I have a generating function of Legendre Polynomials given by: $G(x,r)= \sum_{n=0}^\infty P_n(x)r^n = (1-2rx +r^2)^{-1/2}$
My problem is that I'm asked to find $\int_{-1}^1P_n(x)dx$ but all I have is $P_n(x)r^n$, I'm not sure how to (essentially) remove the $r^n$ so that I have what I need.
I can work out the integral myself I would just like to know how to get from $P_n(x)r^n$ to $P_n(x)$.
One method I thought of was to define some $C_n$ so that:
$\int_{-1}^1P_n(x)dx=C_n$
Then I could integrate both sides but I got confused as to what would be in the $(…)$
$\sum_n C_nr^n=\int_{-1}^1(…)dx$
Any help would be much appreciated
Best Answer
Since $$\int_{-1}^1 G(x,r) \;dx = 2$$ it follows that $$\int_{-1}^1 P_n(x)\;dx = \begin{cases} 2 \qquad \text{if } n = 0 \\ 0 \qquad \text{otherwise} \end{cases}$$