1000 Solved Problems in Modern Physics

70 1 Mathematical Physics

=

√

2

πx

[

1 −

x^2

2!

+

x^4

4!

−···

]

=

√

2

πx

cosx

1.79 The normalization of Legendre polynomials can be obtained byl–foldinte-
gration by parts for the conventional form

Pl(x)=

1

2 ll!

dl

dxl

(x^2 −1)l (Rodrigues’s formula)

∫+ 1

− 1

[Pl(x)]^2 dx=

(

1

2 ll!

) 2 ∫+ 1

− 1

[

dl(x^2 −1)l

dxl

][

dl(x^2 −1)l

dxl

]

dx

=(−1)l(

1

2 ll!

)^2

∫+ 1

− 1

[

d^2 l(x^2 −1)

dx^2 l

]

(x^2 −1)ldx

=(−1)l

(

(2l)!

2 ll!

) 2 ∫+ 1

− 1

(x^2 −1)ldx=

2

2 l+ 1

Putl=nto get the desired result.

The orthogonality can be proved as follows. Legendre’s differential equation

d

dx

[

(1−x^2 )

dPn(x)

dx

]

+n(n+1)Pn(x)=0(1)

can be recast as

[(1−x^2 )Pn′]′=−n(n+1)Pn(x)(2)

[(1−x^2 )Pm′]′=−m(m+1)Pm(x)(3)

Multiply (2) byPmand (3) byPnand subtract the resulting expressions.

Pm[(1−x^2 )Pn′]′−Pn[(1−x^2 )Pm′]′=[m(m+1)−n(n+1)]PmPn (4)

Now, LHS of (4) can be written as

Pm[(1−x^2 )Pn′]′−Pn[(1−x^2 )Pm′]′

=Pm[(1−x^2 )Pn′]′+Pm′[(1−x^2 )Pn′]−Pn[(1−x^2 )Pm′]−Pn[(1−x^2 )Pm′]′

(4) can be integrated

d

dx

[(1−x^2 )(PmPn′−PnPm′)=[m(m+1)−n(n+1)]PmPn

(1−x^2 )

(

PmPn′−PnPm′

)

|^1 − 1 =[m(m+1)−n(n+1)]

∫ 1

− 1

PmPndx

Since (1−x^2 ) vanishes atx =±1, the LHS is zero and the orthogonality

follows.

∫ 1

− 1

Pm(x)Pn(x)dx=0;m

=n