Computing an integral related to Legendre polynomials 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?
 A: 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,
$$\langle P_n , P_n \rangle  = \frac{(n!)^22^{2n+1}}{2n+1}.$$
