5.9 Spherical Coordinates; Legendre Polynomials 341
Next, move the last term to the left-hand member of the equation to find
(n+ 2 )(n− 1 )∫
xPn(x)dx=(
1 −x^2)(
Pn(x)−xPn′(x))
.
This equation can be solved for the integral on the left, provided thatn=1.
(Forn=1, the integration is done directly.)
Summary
∫
Pn(x)dx=−( 1 −x^2 )
n(n+ 1 )P′
n(x), (13)
∫
xPn(x)dx=( 1 −x^2 )
(n+ 2 )(n− 1 )(
Pn(x)−xPn′(x))
(14)
These integration formulas are useful if we can evaluatePn(x)andPn′(x)
easily for anyx. The relations in Eqs. (8) and (9) are useful for this purpose.
We illustrate by findingPn( 0 ).First,notethatPn( 0 )=0 for odd values ofn,
because the Legendre polynomials with odd index are odd functions ofx.For
oddn,Eq.(9)gives
(n+ 1 )Pn+ 1 ( 0 )+nPn− 1 ( 0 )= 0 ,or
Pn+ 1 ( 0 )=−n
n+ 1 Pn−^1 (^0 ).BecauseP 0 ( 0 )=1, we find successively that
P 2 ( 0 )=−^1
2, P 4 ( 0 )=^1 ·^3
2 · 4
, P 6 ( 0 )=−^1 ·^3 ·^5
2 · 4 · 6
,
or in general
Pn( 0 )=(− 1 )n/^21 · 3 ···(n− 1 )
2 · 4 ···n , n=^2 ,^4 ,^6 ,...
Pn( 0 )= 0 , n= 1 , 3 , 5 ,....(15)
Similarly, but not as easily, Eq. (8) can be used to find the values ofPn′( 0 ).It
is simpler to use the relation
Pn′( 0 )=nPn− 1 ( 0 ), (16)which can be derived from Eqs. (8) and (9).