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π
0
2 π *
1 1
sen cos
2 n n′ { n n n n′ n n′ n n n n′ n n′
n+ n′
+
( ) ( ) 2Re ⎡[ ][ ] [ ][ ] ⎤}
= ∫dθr θ θ ∑ E E −− i i a + b ππ +ττ + a + b τπ +πτ
2ρ ⎣
⎦
Só falta agora, definir a integral I3
que é referente aos primeiros e segundos,
′+ 1 + 1 ρ
{{ ( ) ( ) [ ][ ]}
π
2 π
* n n 2i
3 ∫ sen cos
2 ∑ n n′ n′ n′
n n n n n n n n
ρ
0
2
I = dθ r θ θ E E − −i −i e π −τ a τ + b π −a π −b
τ −
π
0
n ′
{() () [ ]
}}
+ 1 n + 1 − 2 i ρ
* * * *
i i e n n
⎡
⎣
an′ n′ bn′ n′ an′ n′ bn′ n′
⎤
⎦
π −τ τ + π − π − τ =
1 1
{ 2 π
sen cos
* 2 Re
2
2 n n′ n′ n′
n n n n n n n n
n′+ n+ iρ
{ ( ) ( ) [ ][ ]}
= ∫ dθ r θ θ ∑ E E − −i −i e π −τ a τ + b π −a π −b
τ =
2ρ π
π
* n′+ 1 n+ 1 2iρ
I3 = ∫ dθsenθcosθ ∑ E ′ 2Re −( − ) ( − ) ×
2 nEn
i i e
2k
0
[ an n n′ bn n n′ an n n′ an n n′ an n n′ bn n n′ bn n n′ bn n n′
]}
× τ π + π π − π π − τ τ + π τ − τ π − π τ + τ τ
{
Usando as soluções dessas integrais em θ demonstradas por Gouesbet,
( n + 1)
( n )
π
2 !
∫ [ πn′ τ n +πnτn′ ] senθcos
θ dθ= δn,
n′
2n+ 1 −1
!
0
e
π
∫[ n n′ n n′
]
0
2 2 2 2
2n ( n+ 1)( n+ 2) 2n( n+ 1) ( n−1)
ππ +ττ senθcos θdθ= se n′ = n+ 1, se n′ = n−1, 0 se n′
≠ n±
1
( 2n+ 1)( 2n+ 3) ( 2n+ 1)( 2n−1)
temos,
( n + )
( )
π n+ 1 n+
1 n n 2n+ 1 2n+
1 2 1!
*
I1 = ∑ ( −i) ( i) ( i) ( − i)
E0E0
2Re⎡⎣a ⎤⎦+
2
nbn
k n( n+ 1) n( n+ 1) 2n+ 1 n−1 !
∑
2 2
π n+ 1 n+ 2 n n+
1 2n+ 1 2n+ 3 2 n ( n+ 1)( n+
2) * *
( −i) () i ()( i − i)
E0E0 ⎡⎣b + 1+ + 1⎤⎦+
2
nbn anan
k n( n+ 1) n+ 1 ( n+ 2) (2n+ 1)(2n+
3)
( )
∑
π n+ 1 n n n−1
2n+ 1 2n−1
2 nn ( + 1) ( n−1)
( −i) ()()( i i −i)
E
2
0E0
k n( n+
1) n− 1 n (2n+ 1)(2n−1)
( )
2 2
⎡⎣bb
+ aa
* *
n n−1 n n−1
⎤⎦
234