Mathematical Methods for Physics and Engineering - Matematica.NET

SPECIAL FUNCTIONSwhich reduces to(x 2 − 1)u (l+2) +2xu (l+1) − l(l +1)u (l) =0.Changing the sign all through, we recover Legendre’s equation (18.1) with u (l) asthe dependent variable. Since, from (18.9), l is an integer and u (l) is regular atx = ±1, we may make the identificationu (l) (x) =c l P l (x), (18.10)for some constant c l that depends on l. To establish the value of c l we notethat the only term in the expression for the lth derivative of (x 2 − 1) l thatdoes not contain a factor x 2 − 1, and therefore does not vanish at x =1,is(2x) l l!(x 2 − 1) 0 . Putting x = 1 in (18.10) and recalling that P l (1) = 1, thereforeshows that c l =2 l l!, thus completing the proof of Rodrigues’ formula (18.9).◮Use Rodrigues’ formula to show thatI l =∫ 1−1P l (x)P l (x) dx = 22l +1 . (18.11)The result is trivially obvious for l = 0 and so we assume l ≥ 1. Then, by Rodrigues’formula,∫1 1[ ][ ]d l (x 2 − 1) l d l (x 2 − 1) lI l =dx.2 2l (l!) 2 −1 dx l dx lRepeated integration by parts, with all boundary terms vanishing, reduces this to∫ 1I l =(−1)l (x 2 − 1) l d2l2 2l (l!) 2 −1 dx 2l (x2 − 1) l dx= (2l)! ∫ 1(1 − x 2 ) l dx.2 2l (l!) 2 −1If we writeK l =∫ 1−1(1 − x 2 ) l dx,then integration by parts (taking a factor 1 as the second part) givesK l =∫ 1Writing 2lx 2 as 2l − 2l(1 − x 2 )weobtainK l =2l∫ 1−1−12lx 2 (1 − x 2 ) l−1 dx.(1 − x 2 ) l−1 dx − 2l∫ 1−1(1 − x 2 ) l dx=2lK l−1 − 2lK land hence the recurrence relation (2l +1)K l =2lK l−1 . We therefore findK l =2l 2l − 22l +12l − 1 ··· 23 K 0 =2 l 2 l l!l!(2l +1)! 2= 22l+1 (l!) 2(2l +1)! ,which, when substituted into the expression for I l , establishes the required result. ◭582