Mathematical Methods for Physics and Engineering : A Comprehensive Guide

SPECIAL FUNCTIONS

to be zero, sinceQm(x) is singular atx=±1, with the result that the general

solution is simply some multiple of one of the associated Legendre functions of

the first kind,Pm(x). We will study the further properties of these functions in

the remainder of this subsection.

Mutual orthogonality

As noted in section 17.4, the associated Legendre equation is of Sturm–Liouville

form (py)′+qy+λρy= 0, withp=1−x^2 ,q=−m^2 /(1−x^2 ),λ=(+1)

andρ= 1, and its natural interval is thus [− 1 ,1]. Since the associated Legendre

functionsPm(x) are regular at the end-pointsx=±1, they must be mutually

orthogonal over this interval for a fixed value ofm,i.e.

∫ 1

− 1

Pm(x)Pkm(x)dx=0 if=k. (18.36)

This result may also be proved directly in a manner similar to that used for demon-

strating the orthogonality of the Legendre polynomialsP(x) in section 18.1.2.

Note that the value ofmmust be the same for the two associated Legendre

functions for (18.36) to hold. The normalisation condition when=kmay be

obtained using the Rodrigues’ formula, as shown in the following example.

Show that

Im≡

∫ 1

− 1

Pm(x)Pm(x)dx=

2

2 +1

(+m)!

(−m)!

. (18.37)

From the definition (18.32) and the Rodrigues’ formula (18.9) forP(x), we may write

Im=

1

22 (!)^2

∫ 1

− 1

[

(1−x^2 )m

d+m(x^2 −1)

dx+m

][

d+m(x^2 −1)

dx+m

]

dx,

where the square brackets identify the factors to be used when integrating by parts.
Performing the integration by parts+mtimes, and noting that all boundary terms
vanish, we obtain

Im=

(−1)+m

22 (!)^2

∫ 1

− 1

(x^2 −1)

d+m

dx+m

[

(1−x^2 )m

d+m(x^2 −1)

dx+m

]

dx.

Using Leibnitz’ theorem, the second factor in the integrand may be written as

d+m

dx+m

[

(1−x^2 )m

d+m(x^2 −1)

dx+m

]

=

∑+m

r=0

(+m)!

r!(+m−r)!

dr(1−x^2 )m

dxr

d^2 +2m−r(x^2 −1)

dx^2 +2m−r

.

Considering the two derivative factors in a term in the summation on the RHS, we
see that the first is non-zero only forr≤ 2 m, whereas the second is non-zero only for
2 +2m−r≤ 2 . Combining these conditions, we find that the only non-zero term in the
sum is that for whichr=2m. Thus, we may write

Im=

(−1)+m

22 (!)^2

(+m)!

(2m)!(−m)!

∫ 1

− 1

(1−x^2 )

d^2 m(1−x^2 )m

dx^2 m

d^2 (1−x^2 )

dx^2 

dx.