Page 52 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 2: Legendre’s Polynomials

Notes
Putting a 1 hx and b h (x 1)

so that a 2 b 2 = (1 hx ) 2 h 2 (x 2 1) 1 2xh h 2
Thus, we have from (i)

1
(1 2xh h 2 ) 1/2 = 1 hx h (x 2 1) cos d
0

1
= 1 h x x 2 1 cos d
0

1
= (1 ht ) d where t x x 2 1 cos
0
2 2
n
n n
or h P n ( ) = (1 ht h t ... h t ...) d
x
0
n
Equating coefficient of h we get
n
n
x
p n ( ) = t d x x 2 1 cos d
0 0
1 2 n
x
P n ( ) = x (x 1) cos d
0
Deductions
(i) Putting x cos in above relation, we get

1
n
P n (cos ) (cos i sin cos ) d .
0
(ii) If we take n = 1 and +ve sign, then we get
1
x
P 1 ( ) x [(x 2 1)]cos d .
0
Laplace’s Second Integral for P (x): When n is a Positive Integer. Show that
n

1 d
x
P n ( ) = n 1
0 2
x (x 1)cos
Proof: From integral calculus, we have

d 2 2
= , where a b . ...(i)
0 a b cos (a 2 b 2 )

Let a xh 1 and b h (x 2 1)

so that a 2 b 2 = 1 2xh h 2

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