Mathematical Physics Vol 1

5.10 Examples

281

Problem 196 Prove the orthogonality of Legendre polynomials starting from Rodrigues formula for Legendre polynomials. P n ( x )= 1 2 n n ! d n d x n ( x 2 − 1 ) n (see [ 35 ]).

Solution It should be proved that

1 Z

I mn =

P m ( x ) P n ( x ) d x = 0 ,

for m̸ = n .

− 1

Starting from Rodrigues formula

d n d x n

1 2 n n !

( x 2 − 1 ) n ,

P n ( x )=

(5.258)

for the integral I mn we obtain

1 Z

I mn =

P m ( x ) P n ( x ) d x =

− 1

1 Z

d n d x n

d m d x m

1 2 n + m n ! m !

( x 2 − 1 ) n

( x 2 − 1 ) m d x .

=

− 1

Let us assume that m < n . By partial integration we obtain

1 2 m + n m ! n ! "

1

d m d x m

d n − 1 d x n − 1 d n − 1 d x n − 1

( x 2 − 1 ) m

( x 2 − 1 ) n

I mn =

− 1

( x 2 − 1 ) n d x 

1 Z

d m + 1 d x m + 1

( x 2 − 1 ) m + 1

 =

− 1

1 Z

d m + 1 d x m + 1

d n − 1 d x n − 1

− 1 2 m + n m ! n !

( x 2 − 1 ) m + 1

( x 2 − 1 ) n d x .

=

− 1

If we repeat partial integration another n − 1 times we obtain

1 Z

d m + n d x m + n

( − 1 ) n 2 m + n m ! n !

I mn = ( x 2 − 1 ) m ( x 2 − 1 ) n d x . As, according to the assumption that m < n , it follows that m + n > 2 m , andas ( x 2 − 1 ) m is a polynomial of degree 2 m it follows that d m + n d x m + n ( x 2 − 1 ) m ≡ 0 , − 1

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