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3 Legendre Polynomials and Legendre Functions
P 2n are even and P 2n+1 are odd polynomials. Some special results are
P 2n (0) =
(−1) n Γ (n + 1/2)
√
πΓ (n + 1)
P n (−1) = (−1)
n
P 2n+1 (0) = 0
P n (1) = 1 , and
|P n (x)| ≤ 1 for − 1 ≤ x ≤ 1 .
Zero Order Legendre Functions of Second Kind
As already mentioned, a general solution of the Legendre differential equation (3.1)
is given by a linear combination of Legendre polynomials P and Legendre functions
Q of second kind. The functions Q n are given by
Q n (x) =
1
2
P n (x) ln
1 + x
1 − x
− W n−1 (x) for − 1 ≤ x ≤ 1
(3.6a)
Q n (z) =
1
2
P n (z) ln
z + 1
z − 1
− W n−1 (z) ,
(3.6b)
with
W n−1 (x) =
n
k=1
1
k
P k−1 (x)P n−k (x) .
(3.6c)
W is a solution of the differential equation [3],
(1 − x
2 )
d 2 W n−1
dx 2 − 2x
dW n−1
dx
+ (n + 1)nW n−1 − 2
dP n (x)
dx
= 0 .
The first few functions are listed below:
Q 0 (x) =
1
2
P 0 (x) ln
1 + x
1 − x
Q 1 (x) =
1
2
P 1 (x) ln
1 + x
1 − x
− 1
Q 2 (x) =
1
2
P 2 (x) ln
1 + x
1 − x
−
3
2
x
Q 3 (x) =
1
2
P 3 (x) ln
1 + x
1 − x
−
5
2
x
2
+
2
3
.
3 Legendre Polynomials and Legendre Functions
P 2n are even and P 2n+1 are odd polynomials. Some special results are
P 2n (0) =
(−1) n Γ (n + 1/2)
√
πΓ (n + 1)
P n (−1) = (−1)
n
P 2n+1 (0) = 0
P n (1) = 1 , and
|P n (x)| ≤ 1 for − 1 ≤ x ≤ 1 .
Zero Order Legendre Functions of Second Kind
As already mentioned, a general solution of the Legendre differential equation (3.1)
is given by a linear combination of Legendre polynomials P and Legendre functions
Q of second kind. The functions Q n are given by
Q n (x) =
1
2
P n (x) ln
1 + x
1 − x
− W n−1 (x) for − 1 ≤ x ≤ 1
(3.6a)
Q n (z) =
1
2
P n (z) ln
z + 1
z − 1
− W n−1 (z) ,
(3.6b)
with
W n−1 (x) =
n
k=1
1
k
P k−1 (x)P n−k (x) .
(3.6c)
W is a solution of the differential equation [3],
(1 − x
2 )
d 2 W n−1
dx 2 − 2x
dW n−1
dx
+ (n + 1)nW n−1 − 2
dP n (x)
dx
= 0 .
The first few functions are listed below:
Q 0 (x) =
1
2
P 0 (x) ln
1 + x
1 − x
Q 1 (x) =
1
2
P 1 (x) ln
1 + x
1 − x
− 1
Q 2 (x) =
1
2
P 2 (x) ln
1 + x
1 − x
−
3
2
x
Q 3 (x) =
1
2
P 3 (x) ln
1 + x
1 − x
−
5
2
x
2
+
2
3
.
