Jtdylktuy

I'm stuck, couldn't figure it out. I appreciate your help.

Show that :
$$frac{1}{2} ln left |frac{1+x}{1-x} right |=sum_{n text { odd}} frac{(2n+1)}{n(n+1)}mathcal {P_n(x)}$$

The $n$-th order Legendre polynomial $P_n$ is even if $n$ is even and odd if $n$ is odd, and these satisfy $LP_n = n(n+1)P_n$, where
$$
Lf = -frac{d}{dx}left((1-x^2)frac{df}{dx}right).
$$
The function $Q_0 = frac{1}{2}lnleft|frac{1+x}{1-x}right|$ is the second classical solution of $Lf=0$, but it does not qualify as an eigenfunction in the classical sense because classical solutions must be bounded near $x=pm 1$. The normalized $P_n$ form an orthonormal basis of $L^2(-1,1)$, which allows the expansion
$$
Q_0(x) = sum_{n=0}^{infty}langle Q_0,P_nrangle P_n.
$$
The Lagrange identity gives
begin{align}
&n(n+1)langle Q_0,P_nrangle \
& =langle Q_0,LP_nrangle \
& = langle Q_0,LP_nrangle-langle LQ_0,P_nrangle \
& = int_{-1}^{1}left(frac{d}{dx}(1-x^2)frac{dQ_0}{dx}right)P_n
-frac{d}{dx}left((1-x^2)frac{dP_n}{dx}right)Q_0dx \
& = int_{-1}^{1}frac{d}{dx}left((1-x^2)left{frac{dQ_0}{dx}P_n-Q_0frac{dP_n}{dx}right}right)dx \
& = (1-x^2)(Q_0' P_n-Q_0P_n')|_{-1}^{1} \
& = (1-x^2)Q_0' P_n|_{-1}^{1}.
end{align}
And
$$
Q_0' = frac{1}{2}left[frac{1}{1+x}+frac{1}{1-x}right]=frac{1}{1-x^2}.
$$
Therefore,
$$
n(n+1)langle Q_0,P_nrangle = P_n|_{-1}^{1}.
$$
Therefore,
$$
Q_0(x) = sum_{n=0}^{infty}frac{P_n(1)-P_n(-1)}{n(n+1)}P_n(x) \
= sum_{n mbox{ odd}}frac{2P_n(1)}{n(n+1)}P_n(x).
$$