Use (5.78e) to show that ∫ 0 1 P ( x ) d x = P l − 1 ( 0 ) − P l + 1 ( 0 ) / ( 2 l + 1 ) . Then use the result of Problem 2 and Chapter$1, Section 13 C to show that ∫ 0 1 P 2 n ( x ) d x = 0 , n > 0 , and ∫ 0 1 P 2 n + 1 ( x ) d x = ( − 1 ) n ( 2 n − 1 ) ! ! 2 n + 1 ( n + 1 ) ! = 1 / 2 n + 1 .
Use (5.78e) to show that ∫ 0 1 P ( x ) d x = P l − 1 ( 0 ) − P l + 1 ( 0 ) / ( 2 l + 1 ) . Then use the result of Problem 2 and Chapter$1, Section 13 C to show that ∫ 0 1 P 2 n ( x ) d x = 0 , n > 0 , and ∫ 0 1 P 2 n + 1 ( x ) d x = ( − 1 ) n ( 2 n − 1 ) ! ! 2 n + 1 ( n + 1 ) ! = 1 / 2 n + 1 .
Use (5.78e) to show that
∫
0
1
P
(
x
)
d
x
=
P
l
−
1
(
0
)
−
P
l
+
1
(
0
)
/
(
2
l
+
1
)
.
Then use the result of Problem 2 and Chapter$1, Section
13
C
to show that
∫
0
1
P
2
n
(
x
)
d
x
=
0
,
n
>
0
,
and
∫
0
1
P
2
n
+
1
(
x
)
d
x
=
(
−
1
)
n
(
2
n
−
1
)
!
!
2
n
+
1
(
n
+
1
)
!
=
1
/
2
n
+
1
.
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