MHB Problem about Rodrigues' formula and Legendre polynomials

[Another1]

using Rodrigues' formula show that $$\int_{-1}^{1} \,{P}_{n}(x){P}_{n}(x)dx = \frac{2}{2n+1}$$

$${P}_{n}(x) = \frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^n$$

my thoughts

$$\int_{-1}^{1} \,{P}_{n}(x){P}_{n}(x)dx = \frac{1}{2^{2n}(n!)^2}\int_{-1}^{1} \,\frac{d^n}{dx^n}(x^2-1)^n\frac{d^n}{dx^n}(x^2-1)^ndx$$

let
$$u = \frac{d^n}{dx^n}(x^2-1)^n$$ and $$du =\frac{d^{n+1}}{dx^{n+1}}(x^2-1)^ndx$$
$$dv = \frac{d^n}{dx^n}(x^2-1)^ndx$$ and $$v = \frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n$$

so
$$uv=\frac{d^n}{dx^n}(x^2-1)^n\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n$$
$$-vdu=-\int_{-1}^{1} \,\frac{d^{n+1}}{dx^{n+1}}(x^2-1)^n\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^ndx$$

$$uv-vdu = \frac{d^n}{dx^n}(x^2-1)^n\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n-\int_{-1}^{1} \,\frac{d^{n+1}}{dx^{n+1}}(x^2-1)^n\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^ndx$$

what should I do?

 

[Sudharaka]

using Rodrigues' formula show that $$\int_{-1}^{1} \,{P}_{n}(x){P}_{n}(x)dx = \frac{2}{2n+1}$$

$${P}_{n}(x) = \frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^n$$

my thoughts

$$\int_{-1}^{1} \,{P}_{n}(x){P}_{n}(x)dx = \frac{1}{2^{2n}(n!)^2}\int_{-1}^{1} \,\frac{d^n}{dx^n}(x^2-1)^n\frac{d^n}{dx^n}(x^2-1)^ndx$$

let
$$u = \frac{d^n}{dx^n}(x^2-1)^n$$ and $$du =\frac{d^{n+1}}{dx^{n+1}}(x^2-1)^ndx$$
$$dv = \frac{d^n}{dx^n}(x^2-1)^ndx$$ and $$v = \frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n$$

so
$$uv=\frac{d^n}{dx^n}(x^2-1)^n\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n$$
$$-vdu=-\int_{-1}^{1} \,\frac{d^{n+1}}{dx^{n+1}}(x^2-1)^n\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^ndx$$

$$uv-vdu = \frac{d^n}{dx^n}(x^2-1)^n\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n-\int_{-1}^{1} \,\frac{d^{n+1}}{dx^{n+1}}(x^2-1)^n\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^ndx$$

what should I do?

Hi Another, :)

Interesting question, thanks. Notice that,

$${P}_{n}(x) = \frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^n=\frac{(2x)n}{2^n n!}\frac{d^{n-1}}{dx^{n-1}}(x^2 - 1)^{n-1}$$

and so on and generally for a differentiation of $r$ times we get,

$${P}_{n}(x) = \frac{(2x)^{r}}{2^n (n-r)!}\frac{d^{n-r}}{dx^{n-r}}(x^2 - 1)^{n-r}$$

No substitute $r=n-1$ and see what you get. I am sure you'll be able to continue from there :)