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\int_{-1}^{1}x^2P_{n-1}(x)P_{n+1}(x)dx = \frac{2n(n+1)}{(2n-1)(2n+1)(2n+3)}

15 Feb 2024
Answer :
Word Count : 179

To solve the integral \(\int_{-1}^{1}x^2P_{n-1}(x)P_{n+1}(x)dx\), where \(P_n(x)\) are the Legendre polynomials, let's use the orthogonality property of Legendre polynomials:

\[\int_{-1}^{1}P_m(x)P_n(x)dx = \frac{2}{2n+1}\delta_{mn}\]

where \(\delta_{mn}\) is the Kronecker delta, which __________ ________ ________ _____ ___ ____ _________ ________ _____ _________ __________.
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