<s>[INST] Integrating odd Legendre polynomials using generating function I must show using generating function of Legendre polynomials, that
\begin{align}
\int_0^1 P_{2n+1}(x)\phantom{1}dx = (-1)^n\frac{(2n)!}{2^{2n+1}n!(n+1)!}
\end{align}
My attempt is to change the generating function $\Phi(u,x)=(1-2ux+u^2)^{-1/2}=\sum_{n=0}^{\infty}u^nP_n(x)$ into infinite series using Newton's generalized binomial theorem
\begin{align}
(1-2ux+u^2)^{-1/2} = \sum_{n=0}^{\infty}(-1)^n\frac{(2n)!}{2^{2n}(n!)^2}(-2ux+u^2)^n=\sum_{n=0}^{\infty}u^nP_n(x)
\end{align}
and then integrating from $0$ to $1$ with respect to $x$. But it seems unsuccessfull. Can you give me some hint to answer this? Or, I must starting from where?
 [/INST] Expand the result of the integral of the generating function with Taylor series, using
$$\sqrt{1+z} = \sum_{n=0}^\infty \binom{\frac{1}{2}}{n}z^n = 1+\sum_{n=0}^\infty \frac{(-1)^n\binom{2n}{n}z^{n+1}}{2^{2n+1}(n+1)}$$
As follows:
\begin{equation}
\begin{split}
I &= \int_0^1 (1-2ux+u^2)^{-1/2}dx\\
&= \left[-\frac{1}{u}\sqrt{1-2ux+u^2}\right]_{x=0}^{x=1}\\
&= -\frac{1}{u}(\sqrt{1-2u+u^2}-\sqrt{1+u^2})\\
&= -\frac{1}{u}(1-u-\sqrt{1+u^2})\\
&= -\frac{1}{u}(1-u-\left(1+\sum_{n=0}^\infty \frac{(-1)^n\binom{2n}{n}u^{2n+2}}{2^{2n+1}(n+1)}\right))\\
&= 1+\sum_{n=0}^\infty \frac{(-1)^n\binom{2n}{n}u^{2n+1}}{2^{2n+1}(n+1)}
\end{split}
\end{equation}
Comparing with $I = \sum_{m=0}^\infty u^m \int_0^1 P_m(x)dx$, we conclude that
$$ \int_0^1 P_{2n+1}(x)dx = \frac{(-1)^n\binom{2n}{n}}{2^{2n+1}(n+1)}=\frac{(-1)^n(2n)!}{2^{2n+1}(n+1)(n!)^2} = \frac{(-1)^n(2n)!}{2^{2n+1}n!(n+1)!}$$
</s>