Mathematical Methods for Physics and Engineering - Matematica.NET

SPECIAL FUNCTIONSto be zero, since Q m l (x) is singular at x = ±1, with the result that the generalsolution is simply some multiple of one of the associated Legendre functions ofthe first kind, Pl m (x). We will study the further properties of these functions inthe remainder of this subsection.Mutual orthogonalityAs noted in section 17.4, the associated Legendre equation is of Sturm–Liouvilleform (py) ′ + qy + λρy = 0, with p =1− x 2 , q = −m 2 /(1 − x 2 ), λ = l(l +1)and ρ = 1, and its natural interval is thus [−1, 1]. Since the associated Legendrefunctions Pl m (x) are regular at the end-points x = ±1, they must be mutuallyorthogonal over this interval for a fixed value of m, i.e.∫ 1−1P m l (x)P m k (x) dx =0 ifl ≠ k. (18.36)This result may also be proved directly in a manner similar to that used for demonstratingthe orthogonality of the Legendre polynomials P l (x) in section 18.1.2.Note that the value of m must be the same for the two associated Legendrefunctions for (18.36) to hold. The normalisation condition when l = k may beobtained using the Rodrigues’ formula, as shown in the following example.◮Show thatI lm ≡∫ 1−1Pl m (x)Pl m (x) dx = 2 (l + m)!2l +1(l − m)! . (18.37)From the definition (18.32) and the Rodrigues’ formula (18.9) for P l (x), we may write∫1 1][ ]I lm =[(1 − x 2 ) m dl+m (x 2 − 1) l d l+m (x 2 − 1) ldx,2 2l (l!) 2 −1dx l+m dx l+mwhere the square brackets identify the factors to be used when integrating by parts.Performing the integration by parts l + m times, and noting that all boundary termsvanish, we obtain∫ 1[]I lm = (−1)l+m (x 2 − 1) l dl+m(1 − x 2 ) m dl+m (x 2 − 1) ldx.2 2l (l!) 2 −1 dx l+m dx l+mUsing Leibnitz’ theorem, the second factor in the integrand may be written as[]d l+m(1 − x 2 ) m dl+m (x 2 − 1) ldx l+m dx l+m∑l+m=r=0(l + m)! d r (1 − x 2 ) m d 2l+2m−r (x 2 − 1) l.r!(l + m − r)! dx r dx 2l+2m−rConsidering the two derivative factors in a term in the summation on the RHS, wesee that the first is non-zero only for r ≤ 2m, whereas the second is non-zero only for2l +2m − r ≤ 2l. Combining these conditions, we find that the only non-zero term in thesum is that for which r =2m. Thus, we may write∫I lm = (−1)l+m (l + m)! 1(1 − x 2 ) l d2m (1 − x 2 ) m d 2l (1 − x 2 ) ldx.2 2l (l!) 2 (2m)!(l − m)!dx 2m dx 2l−1590