I am trying to understand the Integration of the product of two Legendre polynomials.
Given,
$$I=\int_{-1}^{1} P_{l}(x)P_{l'}(x)\mathrm{d}x~$$
Suppose $x = \cos(\theta)$, then our $I$ becomes
$$I=\int_{0}^{\pi} P_{l}(\cos(\theta))P_{l'}(\cos(\theta))\sin(\theta)\mathrm{d}\theta~$$
Since Legendre polynomials are orthogonal to each other, thus at $l \neq l^{'}$, $I = 0.$
So now we consider what happens at $l = l'$:
Our Integral becomes as follows:
$$I=\int_{0}^{\pi} P_{l}(\cos(\theta))P_{l}(\cos(\theta))\sin(\theta)\mathrm{d}\theta~ = \int_{0}^{\pi} (P_{l}(\cos(\theta)))^2\sin(\theta)\mathrm{d}\theta~$$
This is where I get lost. Griffith's Introduction to Electrodynamics, page no. 144, eq. 3.68, tells that for $l=l^{'}$, $I = \frac{2}{2l+1}$, but I am unable to work out that integral. Here is my attempt.
We know from Rodrigues formula that,
$$P_l(x)\equiv \frac{1}{2^l l!}\left(\frac{d}{dx}\right)^l(x^2-1)^l$$
Subsituting $x = \cos(\theta)$ we get,
$$P_l(\cos(\theta))\equiv \frac{1}{2^l l!}\left(\frac{d}{d\theta}\right)^l((\cos(\theta))^2-1)^l$$
Now we substitute $P_l$ in $I$, and we get:
$$ I= \int_{0}^{\pi} \left(\frac{1}{2^l l!}\left(\frac{d}{d\theta}\right)^l((\cos(\theta))^2-1)^l\right)^2\sin(\theta)\mathrm{d}\theta~$$
If $l$ is even (an assumption I am making to make my life easy),
$$ I= \left(\frac{1}{2^l l!}\right)^2\int_{0}^{\pi} \left(\left(\frac{d}{d\theta}\right)^l(\sin(\theta))^{2l}\right)^2\sin(\theta)\mathrm{d}\theta~$$
At this point, it just seems I have complicated the math even more. Can someone help me to reach the final answer? Thanks!
I think that $x=\cos\theta$ is an off-road way, but Rodrigues' formula is a good one:
$$\frac{d^n}{dx^n}P_n(x)=\frac1{2^n n!}\frac{d^{2n}}{dx^{2n}}(x^2-1)^n=\frac{(2n)!}{2^n n!},$$ so that integration by parts ($n$ times) gives $$\int_{-1}^1 P_n^2(x)\,dx\stackrel{\color{Violet}{!}}{=}\frac{(-1)^n}{2^n n!}\int_{-1}^1(x^2-1)^n\frac{d^n}{dx^n}P_n(x)\,dx=\frac{(2n)!}{2^{2n}n!^2}\int_{-1}^1(1-x^2)^n\,dx,$$ and the integral can be computed using $x=1-2t$ and the beta function: $$\int_{-1}^1(1-x^2)^n\,dx=2^{2n+1}\int_0^1 t^n(1-t)^n\,dt=\frac{2^{2n+1}n!^2}{(2n+1)!}.$$