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function (1.50), compute the integral over the azimuth and level items with
the equal numbers
m
in the left-hand and right-hand parts of the equation.
Only the items with equal numbers
m
will be nonzero in the product of the
series (the phase function by the radiance) due to the orthogonality of the
trigonometric functions:
π
2
ϕ
cos
m
2
ϕ
d
ϕ
=
0for
m
1
=
cos
m
1
m
2
,
0
π
=
for
m
1
m
2
,
π
2
ϕ
cos
m
2
ϕ
d
ϕ
=
sin
m
1
0.
0
Finally, obtain:
1
=
ω
0
(
τ
)
B
m
(
p
m
(
µ
)
I
m
(
µ
)
d
µ
τ
µ
µ
0
)
τ
µ
τ
µ
,
,
,
,
,
,
2
(1.66)
−1
+
ω
0
(
τ
)
Sp
m
(
τ
µ
µ
0
)exp(−
τ|µ
0
).
,
,
4
Further from (1.51) the following equation is derived:
τ
µ
µ
0
)
dI
m
(
,
,
µ
=
−
I
m
(
τ
µ
µ
0
)+
B
m
(
τ
µ
µ
0
) ,
,
,
,
,
(1.67)
τ
d
with boundary conditions:
I
m
(0,
µ
µ
0
)
=
µ
0
>
0and
,
0, for
(1.68)
I
m
(
τ
0
,
µ
µ
0
)
=
µ
0
<
0.
,
0for
The following relations are correct for it:
τ
µ
0
)exp
−
τ
d
τ
1
µ
−
I
m
(
τ
µ
µ
0
)
=
B
m
(
τ
,
µ
τ
µ
,
,
,
>
0,
µ
0
(1.69)
τ
0
µ
0
)exp
−
τ
d
τ
−
1
µ
−
I
m
(
B
m
(
τ
,
τ
τ
µ
µ
0
)
=
µ
µ
,
,
,
<
0.
µ
τ
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