# Integration of Legendre Polynomial

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

• The forward shift operator for the Jacobi polynomials (the Legendre polynomials are Jacobi polynomials with $\alpha = \beta = 0$), e.g. homepage.tudelft.nl/11r49/askey/ch1/par8/par8.html, shows that this integral is an integral of a Jacobi polynomial $P^{(4,4)}_{\ell-4}(x)$ times $P_n(x)$. Maybe you can solve this integral? – Noud Oct 8 '15 at 9:28
• Ah, yes! You can use connection coefficients (which I could not find with a simple google search, but I know that they exists) to rewrite the Jacobi polynomials $P_{\ell-4}^{(4,4)}(x)$ as a sum of Legendre polynomials $P_m(x)$ where $m \leq \ell-4$. Then use the orthogonality relation for the Legendre polynomials. Unfortunately I don't have a pen and paper at hand, therefore I can't give you the details (yet). – Noud Oct 8 '15 at 9:31

The facts below should allow you to compute the integral in question in terms of $l$ and $n$:

• The Legendre polynomials $P_0, \dots, P_n$ form a basis for the space of polynomials of degree at most $n$.

• The Legendre polynomials are orthogonal: $$\int_{-1}^{1} P_m(x) P_n(x)\,dx = {2 \over {2n + 1}} \delta_{mn}$$

• $\dfrac{d^4P_l(x)}{dx^4}$ is a polynomial of degree $l-4$ if $l\ge 4$ or the zero polynomial otherwise.

• Yes. In principle, I would be able to replace each $P_l(x)$ by its representation as a polynomial. However, using that would be just a brute force way of calculating thing. I want a simpler solution. – titanium Oct 8 '15 at 3:05
• Maybe I miss your point but it seems that this is useful only for $l < n+3$ . – Start wearing purple Oct 8 '15 at 9:44

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 $l\ge n+4$.

• Thanks. This is really a useful one. I was trying to use the same technique in order to evaluate $$\int_{-1}^1P_l^{(4)}(x)P_n(x) dx$$. I used the same idea as yours to get to the point $$\mathcal{I}_{l,n}=[P_l^{(2)}(x)P_n(x)-P_l^{(1)}(x)P_n^{(1)}(x)+P_l(x)P_n^{(2)}(x)]\rvert_{-1}^{1}=\frac{(l(l+2)(l^2-1)}{8}-\frac{(l(l+1)n(n+1)}{8}+\frac{(n(n+2)(n^2-1)}{8}$$. However, this does not match with the exact solution, when I checked in Mathematica. Do you know where I went wrong? – titanium Oct 8 '15 at 20:07
• @titanium The coefficients should be $\frac14$, $-\frac12$, $\frac14$ instead of $\frac18$, $-\frac18$, $\frac18$. – Start wearing purple Oct 8 '15 at 20:27
• Oh yeah. But, actually I tried that as well. But does not work. – titanium Oct 8 '15 at 20:32
• @titanium We have for example $P_n^{(2)}(1)=\frac{n(n+2) (n^2-1 )}{ 8}$, $P_n^{(1)}(1)=\frac{n(n+1)}{2}$. The lower bound $x=-1$ gives the same contribution due to parity. – Start wearing purple Oct 8 '15 at 20:34
• @titanium It does work, I just checked. Maybe you have made the same misprint in Mathematica as in your comment: $4$th derivative instead of $3$rd? – Start wearing purple Oct 8 '15 at 20:35

Let me see if I can give a solution. The orthogonality relation for the Gegenbauer polynomials is $$\int_{-1}^{1} P_m(x) P_n(x) dx = \frac{2}{2n+1} \delta_{m,n}.$$ The forward shift operator for the Jacobi polynomials, e.g. http://homepage.tudelft.nl/11r49/askey/ch1/par8/par8.html, is given by $$\frac{d}{dx} P_m^{(\alpha,\beta)}(x) = \frac{n+\alpha+\beta+1}{2} P_{m-1}^{(\alpha+1,\beta+1)}(x).$$ Note that the Gegenbauer polynomials are Jacobi polynomial where $\alpha = \beta = 0$. The connection coefficients for the Jacobi polynomials are $$P_n^{(\gamma,\delta)}(x) = \sum_{k=0}^n c_{n,k}(\gamma,\delta;\alpha,\beta) P_k^{(\alpha,\beta)}(x),$$ where \begin{equation*} \begin{split} c_{n,k}(\gamma,\delta;\alpha,\beta) &= \frac{ (\gamma+k+1)_{n-k} (n+\gamma+\delta+1)_k }{ (n-k)! \Gamma(\alpha+\beta+2k+1) } \Gamma(\alpha+\beta+k+1) \\ &\times {}_3 F_2(-n+k, n+k+\gamma+\delta+1, \alpha+k+1; \gamma+k+1, \alpha+\beta+2k+2; 1). \end{split} \end{equation*} Hence \begin{equation*} \begin{split} \frac{d^4}{dx^4} P_{\ell}^{(0,0)} &= \frac{(\ell-2)_4}{2^4} P_{\ell-4}^{(4,4)}(x) = \frac{(\ell-2)_4}{2^4} \sum_{k=0}^{\ell-4} c_{\ell-4,k}(4,4;0,0) P_k^{(0,0)}(x). \end{split} \end{equation*} By the orthogonality relation we have \begin{equation*} \int_{-1}^{1} \frac{d^4}{dx^4} P_{\ell}(x) P_n(x) dx = \frac{(\ell-2)_4}{2^3 (2n+1)} c_{\ell-4,n}(4,4;0,0). \end{equation*} Note that the connection coefficient can be found in Ismail, Classical and Quantum Orthogonal polynomials and that the connection coefficients probably can be simplified a lot. I hope I don't divide by zero somewhere so that you have to intrepid it formally. So the last part is homework. ;)