Image Processing Reference
In-Depth Information
It could now be written that
(
Ki
2
−− =
ε
)
()
K
2
((
K
2
i
ε
)
1 2
/
K
)((
Ki
2
ε
)
12
/
+
K
)
(2.53)
i
ε
KK i
ε

-

-
+
K
+
+
K
2
K
2
K
which has poles at K
′=±+
(
K
(
i
ε/2
K
))
giving
1
2
1
(
e
iK r
e
iKr
)
KdK
Gr
(, )
0
=−
lim
4
π
2
ir
(2.54)
ε

i
K
-
KK i
K

ε
-
ε
0
K
+
+
+
K
−∞
2
2
and then use Cauchy's residue theorem
fz z
( d
=
i
Residues
C
in the upper-half plane (uhp) of the complex plane and the
m
1
1
d
d

-
Residue
==
a
)! lim
(
zafz
)
m
()
1
(
m
1
z
za
Hence, the
1
0
i
ε
+
-
residue =
lim
KK
fz
()

!
2
K
KK i
K
ε
→+
2
Assuming Jordan's lemma holds, namely, that as R → ∞, ∫ uhp = 0.
Therefore, residue = R 1 + R 2 where
eK
Ki
iK r
R
=
lim
1
+
(
ε/
2
KK
)
+
KK i
ε
→+
2
k
e
iK
(
+
(
i
ε
/)) (
2
Kr
K
+
(
i
ε
/
2
K
)
)
R
=−
(2.55)
1
2
Ki
+
(
ε
/
K
)
and as ε → 0, R 1 → − (1/2) e iKr .
eKK
iK r
d
The second integral is −
[
]
[
]
2 2
If we let K ′ = − K ′, then this integral has already been evaluated and so, R 1 = R 2
Ki
+
(
ε
/
KKKi
)
+
+
(
ε
/
KK
)
1
2
1
1
1
4
e
iKr
(2.56)
Gr
(, )
0
=−
2
π
iR
(
+=
R
)
1
2
4
π
2
i r
π
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