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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