Expand function in Legendre polynomials on the interval [-1,1] Expand the following function in Legendre polynomials on the interval [-1,1] :
$$f(x) = |x|$$
The Legendre polynomials $p_n (x)$ are defined by the formula :
$$p_n (x) = \frac {1}{2^n n!} \frac{d^n}{dx^n}(x^2-1)^2$$
for $n=0,1,2,3,...$
My attempt :
we have using the fact that $|x|$ is an even function.
$$a_0 = \frac {2}{\pi}$$
$$ a_n= \frac {2}{π} \int_{-1}^{1}x\cos(nx)\,dx$$
Then what is the next step ?

We know that the Fourier-Legendre series is like 
$$
f(x)=\sum_{n=0}^\infty C_n P_n(x)
$$
where 
$$
C_n=\frac{2n+1}{2} \int_{-1}^{1}f(x)P_n(x)\,dx
$$
So now we are going to calculate the result of 
$$
\frac{2n+1}{2} \int_{-1}^{1}|x|P_n(x)\,dx
$$
As $|x|$ is an even function, and the parity of $P_n(x)$ depends on the parity of $n$, We can write
$$
\int_{-1}^{1}|x|P_n(x)\,dx
$$
as
$$
\int_{-1}^{1}|x|P_{2k}(x)\,dx\ \ k=0,1,2...
$$
and
$$
\int_{-1}^{1}|x|P_{2k}(x)\,dx \\
=\int_{-1}^{0}-xP_{2k}(x)\,dx + \int_{0}^{1}xP_{2k}(x)\,dx\\
=2\int_{0}^{1}xP_{2k}(x)\,dx
$$
As
$$
(n+1)P_{n+1}(x)-x(2n+1)P_{n}(x)+nP_{n-1}(x)=0
$$
we get
$$
2\int_{0}^{1}xP_{2k}(x)\,dx\\
=2(\frac{2k+1}{4k+1}\int_{0}^{1}P_{2k+1}(x)\,dx+\frac{2k}{4k+1}\int_{0}^{1}P_{2k-1}(x)\,dx)
$$
As 
$$
\int_{0}^{1}P_{n}(x)\,dx=\begin{cases}
0& n=2k\\
\frac{(-1)^k (2k-1)!!}{(2k+2)!!}& n=2k+1
\end{cases}
$$
We get
$$
C_{2k}=(2k+1)\frac{(-1)^k (2k-1)!!}{(2k+2)!!}+n\frac{(-1)^{k-1} (2k-3)!!}{(2k)!!}\\
=\begin{cases}
\frac{1}{2}& k=0\\
\frac{(-1)^{k+1} (4k+1)}{2^{2k}(k-1)!}\frac{(2k-2)!}{(k+1)!}& k>0
\end{cases}
$$
So
$$
|x|=\frac{1}{2}+\sum_{k=1}^\infty \frac{(-1)^{k+1} (4k+1)}{2^{2k}(k-1)!}\frac{(2k-2)!}{(k+1)!} P_{2k}(x)
$$
