Sum of Legendre function I'm currently trying to solve the sum
$$
f(x)=\sum\limits_{n=0}^\infty\frac{x^{n+1}}{n+1}P_n(x),
$$
where $P_n(x)$ is the Legendre function of order n.
I also named the sum $f(x)$ since I'm the solution will be a function of $x$. The first thing that can be noticed is that the function $f(x)$ is an uneven function.
My first step was to take the derivative of $f(x)$, this then yields that:
$$
f'(x)=\sum\limits_{n=0}^\infty x^{n}P_n(x)+\sum\limits_{n=1}^\infty\frac{x^{n+1}}{n+1}P'_n(x).
$$
Using the generating function
$$
\frac{1}{\sqrt{1-2hz+h^2}}=\sum\limits_{n=0}^\infty h^nP_n(z)
$$
I'm able to rewrite the first term as:
$$
f'(x)=\frac{1}{\sqrt{1-x^2}}+\sum\limits_{n=1}^\infty\frac{x^{n+1}}{n+1}P'_n(x).
$$
Now for the second term I'm still stuck, I've been trying a few recursive relations, in the hope to maybe get a first order differential equation for $f(x)$. So far I haven't found the right recursive relation yet.
Now I'm wondering if there is maybe a better approach that I should have taken, or that if there is an easy trick to solve the second term?
After playing with the function in Mathematica I figured out that the most probable solution is
$$
f(x)=\mathrm{arctanh}(x),
$$
this gives me a direction, but still I have no clue for the second term. All help is welcome!
 A: Let: 
$$f(x) = \sum _{n=0}^{\infty} \frac{x^{n+1} \, P_{n}(x)}{n+1}$$
As you said:
$$f^{'}(x) = \sum_{n=0}^{\infty} x^{n} \, P_{n}(x) + \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} \, P_{n}^{'}(x)$$
And using the generating function you showed, we get:
$$f^{'}(x) = \frac{1}{\sqrt{1-x^{2}}} + \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1 } \, P_{n}^{'}(x)$$
Now, note that:
$$
{ P }_{ n }(x)'\quad =\quad \frac { \left( n+1 \right) \left( x{ P }_{ n }(x)\quad -\quad { P }_{ n+1 }(x) \right)  }{ 1-{ x }^{ 2 } } 
$$
So we get:
\begin{align}
f^{'}(x) &=\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } }  } +\sum _{ n=0 }^{ \infty  }{ \frac { { x }^{ n+1 } }{ n+1 }  } \frac { \left( n+1 \right) \left( x{ P }_{ n }(x) - { P }_{ n+1 }(x) \right)  }{ 1-{ x }^{ 2 } } \\
&= \frac {1}{\sqrt{1-x^{2}}} + \frac{1}{1-x^{2}} \sum_{n=0}^{\infty}{x^{n+1} \left(x {P}_{n}(x) - {P}_{n+1}(x) \right)} \\
&= \frac{1}{\sqrt{1-x^{2}}} + \frac{1}{1-x^{2}} \sum_{n=0}^{\infty} x^{n+2} \left( P_{n}(x) \right) - \frac{1}{1-x^{2}} \sum_{n=0}^{\infty} x^{n+1} \, P_{n+1}(x) \\
&= \frac{1}{\sqrt{1-x^{2}}} +\frac{x^{2}}{1-x^{2}} \sum_{n=0}^{\infty} x^{n} \left(P_{n}(x) \right) - \frac{1}{1-x^{2}}\left(\sum_{n=0}^{\infty} x^{n} \, P_{n}(x) -1 \right) \\
&= \frac{1}{\sqrt{1-x^{2}}} + \frac{x^{2}}{1-x^{2}} \frac{1}{\sqrt{1-x^{2}}} - \frac{1}{1-x^{2}} \left(\frac{1}{\sqrt{1-x^{2}}} -1\right) \\ 
&= \frac{1}{\sqrt{1-x^{2}}} + \frac{x^{2}-1}{ (1-x^{2}) \, \sqrt{1-x^{2}}} + \frac{1}{1-x^{2}} \\ 
&= \frac{1}{\sqrt{1-x^{2}}} + \frac{-1}{\sqrt{1-x^{2}}} + \frac{1}{1-x^{2}} = \frac {1}{1-x^{2}}
\end{align}
Integration yields:
\begin{align}
f(x) &= \int \frac{dx}{1-x^{2}} = \frac{1}{2} \, \int \left(\frac{1}{1-x} + \frac{1}{1+x} \right) dx \\ 
&= \frac{1}{2} \, \left[\ln(1+x)-\ln(1-x)\right] + C \\
&= \frac{\ln\left(\frac{1+x}{1-x}\right)}{2} + C
\end{align}
and for $x=0$ we have $f(0) = 0$ so we get $C=0$.
Hence our answer is:
$$f(x) = \sum_{n=0}^{\infty}{\frac{{x}^{n+1}}{n+1} \, P_{n}(x)} = \frac{1}{2} \, \ln\left(\frac{1+x}{1-x}\right)$$
A: While the other response is a good exercise in exploiting recurrence relations, a much more direct route is provided by the technique of summing under the integral which converts the series into an integral of the generating function:
$$\begin{align}
f{\left(x\right)}
&=\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}P_{n}{\left(x\right)}\\
&=\sum_{n=0}^{\infty}P_{n}{\left(x\right)}\int_{0}^{x}\mathrm{d}t\,t^{n}\\
&=\int_{0}^{x}\mathrm{d}t\,\sum_{n=0}^{\infty}t^{n}P_{n}{\left(x\right)}\\
&=\int_{0}^{x}\mathrm{d}t\,\frac{1}{\sqrt{1-2xt+t^2}}\\
&=\int_{-x}^{\frac{\sqrt{1-x^2}-1}{x}}\mathrm{d}u\,\frac{2}{1-u^2};~~~\small{\left[\sqrt{1-2xt+t^2}=tu+1\right]}\\
&=2\int_{\frac{1-\sqrt{1-x^2}}{x}}^{x}\frac{\mathrm{d}w}{1-w^2};~~~\small{\left[-u=w\right]}\\
&=2\operatorname{arctanh}{\left(x\right)}-2\operatorname{arctanh}{\left(\frac{1-\sqrt{1-x^2}}{x}\right)}\\
&=\operatorname{arctanh}{\left(x\right)}.\blacksquare\\
\end{align}$$
