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on the other hand,
+
∞
+
∞
f
(
e
−
st
d
t
f
(
lim
s
t
)
=
t
)
d
t
=
f
(+
∞
)
−
f
(
0
)
,
→
0
0
0
where taking the limit inside the integral is allowed by the existence theorem. A
comparison of the two equations yields
0
s f
f
(+
∞
)=
lim
s
(
s
)
.
→
is the limit of
s f
Therefore the final value of
f
0.
Theorems of initial value and final value play an important role in finding
f
(
t
)
(
s
)
as
s
→
(
0
)
f
and
f
(+
∞
)
from the image function
(
s
)
.
1
If
L
[
f
(
t
)] =
, for example, then
s
+
α
s
s
f
(
0
)=
lim
s
+
α
=
1
,
f
(+
∞
)=
lim
s
→
0
+
α
=
0
.
s
s
→
∞
B.2.3 Determine Inverse Image Functions by Calculating Residues
,
s
n
be all singular points of
f
Theorem.
Let
s
1
,
s
2
,
···
(
s
)
. Take a real constant
β
f
such that all the
s
k
are in the half plane
Re
(
s
)
<
β
. Suppose that lim
s
→
∞
(
s
)=
0. The
inverse image function is thus
j
=
1
Res
f
(
s
)
e
st
n
s
j
,
f
(
t
)=
,
where
Res
f
s
j
stands for the residue of
f
e
st
e
st
at
s
(
s
)
,
(
s
)
=
s
j
.
Proof.
Consider the half circle
C
=
L
+
C
R
in Fig. B.1 where all singular points
s
j
are also singular points of
f
e
st
. By the residue theorem,
in
C
(
j
=
1
,
2
, ··· ,
n
)
(
s
)
j
=
1
Res
f
(
s
)
e
st
n
s
j
,
f
e
st
d
s
(
s
)
=
2
π
i
,
C
i.e.,
β
+
i
R
e
st
d
s
j
=
1
Res
f
(
s
)
e
st
n
s
j
.
1
f
f
e
st
d
s
(
)
+
(
)
=
,
s
s
2
π
i
β
−
i
R
C
R
, by applying the Jordan lemma from the theory of functions of complex
variables, we have
As
R
→
+
∞
f
e
st
d
s
lim
R
→
+
∞
(
s
)
=
0
.
C
R
Therefore
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