Geoscience Reference
In-Depth Information
The other variables in matrices
R
and
T
remain continuous in transition
from the left branch cut of contour
Γ
2
m
to the right. Therefore
2
k
A
−
k
2
1
∆
SK
−
1
∆
SK
=
−
k
A
+
k
2
,
ζ
4
/ζ
3
−
iτ
K
2
−
1
∆
SK
−
,
∆
S
∆
SK
−
∆
S
∆
SK
k
0
hY
2
∆
(0)
A
1
∆
SK
=
−
(8.60)
|
sin
I
|
where
∆
SK
,∆
S
and
∆
SK
,
∆
S
are the values of
∆
SK
,
∆
S
on the right-hand
and left-hand parts of the integration contour, respectively. It follows from
(8.48) that the main contribution to the integral is from
s
≤
1. Since
k
=
k
A
+
i
sh
x
,
k
A
k
then at
x
h
with
1weget
|
|
1 . From (7.125) and (8.60) we
obtain
k
A
−
k
2
∆
T
m
≈
∆
T
(0
m
≈
4
i
k
0
hζ
(0
4
/ζ
(0)
τ
K
2
−
3
⎛
⎝
⎞
⎠
i
k
0
hY
(
m
)
k
0
hY
(
m
)
Y
∆
(0)
A
g
ζ
(0)
3
g
ζ
(0)
3
−
−
i
sign
I
×
.
(8.61)
k
0
h
2
ε
a
k
2
Y
2
∆
(0
A
2
k
0
h
2
ε
a
k
2
Y
∆
(0)
A
sign
I
Let now the conductivity be a layer with
σ
g
= const and perfectly con-
ducting underlying half-space at depth
H
. Let also
|
kh
|
1and
|
kd
g
|
1.
Then (7.113)-(7.115) and (7.125)-(7.128) give
ζ
(0)
3
(0)
4
ik
0
hY
(
m
)
Y
(
m
)
g
=1
−
,
=
−
g
cot
(1 +
i
)
H
d
g
.
=
i
1+
i
k
0
d
Y
(
m
)
g
Now the integral in (8.48) can be calculated. Let
√
t
2
∞
Φ
(
ξ
)=
8
it
(1 + 2
it
)
2
exp (
−
π
exp (
iξ
)
−
2
ξt
)
dt,
(8.62)
0
with Re
√
t
2
−
it >
0, then
k
A
−
∞
k
2
sx
)
ds
=
πx
exp (
−
2
h
exp (
−
ik
A
x
)
Φ
(
k
A
x
)
.
k
2
0
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