Page 790 - Handbook Of Integral Equations

The polynomials P n = P n (x) can be calculated recursively using the relations

1 2 2n +1 n
P 0 (x)=1, P 1 (x)= x, P 2 (x)= (3x – 1), ... , P n+1 (x)= xP n (x) – P n–1 (x).
2 n +1 n +1
The first three functions Q n = Q n (x) have the form

2
1 1+ x x 1+ x 3x – 1 1+ x 3
Q 0 (x)= ln , Q 1 (x)= ln – 1, Q 2 (x)= ln – x.
2 1 – x 2 1 – x 4 1 – x 2
The polynomials P n (x) have the implicit representation

[n/2]
m
m
P n (x)=2 –n (–1) C C n x n–2m ,
n 2n–2m
m=0
where [A] is the integer part of a number A.
All zeros of P n (x) are real and lie on the interval –1< x < +1; the functions P n (x) form an
orthogonal system on the interval –1 ≤ x ≤ +1, with

+1 0 if n ≠ m,

P n (x)P m (x) dx = 2
if n = m.
–1
2n +1
The generating function is
∞
1 n
√ = P n (x)s (|s| < 1).
1 – 2sx + s 2
n=0
Integral representations
For n =0, 1, 2, ... ,

π

Γ(ν + n +1) √ ν
n
P (z)= z + cos t z – 1 cos(nt) dt, Re z >0,
2
ν
πΓ(ν +1) 0
π

Γ(ν + n +1)
n n 2 –n/2 n–ν–1 2ν+1
Q (z)=(–1) (z – 1) (z + cos t) (sin t) dt, Re ν > –1,
ν
2 ν+1 Γ(ν +1)
0
Note that z ≠ x, –1< x < 1, in the latter formula.
10.11. Orthogonal Polynomials
All zeros of each of the orthogonal polynomials P n (x) considered in this section are real and
simple. The zeros of the polynomials P n (x) and P n+1 (x) are alternating.
Legendre polynomials
The Legendre polynomials P n = P n (x) satisfy the equation

2

(1 – x )y xx – 2xy + n(n +1)y =0.
x
They are outlined in Section 10.10 of this supplement.

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 775