functions but not for all. The latter case is more
complicated and we do not deal with it in this course. Look at
C
2
: if
z
is on
C
2
then

f
(
z
)
 →
0 as

z

(i.e.,
R
) approaches
∞
, and 1
/

z

s

also goes to zero as

z
 → ∞
. One can formally show that as a consequence
R
C
2
f
(
z
)
/
(
z

s
)
dz
→
0 as
R
→ ∞
, and a detailed proof is given in theoretical courses. So, since
f
(
s
) never
changes as
R
→ ∞
, we get
f
(
s
) =
1
2
πj
Z
C
f
(
z
)
z

s
dz
=
1
2
πj
Z
C
1
f
(
z
)
z

s
dz
+
Z
C
2
f
(
z
)
z

s
dz
→
1
2
πj
Z