# Integrated series identity with Legendre Polynomials

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: $f(x) = \frac{1}{\sqrt{1 + a^2 - 2ax}}$

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);


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

• Mmm... anything without using the elliptical integrals? – Klest Dedja Oct 29 '14 at 9:26

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).