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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.
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.
