Series Legendre Polynomial We have to prove
$$\ln \Big(\frac{x+1}{1-x}\Big)=\sum_{n≥0} \frac{x^{n+1}}{n+1}P_n(x)$$
using the generatrix function of Legendre polynoms.
I don't know if it is useful, but $$\int_{-1}^{1}\frac{1}{1-2tx+x^2}dt=\frac{1}{x}\ln \Big(\frac{x+1}{1-x}\Big)$$
 A: The Generating function for Legendre polynomials is defined as
$$G(x,t):=\sum_{n=0}^\infty t^n P_n(x) $$
Using harmonic functions, we can easily derive:
$$G(x,t)=\frac{1}{\sqrt{1-2tx+t^2}} \quad \quad (1)$$
( See More Here: http://www.ucl.ac.uk/~ucahdrb/MATHM242/OutlineCD2.pdf )
Taking into account $G$'s definition, one should observe that:
$$ \int_0^x G(x,t) dt = \sum_{n=0}^\infty \frac{x^{n+1}}{n+1}P_n(x) \quad \quad (2)$$
Also, from equation $(1)$:
$$\int_0^x G(x,t)dt=\int_0^x \frac{1}{\sqrt{1-2tx+t^2}} dt =\ln\left(\left|\sqrt{t^2-2xt+1}+t-x\right|\right)\Bigg|_0^x=$$
$$\ln\left(\left|\sqrt{1-x^2}\right|\right)-\ln\left(\left|1-x\right|\right)=\ln\left(\left|\sqrt{1+x}\sqrt{1-x}\right|\right)-\ln\left(\left|1-x\right|\right)=\ln\left(\left|\frac{\sqrt{1+x}\sqrt{1-x}}{1-x}\right|\right)$$
$$=\ln\left(\left|\frac{\sqrt{1+x}}{\sqrt{1-x}}\right|\right)=\frac12 \ln \left(\left|\frac{1+x}{1-x}\right|\right) \quad \quad (3)$$
Now, from equations $(2),(3)$:
$$\sum_{n=0}^\infty \frac{x^{n+1}}{n+1}P_n(x)=\frac12 \ln \left(\frac{1+x}{1-x}\right)$$
assuming $(1+x)(1-x)>0$.
