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( 2n
+ 1)
∞
∞
4π 2 * 2 π 2 nn ( + 2) * *
1
= ∑
+ ∑
⎡
+
+
2 0 n n 2 0 +
= + = ( + )
⎣ n n 1 n n 1
k n 1 n( n 1) n 1 k n 1
I E a b E b b a a
⎤⎦
∞
2 π 2 ( n+ 1)( n−1)
* *
+ ∑ E0 ⎡⎣b −1+ −1⎤⎦
=
2
nbn anan
k n
n=
1
Na terceira somatória fazemos n− 1= n→ n= n+ 1, pois ela poderia começar de 2
por causa do n − 1 e obtemos:
( 2n
+ 1)
∞
∞
4π 2 * 2 π 2 nn ( + 2) * *
1
=
0 ∑ Re + ∑
+
+
2 n n 2 0 n n 1 n n+
1
k n= 1 n( n+ 1) n=
1 k ( n+
1)
I E ⎡⎣ a b ⎤⎦ E ⎡⎣b b a a ⎤⎦
∞
2 π 2 ( n+
2)( n)
* *
+ ∑ E0 ⎡⎣b + 1
+
+ 1
⎤⎦
=
2
n
bn an an
k
n=
1
( 2n
+ 1)
( n+
1)
4π 4 π nn ( + 2)
I E ⎡⎣ a b ⎤⎦ E ⎡⎣b b a a ⎤⎦
∞
∞
2 * 2
* *
1
=
2 0 ∑ Re
n n
+
2 0 ∑ Re
n n+ 1+
n n+
1
k n= 1 n( n+ 1) k n=
1 ( n+
1)
Agora para I 2 ,
2
⎧ ∞
2 2
π E0
⎪ ⎧ n n+ 1 n+ 1 n+
2 2n+ 1 2n+ 3 2 n ( n+ 1)( n+
2)
I2 =− ⎨2Re⎨∑( i) ( −i) ( − i) ( i)
[ a
+ 1+ + 1]
+
2
n
bn
2 k ⎪⎩
⎩ n=
1
n( n+ 1) ( n+ 1 )( n+ 2) (2n+ 1)(2n+
3)
∞
∑
n=
1
2 2
n n n n 2n+ 1 2n− 1 2 n( n+ 1) ( n−1)
i −i − i i a + b
nn ( + 1) n− 1 n (2n+ 1)(2n−1)
− 1 + 1
()( ) ( ) ()
( )
2
∞
⎛ 2n
+ 1 ⎞ 2 ( n + 1! )
+ ∑
⎫ ⎪⎪
⎜ ⎟
an
+ bn
⎬⎬ =
n=
1 ⎝nn
( + 1) ⎠ 2n+
1 n − 1 ! ⎪⎪ ⎭⎭
( ) [ ] ⎫
[ ]
n−1 n−1
2
∞ ∞ ∞
π E ⎧
0 ⎪ ⎧ 2 nn ( + 2) 2( n+ 1)( n− 1) ⎛ 2n+
1 ⎞ ⎫⎫ ⎪⎪
=− 2Re
2 ⎨ ⎨∑ [ an+ 1+ bn+ 1] + ∑ [ an−1+ bn−1] + ∑2⎜
⎟[ an + bn]
⎬⎬
2 k ⎩⎪
⎩n= 1 ( n+ 1) n= 1 n n=
1 ⎝n( n+
1) ⎠ ⎭⎭ ⎪⎪
Fazendo na primeira somatória n+ 1= n→ n= n− 1 e na segunda
n− 1= n→ n= n + 1:
2
0
2 2
∞ ∞ ∞
π E ⎪⎧
⎧ nn ( + 2) ( n+ 1)( n− 1) 2n+
1 ⎫⎪⎫
I =− ⎨4Re⎨∑ [ an + bn] + ∑ [ an + bn] + ∑ [ an + bn]
⎬⎬=
2 k ⎪⎩
⎩n= 1 ( n+ 1) n= 1 n n=
1n( n+
1) ⎭⎪⎭
2
∞
π E ⎧
0 ⎪ ⎡nn ( + 2) ( n+ 1)( n− 1) 2n+
1 ⎤ ⎪⎫
=− 4Re
2 ⎨ ∑
[ an
bn]
2 k
⎢ + + + ⎬=
n=
1 ( n+ 1) n n( n+ 1)
⎥
⎪⎩
⎣
⎦ ⎪⎭
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