Using the generating function of Legendre polynomials, show \begin{equation} \sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}P_n(x)=\frac{1}{2}\ln \left(\frac{1+x}{1-x}\right) \end{equation} My attempt
I know the generating function of Legendre polynomials is given by $G (x, t) = \sum_{n = 0}^{\infty} P_n (x) t^n \text{ with } | t | < 1, | x | \leqslant 1. $ Also we can see to $G$ as $G (x, t) = \frac{1}{\sqrt{1 - 2 x t + t^2}} . $ Then \begin{equation} G (x, x) = \frac{1}{\sqrt{1 - x^2}} = \sum_{n = 0}^{\infty} P_n (x) x^n \tag{2}. \end{equation} Now I want to know to where $\underset{n = 0}{\overset{\infty}{\sum}} \dfrac{P_n (x) x^n}{n + 1}$ converges?
Since if we find $f (x)$ from (2) such as $\underset{n = 0}{\overset{\infty}{\sum}} \dfrac{P_n (x) x^n}{n + 1} \rightarrow f (x)$, then $x f (x)$ should be equal to $\frac{1}{2} \ln \left( \frac{1 + x}{1 - x} \right)$.
Any suggestion will be welcome.
Write \begin{equation} f(x,y)=\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}P_n(y) \end{equation}
Differentiate this to obtain $f_x(x,y) = \sum_{n=0}^{\infty} x^n P_n(y)$. This is known to be $\frac{1}{\sqrt{1 - 2 x y + x^2}}$. Integrating we arrive at $f(x,y) = \log(\sqrt{-2yx+x^2+1} -y+x)$. Subtitute $x=y$ to obtain the answer.