An integral involving Legendre polynomial

Question

Show that

$$\int\limits_{-1}^{1} P_n^2(x)dx=\frac{2}{2n+1}$$ where

$$P_n(x)=\frac{1}{n!2^n}.\frac{d^n(x^2-1)^n}{dx^n},\quad n=0,1,2,...$$

Answer 1

Hint. One may use Rodrigue's formula $$ P_n(x)=\frac{1}{n!2^n}\cdot\frac{d^n(x^2-1)^n}{dx^n},\quad n=0,1,2,\cdots, $$ and integrate by parts $n$ times to get $$ \int_{-1}^{1} P_n^2(x)dx=\frac{(-1)^n}{(n!)^22^{2n}}\cdot\int_{-1}^{1} (x^2-1)^n\frac{d^{2n}(x^2-1)^n}{dx^{2n}}\:dx $$ now observing that $$ \frac{d^{2n}(x^2-1)^n}{dx^{2n}}=\frac{d^{2n}}{dx^{2n}}( x^{2n})=(2n)! $$ one has $$ \int_{-1}^{1} P_n^2(x)dx=\frac{(-1)^n(2n)!}{(n!)^22^{2n}}\cdot\int_{-1}^{1} (x^2-1)^n\:dx $$ then we may conclude using the Euler beta integral to get $$ \int_{-1}^{1} P_n^2(x)dx=\frac{(-1)^n(2n)!}{(n!)^22^{2n}}\!\int_{-1}^{1} (x^2-1)^n\:dx=\frac{(-1)^n(2n)!}{(n!)^22^{2n}}\cdot (-1)^n \frac{(n!)^22^{2n+1}}{(2n+1)!}=\frac{2}{2n+1} $$ as announced.

Answer 2

I think Olivier's approach is the usual one, but as an alternative, you may use Rodrigues' formula to prove first $$ \frac{1}{\sqrt{1+t^2-2tx}}=\sum_{n\geq 0}P_n(x) t^n \tag{1} $$ from which it follows that $$ \int_{-1}^{1}\frac{dx}{1+t^2-2tx}=\sum_{n\geq 0}\left(\int_{-1}^{1}P_n(x)^2\,dx\right) t^{2n} \tag{2} $$ since by integration by parts, $\int_{-1}^{1}P_n(x)P_m(x)\,dx = 0$ as soon as $n\neq m$.
Since the LHS of $(2)$ equals $\frac{1}{t}\text{arctanh}(t)$, the claim $$ \int_{-1}^{1}P_n(x)^2\,dx = \frac{2}{2n+1}\tag{3}$$ follows by computing a simple Taylor series.