Why does $\int_{-1}^1 ((1-x^2)(P_m'P_n-P_n'P_m))'\,dx = 0$ for Legendre polynomials?

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I was looking at a proof of the orthogonality of the Legendre polynomials in Lebedev's Special Functions and their Applications:

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I can't understand why the integral of the first term vanishes. I appreciate any help.

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$$ \int_{-1}^1 \Big((1 - x^2)(P_m'P_n - P_n'P_m) \Big)' \mathrm{d}x = \Big[(1 - x^2)(P_m'P_n - P_n'P_m)\Big]_{-1}^1 $$ $$= \Big[(1 - 1^2)(P_m'(1)P_n(1) - P_n'(1)P_m(1))\Big] - \Big[(1 - (-1)^2)(P_m'(-1)P_n(-1) - P_n'(-1)P_m(-1))\Big]$$ $$ = \Big[(0)(P_m'(1)P_n(1) - P_n'(1)P_m(1))\Big] - \Big[(0)(P_m'(-1)P_n(-1) - P_n'(-1)P_m(-1))\Big] $$ $$= 0 - 0 $$ $$= 0 $$