proof (Legendre polynomial )

Question

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

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

Answer 1

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 $Q_0 = \frac{1}{2}\ln\left|\frac{1+x}{1-x}\right|$ 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 $$ Q_0(x) = \sum_{n=0}^{\infty}\langle Q_0,P_n\rangle P_n. $$ The Lagrange identity gives \begin{align} &n(n+1)\langle Q_0,P_n\rangle \\ & =\langle Q_0,LP_n\rangle \\ & = \langle Q_0,LP_n\rangle-\langle LQ_0,P_n\rangle \\ & = \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_0\frac{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, $$ n(n+1)\langle Q_0,P_n\rangle = P_n|_{-1}^{1}. $$ Therefore, $$ Q_0(x) = \sum_{n=0}^{\infty}\frac{P_n(1)-P_n(-1)}{n(n+1)}P_n(x) \\ = \sum_{n \mbox{ odd}}\frac{2P_n(1)}{n(n+1)}P_n(x). $$