Orthogonal Property of Legendre Polynomials

Question

How can I get $$nu_{n} + (n-1)u_{n-1},$$ where $$u_{n} = \int_{-1}^{1} x^{-1}P_{n-1}(x) P_{n}(x)\, \mathrm {d}x\; ? $$

I did many search, and also I did try by myself. But without a success. Is there any way to solve this?

Thanks.

Answer 1

Let $P_n$ be the Legendre-polynomials. I am taking information about them from Wikipedia:

First, I use Bonnet's recursion formula $$ (n+1) P_{n+1}(x) = (2n + 1) x P_n(x) - n P_{n-1}(x) $$ and rewrite it for $n-1$ instead of $n$ (because I want to get rid of $P_{n-2}$ in the expression you want to calculate): $$ n P_{n}(x) = (2n -1) x P_{n-1}(x) - (n-1) P_{n-2}(x) $$ We solve for $P_{n-2}$. $$ P_{n-2}(x)=\frac{(2n -1) x P_{n-1}(x)-nP_{n}(x)}{(n-1)} $$

Now the expression can be calculated. \begin{align*} n u_n+(n-1)u_{n-1} &= \int_{-1}^1 n x^{-1} P_{n-1}(x) P_{n}(x) + (n-1) x^{-1} P_{n-2}(x) P_{n-1}(x) dx \\ &=\int_{-1}^1 n x^{-1} P_{n-1}(x) P_{n}(x) + x^{-1} \bigg( (2n -1) x P_{n-1}(x)-nP_{n}(x)\bigg)P_{n-1}(x) dx \\ &= \int_{-1}^1 n x^{-1} P_{n-1}(x) P_{n}(x) - n x^{-1} P_{n-1}(x) P_{n}(x) +(2n -1) P_{n-1}^2(x) dx \\ &= \int_{-1}^1 (2n -1) P_{n-1}^2(x) dx \end{align*} Now we use the orthonormalization formula $$ \int_{-1}^1 P_i(x) P_j(x) dx = \frac{2}{2i+1} \delta_{ij} $$ for $i=j=n-1$ to get the final result $$ n u_n+(n-1)u_{n-1} = (2n-1)\frac{2}{2(n-1)+1} = 2 $$