Let $\,P_n\left(x\right)\,$ be the Legendre polynomial of degree $n$ and let
$$P_{m+1}\left(0\right)=-\frac{m}{m+1}\,P_{m-1}\left(0\right), \quad \;m=1,2,\ldots$$
If $\displaystyle\, P_n\left(0\right)=-\frac{5}{16},\,$ then $\displaystyle\int_{-1}^{1}P_n^2\left(x\right)\,dx=\,?$
I know $\displaystyle\,\int_{-1}^{1}P_n^2\left(x\right)\,dx=\frac{2}{2n+1}$, but I am unable to find $n$ using the given conditions. Please someone help.
Thanks .
Simply compute the first few values $P_m(0)$ from the trival one $P_0\equiv 1$: $$P_2(0)=\left(-\frac{1}{2}\right)\times 1$$ $$P_4(0)=\left(-\frac{3}{4}\right)\left(-\frac{1}{2}\right)\times 1= \frac{3}{8}$$ $$P_6(0)=\left(-\frac{5}{6}\right)\left(-\frac{3}{4}\right)\left(-\frac{1}{2}\right)\times 1= -\frac{5}{16}$$ So $n=6$ and this is the only integer with the given value, because the sequence $|P_{2m}(0)|$ is strictly decreasing. Therefore the result is $$\int_{-1}^{1}P_n^2\left(x\right)\,dx=\frac{2}{2\times 6 +1 }=\frac{2}{13}$$