The Legendre function for n = 1 is y 2 ( x ) = P 1 ( x ) = x and y 1 ( x ) = 1 − 1 2 x ln 1 + x 1 − x . Result used: Equation (6) is, y 1 ( x ) = 1 − n ( n + 1 ) 2 ! x 2 + ( n − 2 ) n ( n + 1 ) ( n + 3 ) 4 ! x 4 − ⋯ Equation (7) is, y 2 ( x ) = x − ( n − 1 ) ( n + 2 ) 3 ! x 3 + ( n − 3 ) ( n − 1 ) ( n + 2 ) ( n + 4 ) 5 ! x 5 − ⋯ Equation ( 1 1 ′ ) is, P 0 ( x ) = 1 , P 1 ( x ) = x , P 2 ( x ) = 1 2 ( 3 x 2 − 1 ) , P 3 ( x ) = 1 2 ( 5 x 3 − 3 x ) P 4 ( x ) = 1 8 ( 35 x 4 − 30 x 2 + 3 ) , P 5 ( x ) = 1 8 ( 63 x 5 − 70 x 3 + 15 x ) Calculation: First substitute n = 1 in equation (7) to obtain the Legendre function y 2 ( x ) for n = 1 as shown below. y 2 ( x ) = x − ( n − 1 ) ( n + 2 ) 3 ! x 3 + ( n − 3 ) ( n − 1 ) ( n + 2 ) ( n + 4 ) 5 ! x 5 − ⋯ = x − ( 1 − 1 ) ( 1 + 2 ) 3 ! x 3 + ( 1 − 3 ) ( 1 − 1 ) ( 1 + 2 ) ( 1 + 4 ) 5 ! x 5 − ⋯ = x − ( 0 ) ( 3 ) 3 ! x 3 + ( − 2 ) ( 0 ) ( 3 ) ( 5 ) 5 ! x 5 − ⋯ = x − 0 + 0 − ⋯ = x From equation ( 1 1 ′ ) it is clear that, P 1 ( x ) = x . Therefore, the Legendre function y 2 ( x ) for n = 1 is y 2 ( x ) = P 1 ( x ) = x . Now substitute n = 1 in equation (6) to obtain the Legendre function y 1 ( x ) for n = 1 as shown below. y 1 ( x ) = { 1 − n ( n + 1 ) 2 ! x 2 + ( n − 2 ) n ( n + 1 ) ( n + 3 ) 4 ! x 4 − ( n − 4 ) ( n − 2 ) n ( n + 1 ) ( n + 3 ) ( n + 5 ) 6 ! x 6 + ⋯ } = { 1 − ( 1 ) ( 1 + 1 ) 2 ! x 2 + ( 1 ) ( 1 − 2 ) ( 1 + 1 ) ( 1 + 3 ) 4 ! x 4 − ( 1 − 4 ) ( 1 − 2 ) ( 1 ) ( 1 + 1 ) ( 1 + 3 ) ( 1 + 5 ) 6 ! x 6 + ⋯ } = 1 − ( 1 ) ( 2 ) 2 ! x 2 + ( 1 ) ( − 1 ) ( 2 ) ( 4 ) 4 ! x 4 − ( − 3 ) ( − 1 ) ( 1 ) ( 2 ) ( 4 ) ( 6 ) 6 ! x 6 + ⋯ = 1 − 2 2 x 2 − 8 24 x 4 − 144 720 x 6 − ⋯ On further simplifications, y 1 ( x ) = 1 − x 2 − 1 3 x 4 − 1 5 x 6 − ⋯ = 1 − ( x 2 + 1 3 x 4 + 1 5 x 6 + ⋯ ) = 1 − x ( x + 1 3 x 3 + 1 5 x 5 + ⋯ ) Obtain the expansion of 1 2 ln ( 1 + x ) ( 1 − x ) as follows. 1 2 ln ( 1 + x ) ( 1 − x ) = 1 2 ( ln ( 1 + x ) − ln ( 1 − x ) ) = 1 2 ( ( x − x 2 2 + x 3 3 − x 4 4 ⋯ ) − ( − x − x 2 2 − x 3 3 − x 4 4 ⋯ ) ) = 1 2 ( 2 x + 2 x 3 3 + 2 x 5 5 + ⋯ ) = x + x 3 3 + x 5 5 + ⋯ Therefore, the Legendre function y 1 ( x ) for n = 1 becomes, y 1 ( x ) = 1 − x ( x + 1 3 x 3 + 1 5 x 5 + ⋯ ) = 1 − x ( 1 2 ln ( 1 + x ) ( 1 − x ) ) = 1 − 1 2 x ln 1 + x 1 − x Hence the Legendre function y 1 ( x ) for n = 1 is y 1 ( x ) = 1 − 1 2 x ln 1 + x 1 − x .

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