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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.

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    $\begingroup$ 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). $\endgroup$ – Jack D'Aurizio Sep 20 '19 at 19:14
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Part of the difficulty of dealing with these type problems is to know what is considered 'known.' A previous answer used the well-known recursion as a way to establish a proof. This one depends on the (also well-known) generating function,

$$ (G.F.) \quad \sum_{n=0}^\infty y^n P_n(x) = (1-2x \ y + y^2)^{-1/2} $$

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.

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