# legendre Polynomial

prove:

$\int_{-1}^1 (x^2P_{n+1}(x)+P_{n-1}(x))\ dx$=$2n(n+1)\over {(2n-1)(2n+1)(2n+3)}$

I think to use the formula:

$(n+1)P_{n+1}(x)=(2n+1)xP_n(x)-nP_{n-1}(x)$

Then multiply the L.H.S. and R.H.S. by $x^2P_{n-1}(x)$ and integrate the two sides, so we get:

$\int_{-1}^1 (n+1)x^2P_{n+1}(x)\ dx=\int_{-1}^1 (2n+1)x^3P_n(x)P_{n-1}(x) \ dx-\int_{-1}^1 nx^2P^2_{n-1}(x) \ dx$

True? And if it is true how can I find the R.H.S. ?

• Prove what? I don't see any statement to prove, just an integral. – Robert Israel Apr 5 '16 at 22:14
• sorry, I edited this, Thanks. – Dima Apr 5 '16 at 22:16
• Are you familiar with the technique known as Integration By Parts? And also, do you know what $\int P_{n}(x) dx$ equals indefinitely? – Xoque55 Apr 5 '16 at 22:27
• You should study the orthogonality relation of Legendre polynomials. Anyway, your statement is wrong. Maybe some square is missing? – Jack D'Aurizio Apr 5 '16 at 23:47

There seems to be a typo in the original question. I am assuming it is $$\int_{-1}^1x^2P_{n+1}(x)\cdot P_{n-1}(x)dx=\frac{2n(n+1)}{(2n-1)(2n+1)(2n+3)}$$ With that correction, my first try would have been to apply the Rodrigues formula and integrate by parts $n+1$ times, and while that works, it seems easier to apply the recurrence relation for the Legendre polynomials, $$(n+1)P_{n+1}(x)=(2n+1)xP_n(x)-nP_{n-1}(x)$$ To get $$xP_{n+1}(x)=\frac{(n+2)}{(2n+3)}P_{n+2}(x)+\frac{(n+1)}{(2n+3)}P_n(x)$$ and $$xP_{n-1}(x)=\frac{n}{(2n-1)}P_n(x)+\frac{(n-1)}{(2n-1)}P_{n-1}(x)$$ And then apply the orthogonality relation for the Legendre polynomials $$\int_{-1}^1P_n(x)P_m(x)dx=\frac2{(2n+1)}\delta_{nm}$$ So now only the $P_n(x)$ terms survive orthogonality and we have $$\int_{-1}^1x^2P_{n+1}(x)\cdot P_{n-1}(x)dx=\frac{(n+1)}{(2n+3)}\cdot\frac{n}{(2n-1)}\cdot\frac2{(2n+1)}=\frac{2n(n+1)}{(2n-1)(2n+1)(2n+3)}$$
The sequence $\{P_n(x)\}_{n\geq 0}$ has the property that $\partial P_n(x)=n$ and $$\int_{-1}^{1}P_n(x)\,P_m(x)\,dx = \frac{2\cdot \delta(n,m)}{2n+1}.$$
Since $P_0(x)=1, P_1(x)=x$ and $P_2(x)=\frac{3}{2}x^2-\frac{1}{2}$, assuming $n\geq 1$:
$$\begin{eqnarray*} \int_{-1}^{1}\left(x^2 P_{n+1}(x)+P_{n-1}(x)\right)\,dx &=&2\cdot\delta(0,n-1)+\int_{-1}^{1}\frac{2P_2(x)+P_0(x)}{3}\cdot P_{n+1}(x)\,dx \\&=&2\cdot \delta(n,1)+\frac{4}{15}\cdot\delta(2,n+1)+\frac{2}{3}\cdot\delta(0,n+1)\end{eqnarray*}$$ hence your integral equals $2$ is $n=0$, $\frac{34}{15}$ if $n=1$ and $0$ otherwise.