# Legendre Polynomial as an integral

How can I prove the following property of the Legendre's polynomial: $$\frac{1}{\pi}\int_{0}^{\pi}\left\{x\pm\sqrt{x^{2}-1}\cos\theta\right\}^{n}d\theta = P_{n}(x)$$

I've tried differentiating under the integral sign but with not much progress.I also tried replacing the cosine with an equivalent complex number. Any kind of help or hint would be appreciated.

• You may just prove that both sides agree at $n\in\{0,1\}$ and that they fulfill the same recurrence relation (Bonnet's recursion formula). – Jack D'Aurizio Sep 20 '19 at 19:14

$$(G.F.) \quad \sum_{n=0}^\infty y^n P_n(x) = (1-2x \ y + y^2)^{-1/2}$$
For the integral $$I_n(x)$$, use the binomial expansion and switch $$\Sigma$$ and $$\int$$: $$I_n(x)=\sum_{k=0}^n x^{n-k}(x^2-1)^{k/2}\binom{n}{k} \frac{1}{\pi} \int_{0}^\pi \cos^k{t} \ dt$$ Use the well-known integral ID $$\frac{1}{\pi} \int_{0}^\pi \cos^k{t} \ dt = \frac{1 + (-1)^k}{2} \binom{k}{k/2} 2^{-k}$$ so that $$I_n(x) = \sum_{k=0}^\infty x^{n-2k}(x^2-1)^k \binom{n}{2k} \binom{2k}{k} 2^{-2k}$$ Note the limit has been extended to $$\infty$$ because once $$2k$$ exceeds $$n$$ the first binomial will made each term indentically zero. Now show that $$I_n(x)$$ obeys (G.F.):
$$\sum_{n=0}^\infty I_n(x) y^n = \sum_{k=0}^\infty x^{-2k}(x^2-1)^k\binom{2k}{k} 2^{-2k} \sum_{n=0}^\infty (x \ y)^n\binom{n}{2k}$$
The innermost sum is $$(xy)^{2k} (1-xy)^{-(2k+1)},$$ and of course $$\sum_{k=0}^\infty (z/4)^k \binom{2k}{k} = (1-z)^{-1/2}.$$ Algebra completes the proof. To be pedantic, you need to have regions over which $$x$$ and $$y$$ don't cause the series to diverge, but the proof presented works in formal power series.