<s>[INST] $\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}P_n(x)=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$ 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.
 [/INST] 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.
</s>