Integration of Legendre Polynomial I need to evaluate the following integral
\begin{equation} \int_{-1}^1 \frac{d^4P_l(x)}{dx^4}P_n(x)dx\end{equation}. Of course the answer I need is in terms of $l$ and $n$. Does anyone have any idea how to proceed?

Let us denote
$$\mathcal{I}_{l,n}:=\int_{-1}^1P_l^{(4)}(x)P_n(x)\,dx.$$
If $l\leq n+3$, then $P_l^{(4)}(x)$ is a polynomial of degree at most $n-1$ and $\mathcal{I}_{l,n}$ vanishes because of orthogonality of Legendre polynomials. It also vanishes when $n$ and $l$ have different parity. Hence in the following it will be assumed that $l\geq n+4$ and $l=n\;\operatorname{mod}\;2$.
Integrating by parts, we can rewrite the integral as
$$\int_{-1}^1P_l(x)P_n^{(4)}(x)\,dx+\text{boundary terms}.$$
The same argument as above shows that the integral in this expression vanishes for $l\ge n+5$, and therefore the answer is completely determined by boundary terms:
\begin{align}
\nonumber\mathcal{I_{l,n}}&=\left[\color{red}{P_l^{(3)}(x)P_n(x)}-\color{blue}{P_l^{(2)}(x)P_n^{(1)}(x)}+\color{green}{P_l^{(1)}(x)P_n^{(2)}(x)}-\color{magenta}{P_l(x)P_n^{(3)}(x)}\right]\biggl|_{-1}^{\;1}=\\
\nonumber&=\color{red}{\frac{l(l+3)\left(l^2-1\right)\left(l^2-4\right)}{24}}-
\color{blue}{\frac{l(l+2)\left(l^2-1\right)n(n+1)}{8}}+\\
\nonumber&\;+\color{green}{\frac{l(l+1)n(n+2)\left(n^2-1\right)}{8}}-
\color{magenta}{\frac{n(n+3)\left(n^2-1\right)\left(n^2-4\right)}{24}}=\\
&=\frac{\left(l-n\right)\left(\left(l+n\right)^2-1\right)
\left(\left(l-n\right)^2-4\right)(l+n+3)}{24}.\tag{$\spadesuit$}
\end{align}
Therefore the only remaining case to consider is $l=n+4$. Here the answer may be computed using Rodrigues formula. Replacing expressions for Legendre polynomials into the initial integral and integrating by parts, we get 
$$\mathcal{I}_{n+4,n}=2(2n+3)(2n+5)(2n+7).$$
But this is compatible with the previous result ($\spadesuit$), which may therefore be used for all $l\ge n+4$.
