Legendre Polynomials Recursion Problem

Question

Using the recurrence equation for Legendre Polynomials: $$(k+1)P_{k+1}(x)=(2k+1)xP_k(x)-k P_{k-1}(x) \text{ , } k \in \mathbb{N}$$ Compute the Integral: $$ \int_{-1}^1xP_k(x)P_{k+1}(x)dx $$ I am completely at lost as how to proceed. Any ideas?

Answer 1

Although the hints above should be enough to solve this yourself, I will give the solution here:

$\newcommand{\partial}[1]{\left[#1\right]}$ $\newcommand{\bracket}[1]{\left(#1\right)}$ \begin{equation} xP_k(x)=\frac{k+1}{2k+1}P_{k+1}(x) + \frac{k}{2k+1}P_{k-1}(x) \end{equation} so \begin{equation} \begin{split} I&=\int_{-1}^1 xP_k(x)P_{k+1}(x)dx \\ &=\frac{1}{2k+1}\int_{-1}^1 [(k+1)P_{k+1}(x) + kP_{k-1}(x)] P_{k+1}(x)dx \\ &=\frac{1}{2k+1}\bracket{(k+1)\int_{-1}^1 P_{k+1}^2(x)dx + k\int_{-1}^1 P_{k-1}(x)P_{k+1}(x)dx} \\ &=\frac{1}{2k+1}\bracket{(k+1)\cdot \frac{2}{2(k+1)+1} + k\cdot 0} \\ &=\frac{2k+2}{(2k+1)(2k+3)} \end{split} \end{equation}

Of course, this derivation requires $k>=1$ because otherwise $P_{k-1}(x)$ has no meaning. For $k=0$ though, the integral evaluates to \begin{equation} \begin{split} I&=\int_{-1}^1 xP_0(x)P_1(x)dx \\ &=\int_{-1}^1 x^2dx \\ &=\partial{\frac{x^3}{3}}_{-1}^{+1} \\ &=\frac{2}{3} \end{split} \end{equation}

so the formula above also holds for $k=0$.