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- Jan 26, 2012

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**Problem**: For $n\geq 0$, let $P_n(x)$ denote the Legendre polynomial. Show that

\[\int_{-1}^1 P_m(x)P_n(x)\,dx = \begin{cases}0 & m\neq n\\ \dfrac{2}{2n+1} & m=n\end{cases}\]

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**Hints for the $m\neq n$ case**:

and rewrite it in the form

\[[(1-x^2)y^{\prime}]^{\prime} = -\lambda(\lambda+1)y.\tag{1}\] Then use the fact that $P_m(x)$ and $P_n(x)$ are solutions to this equation with $\lambda=m$ and $\lambda=n$ respectively, and substitute them into $(1)$. From there, find a way to combine these two equations in order to make $P_m(x)P_n(x)$ appear and then integrate to get the result.

**Hints for the $m=n$ case**:

\[(n+1)P_{n+1}(x)+nP_{n-1}(x) = (2n+1)xP_n(x)\]

and

\[nP_n(x) + (n-1)P_{n-2}(x) = (2n-1)xP_{n-1}(x)\]

and then use the $m\neq n$ integral result to come up with a recurrence relation involving $\displaystyle\int_{-1}^1 P_n^2(x)\,dx$ and $\displaystyle\int_{-1}^1 P_{n-1}^2(x)\,dx$. Finally, use the fact that $P_0(x)=1$ and induction to show that $\displaystyle\int_{-1}^1 P_n^2(x)\,dx = \frac{2}{2n+1}$

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