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The Legendre Polynomials can be defined in many different ways and have several properties. Many of these can be found in books or in the net, but I couldn't find this one anywhere:

Prove that:

$$\frac{1}{2\pi} \int_0^{2\pi} \sum_{n = 0}^{+\infty} P_n(cos\phi)d\phi = \sum_{n = 0}^{+\infty} \lvert P_n(0) \rvert ^2.$$

Where $P_n(x)$ is the $n$-th Legendre Polynomial.

I put th "power series" as a tag because of the definition of the Legendre Polynomials: by definition this is their generating function: $ f(x) = \frac{1}{\sqrt{1 + a^2 - 2ax}}$

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From looking at the first few $n$, it seems that in fact you don't need the sum:

$$ \dfrac{1}{2\pi} \int_0^{2\pi} P_n(\cos \phi)\; d\phi = P_n(0)^2 $$

where in fact for odd $n$, both sides are $0$.

The right sides are easy to find using Bonnet's recursion formula

$$ (n+1) P_{n+1}(x) = (2n+1) x P_n(x) - n P_{n-1}(x) $$

I'm not sure how best to get the left sides. In Maple you can use the generating function as follows:

int(1/sqrt(1+a^2-2*a*cos(phi)),phi=0..2*Pi) assuming a > 0,a<1;

$${\frac {4}{a+1}{\rm EllipticK} \left( 2\,{\frac {\sqrt {a}}{a+1}} \right) } $$

> convert(%, FPS, a);

$$\sum _{k=0}^{\infty }2\,{\frac {\pi \, \left( \left( 2\,k \right) ! \right) ^{2}{16}^{-k}{a}^{2\,k}}{ \left( k! \right) ^{4}}} $$

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  • $\begingroup$ Mmm... anything without using the elliptical integrals? $\endgroup$ – Klest Dedja Oct 29 '14 at 9:26
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Using the recursive formula for the Legendre polynomials (see Robert Israel's answer or https://en.wikipedia.org/wiki/Legendre_polynomials), it's easy to show by induction that $P_{2n}(0) = (-\frac{1}{4})^n\binom{2n}{n}$. By Stirling's approximation formula (https://en.wikipedia.org/wiki/Stirling%27s_approximation), $|P_{2n}(0)|$ is bounded below by $C/\sqrt{n}$ for some positive $C$. Thus, the right-hand side sum diverges (by comparing to a harmonic series).

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