Computing an integral related to Legendre polynomials

Question

I have a series of Functions $ P_n(x)=\frac{\mathrm d^n}{\mathrm dx^n} (x^2-1)^n$ I want to calculate the value of this Integral:

$$ \int_ {-1}^1 P_n(x) P_n(x)\mathrm dx $$

I tried with partial Integration but just get stuck. Is it even possible?

Answer 1

The following derivation is self-contained, in the sense that it does not leverage properties of Legendre polynomials.

To facilitate notations, let $Q_n(X) = (X^2-1)^n$ and let $Q_n^{(n)}$ denote the $n$-th derivative of $Q_n$, so that $P_n = Q_n^{(n)}$.

Let $n\geq 1$ and note that $$\langle P_n , P_n \rangle = \int_{-1}^1 Q_n^{(n)}(x) Q_n^{(n)}(x) dx = \Big[Q_n^{(n)}(x) Q_n^{(n-1)}(x)\Big]_{-1}^1 - \int_{-1}^1 Q_n^{(n+1)}(x) Q_n^{(n-1)}(x) dx.$$

Since $n\geq 1$, $1$ and $-1$ are roots of $Q_n$, each with multiplicity $n$ hence $$Q_n^{(n-1)}(-1) = Q_n^{(n-1)}(1)=0.$$

If $n=1$ we stop here, otherwise we integrate by parts repeatedly until $Q_n^{(n-n)}$ appears and we find $$\langle P_n , P_n \rangle = (-1)^n \int_{-1}^1 Q_n^{(2n)}(x) Q_n^{(n-n)}(x) dx,$$ since for every $k\in \{2,\ldots,n\}$, $0\leq n-k\leq n-1 \implies Q_n^{(n-k)}(-1) = Q_n^{(n-k)}(1)=0$.

The degree of $Q_n$ is $2n$, thus $Q_n^{(2n)}$ is a constant polynomial: $Q_n^{(2n)} = (2n)!$, hence $$\langle P_n , P_n \rangle = (-1)^n (2n)! \int_{-1}^1 Q_n^{(0)}(x) dx = (-1)^n (2n)! \int_{-1}^1 Q_n(x) dx = (2n)! \int_{-1}^1 (1+x)^n (1-x)^n dx. $$

Integrating by parts yet again, $$ \begin{align} \int_{-1}^1 (1+x)^n (1-x)^n dx &= \frac n{n+1} \int_{-1}^1 (1+x)^{n+1} (1-x)^{n-1} dx \\& = \ldots = \frac n{n+1} \frac {n-1}{n+2} \ldots \frac 1{2n} \int_{-1}^1 (1+x)^{2n} (1-x)^{n-n} dx \\ &= \frac{(n!)^2}{(2n)!} \frac{2^{2n+1}}{2n+1}. \end{align} $$ At last, $$\langle P_n , P_n \rangle = \frac{(n!)^22^{2n+1}}{2n+1}.$$